A theory for generalized morphisms and beyond
Abstract.
Some sorts of generalized morphisms are defined from very basic mathematical objects such as sets, functions, and partial functions. A wide range of mathematical notions such as continuous functions between topological spaces, ring homomorphisms, module homomorphisms, group homomorphisms, and covariant functors between categories can be characterized in terms of the generalized morphisms. We show that the inverse of any bijective generalized morphism is also a generalized morphism (of the same kind), and hence a generalized isomorphism can be defined as a bijective generalized morphism.
Galois correspondences are established and studied, not only for the Galois groups of the generalized automorphisms, but also for the “Galois monoids” of the generalized endomorphisms.
Ways to construct the generalized morphisms and the generalized isomorphisms are studied.
New interpretations on solvability of polynomials and solvability of homogeneous linear differential equations are introduced, and these ideas are roughly generalized for “general” equation solving in terms of our theory for the generalized morphisms.
Some more results are presented. For example, we generalize the algebraic notions of transcendental elements over a field and purely transcendental field extensions, we obtain an isomorphism theorem that generalizes the first isomorphism theorems (for groups, rings, and modules), and we show that a part of our theory is closely related to dynamical systems.
Key words and phrases:
Generalized morphisms; Galois theory; Galois correspondence; Constructions of morphisms; Solvability of equations; Isomorphism theorem; Transcendental elements2020 Mathematics Subject Classification:
Primary 08A99, 08A35; Secondary 12H05, 18A99, 12F10, 13B05, 12F20, 20B25, 20B271. Introduction
The notions of generalized morphisms introduced in this paper develop from very basic mathematical objects such as sets, functions, and partial functions. As a consequence, many familiar mathematical notions are covered by the generalized morphisms. Specifically, continuous functions between topological spaces, ring homomorphisms, module homomorphisms, group homomorphisms, and covariant functors between categories can be characterized in terms of the generalized morphisms. Indeed, in these examples, the generalized morphisms preserve the structures of the mathematical objects involved. Basically, this is because we define the generalized morphisms to be commutative with the intrinsic operations on the mathematical objects involved. Moreover, we show that the inverse of any bijective generalized morphism is also a generalized morphism (of the same kind), and hence a generalized isomorphism can be defined as a bijective generalized morphism.
The notion of generalized morphisms in our theory is not the same as the notion of morphisms in category theory. In category theory, a morphism between two objects does not necessarily preserve the structure of objects, while a generalized morphism in our theory does. Nevertheless, since covariant functors preserve the structures of categories, all covariant functors can be characterized in terms of the generalized morphisms.
One may have encountered concepts similar to the notions of our generalized morphisms, e.g. the notion of homomorphisms between structures in the field of mathematical logic. However, we have not found any research with any content similar to ours.
Besides introducing the new notions, we establish Galois correspondences for the Galois groups of the generalized automorphisms as well as for the “Galois monoids” of the generalized endomorphisms. As is well known, in the Galois theory for infinite algebraic field extensions, Krull topology is put on Galois groups, and in differential Galois theory, differential Galois groups are endowed with Zariski topology. Then the fundamental theorem in each of the two theories tells us that there exists a Galois correspondence between the set of all closed subgroups of the Galois group of a Galois field extension (or a Picard-Vessiot extension) and the set of all intermediate (differential) fields. Now for our theory, we first study the lattice structures related to Galois correspondences. Then, we can determine under what conditions we can characterize Galois correspondences with topology (as is done in the above two Galois theories).
Moreover, we study ways to construct the generalized morphisms and the generalized isomorphisms, which we think to be important because the morphisms and isomorphisms preserve the structures of the mathematical objects involved.
In Part I of this paper, which consists of Sections 2 to 7, we address the above issues for the case where only single-variable functions are involved, and in Part II, we address the above issues for the case where multivariable functions or partial functions are involved.
In Part III, new interpretations on solvability of polynomials and solvability of homogeneous linear differential equations are introduced. The key idea is to “decompose” an equation into equations such that each solution of can somehow be expressed in terms of the solutions of , where for example in the case of a separable polynomial in , the Galois group of each in is simple. Moreover, this idea is roughly generalized for “general” equation solving in terms of the theory developed in Parts I and II.
Some more results are presented in Part IV. For example, in Section 19, we obtain analogues of the well-known fact that the Galois group of an irreducible polynomial acts transitively on its roots, and in Section 21, we generalize the algebraic notions of transcendental elements over a field and purely transcendental field extensions. Moreover, in Section 22, we obtain an isomorphism theorem which generalizes the first isomorphism theorems (for groups, rings, and modules). And in Section LABEL:App_to_dyn, we show a close relation between a part of our theory and dynamical systems.
To help the reader better understand the structure of the paper, we simplify the table of contents as follows.
Part I. Theory for the case of unary functions
Section 2. Basic notions and properties
Section 3. Galois correspondences
Section 4. Lattice structures of objects arising in Galois correspondences on a -space
Section 5. Topologies employed to construct Galois correspondences
Section 6. Generalized morphisms and isomorphisms from a -space to a -space
Section 7. Constructions of the generalized morphisms and isomorphisms
Part II. Theory for the case of multivariable total or partial functions
Section 8. Basic notions for the case of multivariable (total) functions
Section 9. Basic notions for the case of (multivariable) partial functions
Section 10. Basic properties and more notions
Section 11. Galois correspondences
Section 12. Lattice structures of objects arising in Galois correspondences on a -space
Section 13. Topologies employed to construct Galois correspondences
Section 14. Constructions of the generalized morphisms and isomorphisms
Part III. Solvability of equations
Section 15. A solvability of polynomial equations
Section 16. A solvability of homogeneous linear differential equations
Section 17. A possible strategy for equation solving
Part IV. Other topics and future research
Section 18. Dualities of operator semigroups
Section 19. Fixed sets and transitive actions of and
Section 20. Two questions on Galois -extensions and normal subgroups of Galois -groups
Section 21. -transcendental elements and -transcendental subsets
Section 22. A generalized first isomorphism theorem
Section 23. On topological spaces
Section LABEL:App_to_dyn. On dynamical systems
Section LABEL:Other_topics. Other topics for future research
The contents of Part I are roughly described section by section as follows.
Let be a set of functions from a set to . If the composite of any two elements in still lies in , then we call an operator semigroup on (cf. Definition 2.1.1). Let . Then we call the -space generated by , where denotes the image of the restriction of to (cf. Definition 2.2.1). -spaces represent mathematical objects whose structures are preserved under generalized morphisms. A critical notion in this theory is -morphism, which is defined as a map from a -space to a -space such that commutes with every element of ; that is, and , (cf. Definition 2.3.1). From these simple definitions much follows. For example, any ring homomorphism between and with field fixed pointwisely can be characterized in terms of a -morphism (cf. Proposition 2.3.2), and a map from a topological space to itself is continuous if and only if the map induces a -morphism (cf. Proposition 2.3.5). In Subsection 2.4, we show that the inverse of any bijective -morphism from a -space to a -space is a -morphism from to , and hence a -isomorphism can be defined as a bijective -morphism.
In Section 3, we establish the Galois correspondence between the Galois groups of a -space and the fixed subsets of under the actions of the -automorphisms of (cf. Corollary 3.2.6). Moreover, we shall find the Galois correspondence between the “Galois monoids” of a -space and the fixed subsets of under the actions of the -endomorphisms of (cf. Corollary 3.2.5).
To better understand these Galois correspondences, in Section 4, we study the lattice structures of those objects which arise in Corollaries 3.2.5 and 3.2.6. Results in Section 4 will be employed in Section 5.
In the fundamental theorem for infinite algebraic field extensions (resp. for differential field extensions), the Galois correspondences are characterized with Krull topology (resp. Zariski topology) (see e.g. [1, 3, 4, 5, 9]). In Section 5, we shall see when and how we can characterize Galois correspondences with topology.
Section 6 introduces the notions of -morphism and -isomorphism. As a generalization of the notion of -morphism, a -morphism (cf. Definitions 6.1.1 and 6.1.8) is from a -space to a -space, where both and are operator semigroups and . We shall find that some more familiar mathematical notions can be characterized by -morphisms (cf. Propositions 6.1.2, 6.1.3 and 6.1.10). In Subsection 6.2, we show that the inverse of any bijective -morphism from a -space to a -space is a -morphism from to , and thus a -isomorphism can be defined as a bijective -morphism.
Main results in Section 7 are about constructions of -morphisms and -morphisms. Roughly speaking, for a -morphism or -morphism from a -space , Subsections 7.1, 7.2 and 7.3 discuss how to construct the morphism by a map from a set which generates (i.e. ). Moreover, in Subsection 7.4, we shall introduce a construction of -morphisms in terms of topology.
The contents of Part II are described as follows.
To make our theory more general, in Section 8, we generalize the notion of operator semigroup to incorporate functions of more than one variable (cf. Definition 8.1.2). And correspondingly, the concepts of -spaces, -morphisms and -morphisms are generalized (cf. Definitions 8.2.1, 8.3.1, and 8.5.4). Then we shall find that ring homomorphisms, module homomorphisms, and group homomorphisms can be characterized in terms of -morphisms (cf. Propositions 8.3.4 to 8.3.7) or -morphisms (cf. Propositions 8.5.7 to 8.5.11).
To further generalize our theory, in Section 9, we allow operators in to be partial functions (cf. Definition 9.1.4). And correspondingly, the notions of -spaces, -morphisms and -morphisms are generalized (cf. Definitions 9.2.1, 9.3.1, and 9.5.5). Because operators in are now allowed to be partial functions, we can generate fields, differential fields, and categories as -spaces in a natural way (cf. Examples 9.2.2 to 9.2.4). Then we shall find that ring homomorphisms between fields, differential ring homomorphisms between differential fields, and covariant functors between categories can be characterized in terms of -morphisms (cf. Propositions 9.3.4 and 9.3.6) or -morphisms (cf. Propositions 9.5.8 and 9.5.9).
For the case of (partial) functions of more than one variable, results obtained in Sections 2, 3, 4, 5 and 7 are generalized in Sections 10, 11, 12, 13 and 14, respectively. (Note that results obtained in Section 6 are generalized in Sections 8 and 9.)
Part III addresses the following issues.
A main goal of the classical Galois theory is to study the solvability by radicals of polynomial equations. In Section 15, however, we shall introduce another understanding of solvability of polynomial equations, which somehow corresponds to a composition series of the Galois group of the polynomial. And in Subsection 15.2 we shall explain why we introduce this notion of solvability. Roughly speaking, the central idea is that a separable polynomial can be “decomposed” into polynomials such that the Galois group of each is simple and any root of can be expressed in terms of the roots of (cf. Corollary 15.2.5).
Analogously, in Section 16, we shall introduce a new interpretation of solvability of homogeneous linear differential equations, which somehow corresponds to a “composition Zariski-closed series” of the differential Galois group of the equation. In Subsection 16.3 we shall explain why we introduce this notion of solvability. As what we did in Section 15, our main approach is to “decompose” a homogeneous linear differential equation into homogeneous linear differential equations so that any solution of can be expressed in terms of the solutions of (cf. Corollary 16.3.6).
Section 17 generalizes some results in Sections 15 and 16, and it roughly describes in terms of our theory a possible strategy for equation solving.
In Part IV we give some more results which we think to be important or deserve deeper research.
Section 18 is about dualities of operator semigroups. Let be an operator semigroup and let be a -space. We denote by (resp. ) the set of all -endomorphisms of (resp. the set of all -automorphisms of ). Then, roughly speaking, there is a duality between and the maximum operator semigroup which “accommodates” (i.e. , would still be a -endomorphism of if were extended to the maximum). Analogously, there is a duality between and the maximum operator semigroup which “accommodates” .
In Section 19, we shall talk about fixed subsets of a -space under the actions of and . And we shall obtain some transitive properties of the actions of and on , which are analogues of the well-known fact that the Galois group of an irreducible polynomial acts transitively on its roots.
Section 20 addresses an analogue of the issue which is normally the last part of the fundamental theorem of a Galois theory: the correspondence between the normal subgroups of the Galois group and the Galois extensions of the base field.
In Section 21, the algebraic notions of transcendental elements over a field and purely transcendental field extensions are generalized.
In Section 22, we shall obtain an isomorphism theorem which generalizes the first isomorphism theorems (for groups, rings, and modules).
In Section 23, we shall introduce for topological spaces some notions and properties related to our theory. And we shall employ Corollaries 3.2.5 and 3.2.6 to obtain Galois correspondences on topological spaces.
Let be a dynamical system, where is a monoid, is the phase space and is the evolution function. In Section LABEL:App_to_dyn, it is shown that induces in a natural way an operator semigroup on such that contains the identity function and is a -space. Conversely, for any operator semigroup which contains the identity function, any -space induces a dynamical system in a natural way. Hence we can apply our theory for operator semigroups to dynamical systems. In particular, we shall obtain Galois correspondences on dynamical systems.
Section LABEL:Other_topics gives some more suggestions on future research.
Part I. THEORY FOR THE CASE OF UNARY FUNCTIONS
In Part I of the article, we introduce the generalized morphisms where all functions involved are unary. The generalized morphisms preserve structures of mathematical objects which we call -spaces, where stands for a semigroup of functions with composition of functions as the binary operation.
2. Basic notions and properties
2.1. Operator semigroup
Sets which are closed under some operations are ubiquitous in mathematics. The following notion is related to this observation.
Definition 2.1.1.
Let be a set and let be a set of functions from to . If , the composite , then we call an operator semigroup on , and we call the domain of .
Remark.
It is clear that , possibly empty, is a semigroup with composition of functions as the binary operation.
Example 2.1.2.
Let be a subfield of a field and let be the ring of all polynomials (in one variable) over . Let
(In particular, is a constant (polynomial) function if is a constant polynomial.) Then it is not hard to see that is an operator semigroup on . Note that the map which induces , i.e. given by , is surjective, but it is not necessarily injective.
Remark.
Of course, there are other ways to define an operator semigroup on a field . For example, let consist of only the identity function on and let be the set of all ring homomorphisms from to , then both and are operator semigroups on . But unlike in Example 2.1.2, neither nor could lead us to any desirable result (in the remaining part of the article).
In fact and roughly speaking, to obtain desirable results, an operator semigroup should represent the intrinsic operations on the mathematical object involved.
Notation 2.1.3.
Throughout the paper, we denote by Id the identity function (on some set which is clear from the context).
As Example 2.1.2, the following two will lead us to desirable results.
Example 2.1.4.
Let be a topological space and let be the power set of . Let
where is the closure of . Then has only two elements and it is an operator semigroup on .
Example 2.1.5.
Recall that a differential ring is a commutative ring with identity endowed with a derivation . Then , where , is an operator semigroup on .
Definition 2.1.6.
Let be a set and let be a set of functions from to . We call the intersection of all operator semigroups (on ) containing the operator semigroup on generated by , and we denote it by . If , then we may denote by for brevity.
Remark.
is the smallest operator semigroup (on ) which contains .
2.2. -spaces and quasi--spaces
Operator semigroups are used to generate -spaces defined as follows.
Definition 2.2.1.
Let be an operator semigroup on and let . We call } the -space generated by and denote by . If , then we also denote by .
Moreover, if is also a -space generated by some subset of , then we say that is a -subspace of and write or .
Remark.
may have generators other than , and we may just call a -space.
Trivially, is a -space and is a -space.
-spaces are mathematical objects whose structures are preserved under the generalized morphisms (to be defined later), and this claim will be shown later on the following examples.
Example 2.2.2.
Let be a subfield of a field . Let be the operator semigroup defined in Example 2.1.2. Then , is the ring . In particular, if is an algebraic field extension, then , is the field .
Example 2.2.3.
Let be a topological space and let be the power set of . If is defined as in Example 2.1.4, then is a -space.
Example 2.2.4.
Let be a differential ring with a derivation . Let be the operator semigroup defined in Example 2.1.5. Recall that an ideal of is a differential ideal if implies that . Thus an ideal of is a differential ideal of if and only if the -space . We will talk more about differential rings in Sections 6, 7 and 9.
We are going to develop some properties of the new notions. In Sections 2 to 7, unless otherwise specified, denotes a set and denotes an operator semigroup on .
First, the following example is used to prove the proposition right after it.
Example 2.2.5.
Let given by . Then is an operator semigroup on , and
Proposition 2.2.6.
-
(a)
A generator set of a -space may have no intersection with the -space.
-
(b)
It is possible that no subset of a -space may generate the -space.
-
(c)
is not necessarily a -space.
-
(d)
A subset of is not necessarily a -space even if it satisfies: , and .
Proof.
For (a): In Example 2.2.5,
For (b): In Example 2.2.5, no subset of may generate .
For (c): In Example 2.2.5, , and hence is not a -space.
For (d): In Example 2.2.5, satisfies: , and , but is not a -space. ∎
However, any -space must satisfy: , and , as shown below.
Proposition 2.2.7.
Let be a -space. Then and , . Hence , , i.e. is -subspace of (Definition 2.2.1). In particular, .
Proof.
In light of Proposition 2.2.7 and (d) in Proposition 2.2.6, we introduce the following concept, whose use will be clear later.
Definition 2.2.8.
Let . If , then we call a quasi--space. Moreover, if and both and are quasi--spaces, then we call a quasi--subspace of and write or .
By Proposition 2.2.7, the following is immediate.
Proposition 2.2.9.
Any -space is a quasi--space.
However, the converse of Proposition 2.2.9 is not always true, as is implied by (d) in Proposition 2.2.6, unless we add a condition as follows.
Proposition 2.2.10.
Let be a quasi--space. If , then is a -space.
Remark.
If , then .
Proposition 2.2.10 implies that, if and is a quasi--space, then is a -space, and hence in this case -space and quasi--space are actually the same notion. In fact, we shall see that in most of our examples.
Notation 2.2.11.
In this article, denotes an index set unless otherwise specified.
Proposition 2.2.12.
The intersection of any family of quasi--spaces is a quasi--space.
Proof.
Let be a family of quasi--spaces and let . Then , (since ). And by Definition 2.2.8, , . Hence , . Therefore, . ∎
Proposition 2.2.13.
The union of any family of quasi--spaces is a quasi--space.
For -spaces, we have
Proposition 2.2.14.
The union of any family of -spaces is a -space.
Proof.
Let be a collection of -spaces and let . By Definition 2.2.1, , we can suppose , where . Then is a -space. ∎
However, the intersection of a set of -spaces is not necessarily a -space, as shown below.
Example 2.2.15.
Let
.
Let be given by , let be given by , let (Definition 2.1.6), and let . Then each element of is nonzero (because ) and has the form where and each (because each element of has the form). Assume that is a -space , where . Because , we can tell that either or . However, it follows that either or , a contradiction.
Nevertheless, we have
Proposition 2.2.16.
Let be the intersection of any family of -spaces. Then is a quasi--space. Moreover, if or more generally, , then is a -space.
2.3. -morphisms
A generalized morphism which we call a -morphism between -spaces is only required to satisfy a simple but important condition.
Definition 2.3.1.
Let be an operator semigroup and let be a map from a -space to a -space. If , commutes with on , i.e. , and , then is called a -morphism.
Remark.
-
(1)
By Proposition 2.2.7, and , , and hence is well-defined.
-
(2)
Note that if and , then is not defined unless .
-
(3)
We regard the empty map as a -morphism from the -space to .
The following four results show that the notion of -morphism can characterize some familiar notions in mathematics.
Proposition 2.3.2.
Let be a subfield of a field . Let be the operator semigroup defined in Example 2.1.2. Then , a map is a ring homomorphism with fixed pointwisely if and only if is a -morphism.
Proof.
Clearly, and are the rings and , respectively.
1. Sufficiency
Suppose that is a -morphism. Then by Definition 2.3.1, , and .
Considering the case where is a constant polynomial function given by , we can tell that , , i.e. is fixed pointwisely under .
Let . Then with and . Hence
( is defined in Example 2.1.2)
(since is a -morphism)
(since is a -morphism)
.
And
(since is a -morphism)
(since is a -morphism)
.
Therefore, is a ring homomorphism with fixed pointwisely.
2. Necessity
Suppose that is a ring homomorphism with fixed pointwisely. Then
, , and
, and .
Let and let . Because is a polynomial function, it is not hard to see that , as desired. ∎
Remark.
Then for ring isomorphisms, we have
Corollary 2.3.3.
Let be a subfield of a field , let be the operator semigroup defined in Example 2.1.2, and let be algebraic over with , where denotes the degree of the minimal polynomial of over . Then a map is a ring isomorphism with fixed pointwisely if and only if it is a -morphism.
Proof.
By Proposition 2.3.2, is a ring homomorphism with fixed pointwisely if and only if it is a -morphism. So it suffices to show that the ring homomorphism is bijective.
Since is algebraic over , is a field. It follows that the ring homomorphism must be injective. Moreover, since , must be surjective. ∎
Recall that every finite field extension is algebraic and every finite separable field extension is simple. So by Corollary 2.3.3, we immediately have the following, which is for later use.
Corollary 2.3.4.
Let be a finite separable field extension. Let be the operator semigroup defined in Example 2.1.2. Then is a -space generated by a single element of . Moreover, a map is a field automorphism with fixed pointwisely if and only if it is a -morphism.
Note that every ring homomorphism discussed above involves only one polynomial ring ( in Example 2.1.2). We will treat ring homomorphisms which involve two (distinct) polynomial rings in Subsection 6.1. Moreover, ring homomorphisms involving polynomial rings in more than one variable will be treated in Sections 8 and 9.
The following may be a little surprising.
Proposition 2.3.5.
Let be a topological space, let be the power set of , and let , given by . Let a map induce a map as follows.
that is closed in , let , where with a slight abuse of notation denotes the set . And which is not closed in , let .
Then is continuous if and only if is a -morphism.
Remark.
If , then , and hence .
Proof.
1. Necessity
Suppose that is continuous. To prove that is a -morphism, it suffices to show that , and . The equation holds when . So we only need to show that , .
If , by the definitions of and ,
as desired.
Since is continuous, (see e.g. [6]). Hence . On the other hand, because . So . Thus if , then
as desired.
2. Sufficiency
Suppose that is a -morphism. To prove that is continuous, we only need to show that , with .
Assume such that and . Let . Then and (because otherwise ). Hence
but
because Therefore, , which is contrary to the assumption that is a -morphism. ∎
The map in Proposition 2.3.5 is from a topological space to itself. We will treat continuous functions between different topological spaces in Subsection 6.1.
Several properties of -morphisms are as follows. We shall employ them later.
Proposition 2.3.6.
Proof.
Moreover, if or , then by Proposition 2.2.10, is a -space, and thus . ∎
Definition 2.3.7.
Let be a -space. We call a -morphism from to a -endomorphism of , and we denote by End the collection of all -endomorphisms of .
Proposition 2.3.8.
Let be a -space. Then constitutes a monoid, which we still denote by , with composition of functions as the binary operation.
Remark.
We regard the empty function as the identity function Id. So .
Proof.
The identity map on lies in , and so we only need to show that is a semigroup with composition of functions as the binary operation.
Let , let and let . Then
So commutes with on , and hence .
Composition of functions is associative. Therefore, is a monoid with composition of functions as the binary operation. ∎
Then the following is obvious.
Proposition 2.3.9.
Let be a -space. Then the intersection of any family of submonoids of is again a submonoid of .
2.4. -isomorphisms
Definition 2.4.1.
Let , and be -spaces.
Let be a -morphism from to . If is bijective, then we call a -isomorphism from to . We denote by Iso the family of all -isomorphisms from to .
Moreover, if is a -isomorphism from to itself, then we call a -automorphism of . We write Aut for the collection of all -automorphisms of .
To justify Definition 2.4.1, we have
Proposition 2.4.2.
Let and be -spaces, let and let be the inverse map of . Then .
Proof.
By Definition 2.3.1, and , , and so . Then and , . Hence is a -morphism from to , and so . ∎
Proposition 2.4.3.
Let be a -space. Then constitutes a group, which we still denote by , with composition of functions as the binary operation.
Proof.
Let and let be the inverse map of . By Proposition 2.4.2, .
Let . Then by Proposition 2.3.8, . Hence because both and are bijective.
Moreover, composition of functions is associative and the identity map on lies in . Therefore, is a group with composition of functions as the binary operation. ∎
Then the following, which is for later use, is obvious.
Proposition 2.4.4.
Let be a -space. Then the intersection of any family of subgroups of is again a subgroup of .
2.5. Galois -extensions
The remaining part of Section 2 will be used later to study e.g. Galois correspondences. The concept of Galois extensions in the classical Galois theory is generalized as follows.
Definition 2.5.1.
Let be a -space and let . Let
We call a Galois -extension of , and we say that is fixed pointwisely under the action of on .
More generally, let . Then we denote by the set .
In particular, if , then we use instead of for brevity.
Remark.
.
Proposition 2.5.2.
Let be a -space and let be a subset of . Then . Moreover, if or , then .
Proof.
Let and . Then , (by Definition 2.5.1), and so . Thus , and so . Hence is a quasi--space by Definition 2.2.8, and thus .
Moreover, if or , then by Proposition 2.2.10, is a -space, and hence . ∎
However, in Proposition 2.5.2, if , then it is possible that is not a -space. We shall prove this claim right after Example 7.1.10.
The following is obvious.
Proposition 2.5.3.
Let be a -space and let . Then .
2.6. Galois -monoids and Galois -groups
The concept of Galois group in the classical Galois theory is generalized to two notions as follows.
Definition 2.6.1.
Let be a -space and let .
The Galois -monoid of over , denoted by , is the set .
The Galois -group of over , denoted by , is the set .
Remark.
Trivially, because we regard the empty function as the identity function .
To justify Definition 2.6.1, the following two results are needed, of which the first one generalizes Proposition 2.3.8.
Proposition 2.6.2.
Let be a -space and let . Then is a submonoid of with composition of functions as the binary operation.
Proof.
Because , by Proposition 2.3.8, we only need to show that is closed under composition of functions.
Analogously, Proposition 2.4.3 is generalized as follows, where the proof is omitted.
Proposition 2.6.3.
Let be a -space and let . Then is a subgroup of with composition of functions as the binary operation.
We will employ the following three propositions later, of which the proofs are obvious.
Proposition 2.6.4.
Let be a -space. Then .
Proposition 2.6.5.
Let be a -space. Then , , and , .
Proposition 2.6.6.
Let be a -space. Then , and .
2.7. Generated submonoids of and subgroups of
We will use the following notions and results later.
Definition 2.7.1.
Let be a -space.
If , then the intersection of all submonoids of containing , denoted by , is called the submonoid of generated by X.
If , then the intersection of all subgroups of containing , denoted by , is called the subgroup of generated by .
Remark.
Proposition 2.7.2.
Let be a -space and let . Suppose that , . Then .
Proof.
Since and every subgroup of is a submonoid of , the set of all subgroups of containing is a subset of the set of all submonoids of containing . Then by Definition 2.7.1 we can tell that . So it is sufficient to prove that .
If is empty, then Id on , as desired.
Suppose that is not empty. Let . Since is the group generated by , , and such that (which is a product of group elements). Because it is assumed that , , we can tell that any submonoid of containing also contains . Hence , as desired. ∎
3. Galois correspondences
In this section, we shall develop two main results on Galois correspondences. Roughly speaking, the first one shows the Galois correspondence between the Galois -monoids (Definition 2.6.1) of a -space and the fixed subsets of under actions of -endomorphisms of , and the second one shows the Galois correspondence between the Galois -groups (Definition 2.6.1) of a -space and the fixed subsets of under actions of -automorphisms of .
For a -space and , by Proposition 2.5.2, is a quasi--subspace of , rather than a -subspace of unless or . This fact implies that, quasi--subspaces, rather than -subspaces, of a -space may play a major role in establishing Galois correspondences, though Proposition 2.2.10 tells us that in the case such as , quasi--space and -space are actually the same notion.
Notation 3.0.1.
Let be a -space and let . We denote by the set of all quasi--subspaces of , and we denote by the set of all intermediate quasi--spaces between and , i.e. the set of all quasi--spaces with .
Remark.
-
(1)
By Proposition 2.2.10, both and are sets of -spaces if .
-
(2)
.
Notation 3.0.2.
We denote by the set of all subgroups of a group , and we denote by the set of all submonoids of a monoid .
3.1. Two examples
Let be a -space and let be a subset of . Then by Definition 2.5.1, is a Galois -extension of . Let’s check whether there always exists a correspondence between and as in the classical Galois field theory (see Definition 2.6.1 for the notation ).
Example 3.1.1.
Let given by . Then is an operator semigroup on , and . Therefore, is a -space and the only (quasi-)-subspaces of are and .
Let . Then , commutes with on . Thus for and given by ,
which implies that is a translation along the real axis. Hence and . On the other hand, we can tell . Thus, . Then is a Galois -extension of and (since the only (quasi-)-subspaces of are and ).
However, has an infinite number of subgroups, so
The preceding example shows that the number of all intermediate (quasi-)-spaces of a Galois -extension may be strictly smaller than that of all subgroups of the corresponding Galois -group. The converse is also possible, as shown below.
Example 3.1.2.
Let Id. Then is an operator semigroup on any set and any subset of is a -space. Obviously any map from to belongs to , and so any bijective map from to lies in .
Suppose . Then , which is the symmetric group on letters. Then is a Galois -extension of , is the power set of , and
Then
Example 3.1.2 shows that the number of all intermediate (quasi-)-spaces of a Galois -extension may be strictly larger than that of all subgroups of the corresponding Galois -group.
However, if we focus on the fixed subsets of a -space under actions of -automorphisms of and on the subgroups (of ) which are Galois -groups, then we will see a Galois correspondence. Moreover, we shall find a Galois correspondence for Galois -monoids as well.
3.2. Galois correspondences
First, we need to define the fixed subsets of a -space under actions of -endomorphisms (or -automorphisms) of as follows.
Definition 3.2.1.
Let be a -space and let . Then
and
Remark.
The following characterizes and .
Lemma 3.2.2.
Let be a -space and let . Then
and
Proof.
1. Sufficiency
If , then by Definition 3.2.1.
If , then by Definition 3.2.1.
2. Necessity
On the other hand, by Proposition 2.6.4.
Therefore, .
If , analogously we can show . ∎
Second, we define
Definition 3.2.3.
Let be a -space and let . Then
and
Remark.
The following characterizes and .
Lemma 3.2.4.
Let be a -space, let , and let . Then
and
Proof.
1. Sufficiency
If , then by Definition 3.2.3.
If , then by Definition 3.2.3.
2. Necessity
On the other hand, by Proposition 2.6.5.
Therefore, .
Analogously, if , we can show . ∎
Now two main results of this section follow. Roughly speaking, the first one shows the Galois correspondence between the Galois -monoids of a -space and the fixed subsets of under actions of -endomorphisms of .
Corollary 3.2.5.
Let be a -space and let . Then the correspondences
and
define inclusion-inverting mutually inverse bijective maps i.e. is an inclusion-reversing bijective map whose inverse is also an inclusion-reversing bijective map between and .
Proof.
Analogously, we have the following, which, roughly speaking, shows the Galois correspondence between the Galois -groups of a -space and the fixed subsets of under actions of -automorphisms of .
Corollary 3.2.6.
Let be a -space and let . Then the correspondences
and
define inclusion-inverting mutually inverse bijective maps between and .
Proof.
The same as that of Corollary 3.2.5. ∎
As a matter of fact, Corollaries 3.2.5 and 3.2.6 tell us, essentially, what Galois correspondences are: They tell us where to find Galois correspondences in any specific case. In fact, with examples we gave (e.g. Proposition 2.3.5) or will give for -morphisms, we can tell from Corollaries 3.2.5 and 3.2.6 that Galois correspondences are ubiquitous.
In Section 11, we shall prove that Corollaries 3.2.5 and 3.2.6 still hold for the case where elements of are allowed to be functions in more than one variable and/or even partial.
As is well known, in the Galois theory for infinite algebraic field extensions, Krull topology is put on Galois groups, and in differential Galois theory, differential Galois groups are endowed with Zariski topology. Then the fundamental theorem in each of the two theories tells us that there exists a Galois correspondence between the set of all closed subgroups of the Galois group of a Galois field extension (or a Picard-Vessiot extension) and the set of all intermediate (differential) fields.
Now for Corollary 3.2.5, can we define a topology on and a topology on such that is the set of all closed quasi--subspaces (of ) containing and is the set of all closed submonoids of ? If such topologies exist, then we can characterize the Galois correspondence between and in terms of closed quasi--subspaces of and closed submonoids of .
And for Corollary 3.2.6, can we define a topology on and a topology on with being the set of all closed quasi--subspaces (of ) containing and being the set of all closed subgroups of ? If such topologies exist, then we can characterize the Galois correspondence between and in terms of closed quasi--subspaces of and closed subgroups of .
Before we can answer the above questions in Section 5, we need to address a related issue in Section 4, which is about the lattice structures of the sets involved in Corollaries 3.2.5 and 3.2.6.
There is one more thing which we have not talked about for Galois correspondences. It is normally the last part of the fundamental theorem of a Galois theory, i.e. the correspondence between the normal subgroups of a Galois group and the Galois extensions of the base field. We will address this issue in Section 20.
4. Lattice structures of objects arising in Galois correspondences on a -space
In this section, we shall study lattice structures of objects which arise in Corollaries 3.2.5 and 3.2.6. The value of this section, though being a preparation for the next section, asserts itself.
Recall that a lattice is a (nonempty) partially ordered set in which every pair of elements has a meet (greatest lower bound) and a join (least upper bound), and is a complete lattice if any subset of it has a greatest lower bound and a least upper bound. Note that if lattice is complete, then and (see e.g. [2]).
Notation 4.0.1.
To abbreviate our notations, by , and we mean , and , respectively.
4.1. , and
Proposition 4.1.1.
Let be a -space. Then Notation 3.0.1 is a lattice with inclusion as the binary relation if , and . Moreover, is a complete lattice if let and let , .
Proof.
Proposition 4.1.2.
For (Notation 3.0.2), we also have
Proposition 4.1.3.
Let be a -space. Then is a lattice with inclusion as the binary relation if and Definition 2.7.1, . Moreover, if let and let , , then is a complete lattice.
Moreover, we have
Proposition 4.1.4.
Let be a -space. Then is a complete sublattice of the complete lattice defined in Proposition 4.1.2.
Proof.
4.2. and
Lemma 4.2.1.
Let be a -space and let be a set of subsets of . Then .
Proof.
∎
Lemma 4.2.1 implies
Proposition 4.2.2.
is a complete -sublattice of the complete lattice defined in Proposition 4.1.1.
Proof.
Moreover, by Proposition 2.5.2, .
Therefore, is a complete -sublattice of the complete lattice defined in Proposition 4.1.1. ∎
However, is not necessarily a -sublattice of the lattice defined in Proposition 4.1.1. In fact, it is possible that is not even a -semilattice if , , as shown in the following example.
Example 4.2.3.
Let be the operator semigroup on induced by as defined in Example 2.1.2. That is, let
Let a -space . Then and is a finite Galois field extension (see e.g. [7]).
By the way, note that in the preceding example, is not a subfield of . This shows that the notion of -space is not equivalent to the notion of field in this case. Basically this is because all elements of defined in Example 2.1.2 are unary functions. We will solve this problem in Section 9 (cf. Example 9.2.2).
Because , the following is a straightforward consequence of Lemma 4.2.1.
Lemma 4.2.4.
Let be a -space and let be a set of subsets of . Then .
Analogously, Lemma 4.2.4 implies
Proposition 4.2.5.
is a complete -sublattice of the complete lattice defined in Proposition 4.1.1.
4.3. and
Lemma 4.3.1.
Let be a -space and let be a set of subsets of . Then
Proof.
Lemma 4.3.1 implies
Proposition 4.3.2.
is a complete -sublattice of the complete lattice defined in Proposition 4.1.2.
Proof.
By Definition 3.2.3, each element of has the form , where . Then by Lemma 4.3.1,
Moreover, by Proposition 2.6.2, .
Therefore, is a complete -sublattice of the complete lattice defined in Proposition 4.1.2. ∎
Nevertheless, the following example tells us that is not necessarily a -sublattice of the lattice defined in Proposition 4.1.2. Indeed, if , , then it is possible that is not even a -semilattice.
Example 4.3.3.
Let and let {Id on }. Then is a -space and is the set of all maps from to . Let
and
Then as monoids, by Definition 3.2.3. And it is not hard to see that .
Thus by Proposition 4.1.2, .
However, since the transposition of and does not lie in ,
Hence by Lemma 3.2.4, .
Therefore, is not a -sublattice of the lattice defined in Proposition 4.1.2. In fact, is not even a -semilattice if , .
By a similar argument as in the proof of Lemma 4.3.1, we can show the following, whose proof is omitted.
Lemma 4.3.4.
Let be a -space and let be a set of subsets of . Then .
Proposition 4.3.5.
is a complete -sublattice of the complete lattice defined in Proposition 4.1.3.
Proof.
By Definition 3.2.3, each element of has the form , where . Then by Lemma 4.3.4,
Moreover, by Proposition 2.6.3, .
Therefore, is a complete -sublattice of the complete lattice defined in Proposition 4.1.3. ∎
However, the following example shows that is not necessarily a -sublattice of the lattice defined in Proposition 4.1.3. Indeed, if , , then it is possible that is not even a -semilattice.
5. Topologies employed to construct Galois correspondences
As was promised near the end of Section 3, now we are going to answer the following questions.
To construct the Galois correspondence described in Corollary 3.2.5 without resort to Definition 3.2.1 or 3.2.3, can we define a topology on and a topology on such that is the set of all closed quasi--subspaces (of ) containing and is the set of all closed submonoids of ? This question will be answered in Subsections 5.1 and 5.3.
And for Corollary 3.2.6, can we define a topology on and a topology on with being the set of all closed quasi--subspaces (of ) containing and being the set of all closed subgroups of ? This question will be answered in Subsections 5.2 and 5.4.
5.1. Topologies on a -space for
Lemma 5.1.1.
Let be a -space with . Then the topology on generated by a subbasis is the smallest coarsest topology on such that the collection of all closed sets contains .
Proof.
Since the union of all elements of is , qualifies as a subbasis for a topology on . Hence the topology generated by equals the intersection of all topologies on that contain (see e.g. [6]), and so is the smallest topology on containing . Thus by the definition of , is the smallest topology on such that the collection of all closed sets contains . ∎
For a -space with , to characterize the Galois correspondence depicted in Corollary 3.2.5 in terms of closed quasi--subspaces of , i.e. to replace in Corollary 3.2.5 with the set of all closed quasi--subspaces (of ) containing , we need a topology on such that the following equation is satisfied.
(5.1) |
Remark.
The reason why we require the above equality rather than the following one is explained as follows.
(5.2) |
Suppose . Then and is closed. Hence lies in the right side of (5.2). But it is possible that (e.g. when and ), and thus in this case we cannot replace with the right side of (5.2).
The following is to determine whether there exists a topology on such that Equation 5.1 is satisfied: If defined as follows is nonempty, then we can replace with the set of all nonempty closed quasi--subspaces (of ) containing .
Theorem 5.1.2.
Let be a -space and let . Let
and let
where is defined in Lemma 5.1.1. Then and the following statements are equivalent:
Proof.
By Lemma 5.1.1, is the smallest topology on such that the collection of all closed sets contains . Suppose . Then all elements of are closed in , and so . Hence . Thus .
(i) (ii): Let . To prove (ii), we only need to consider the case where . Then by the definition of ,
By Proposition 2.2.13, . Hence
which equals by (5.1). Therefore, , and so (ii) is true.
(ii) (iii): By Definition 3.2.1 and Proposition 2.5.2, . Then from Proposition 4.1.1, we can tell that (iii) is true if (ii) is true.
(iii) (iv): By the definition of in Lemma 5.1.1, all elements of are closed in , and so by Definition 3.2.1 and Proposition 2.5.2,
On the other hand, we show
as follows.
Let such that is nonempty and closed in . Then we need to show .
Since is the collection of all unions of finite intersections of elements of , by DeMorgan’s Laws, the collection of all closed sets in is the family of all intersections of finite unions of elements of . Then is an intersection of finite unions of elements of . Because (iii) is supposed to be true, any finite union of elements of belongs to . So is an intersection of elements of . Then by Lemma 4.2.1, , as desired. Hence
Therefore,
Thus . We showed at the beginning of our proof, and so by the definition of , is the coarsest topology in . Then by the definition of , is the coarsest topology on such that Equation 5.1 is satisfied.
(iv) (i): Obvious. ∎
By Theorem 5.1.2, when , we can put a topology such as on such that Equation (5.1) is satisfied. Then in Corollary 3.2.5 can be replaced by { is closed and nonempty} if , and it can be replaced by { is closed} if .
However, Example 4.2.3 tells us that statement (ii) (and (iii)) in Theorem 5.1.2 is not always true. Hence it is possible that . When this occurs, we cannot endow with any topology such that Equation (5.1) is satisfied. Then in this case, there is no topology on to characterize the Galois correspondence depicted in Corollary 3.2.5 in terms of closed subsets of .
5.2. Topologies on a -space for
To replace in Corollary 3.2.6, we have
Lemma 5.2.1.
Let be a -space with . Then the topology on generated by a subbasis is the smallest topology on such that the collection of all closed sets contains .
Proof.
Almost the same as the proof of Lemma 5.1.1 except that , and are replaced by , and , respectively. ∎
For a -space with , to characterize the Galois correspondence depicted in Corollary 3.2.6 in terms of closed quasi--subspaces of , we need a topology on such that the following equation is satisfied.
(5.3) |
Theorem 5.2.2.
Let be a -space and let . Let
and let
where is defined in Lemma 5.2.1. Then and the following statements are equivalent:
Proof.
It follows that when , we can give a topology such as on such that Equation (5.3) is satisfied. Then in Corollary 3.2.6 can be substituted with { is closed and nonempty} if , and it can be substituted with { is closed} if .
Recall that at the end of Subsection 5.1, we use Example 4.2.3 to show that statement (ii) (and (iii)) in Theorem 5.1.2 is not always true. Because in Example 4.2.3, , statement (ii) (and (iii)) in Theorem 5.2.2 is not necessarily true, either. Hence it is possible that . When this occurs, we cannot give any topology on such that Equation (5.3) is satisfied. Then in this case, there does not exist any topology on such that the Galois correspondence depicted in Corollary 3.2.6 can be characterized in terms of closed subsets of .
5.3. Topologies on
To replace in Corollary 3.2.5, we need
Lemma 5.3.1.
Let be a -space with . Then the topology on generated by a subbasis
is the smallest topology on such that the collection of all closed sets contains .
Proof.
Since the union of all elements of is , qualifies as a subbasis for a topology on . Hence the topology generated by equals the intersection of all topologies on that contain (see e.g. [6]), and so is the smallest topology on containing . Thus by the definition of , is the smallest topology on such that the collection of all closed sets contains . ∎
For a -space with , to characterize the Galois correspondence depicted in Corollary 3.2.5 in terms of closed submonoids of , we need a topology on such that the following equation is satisfied.
(5.4) |
Remark.
See Notation 3.0.2 for .
Theorem 5.3.2.
Let be a -space and let . Let
and let
where is defined in Lemma 5.3.1. Then and the following statements are equivalent:
-
(i)
.
-
(ii)
For any intersection of finite unions of elements of , implies .
-
(iii)
; that is, is the coarsest topology on such that Equation 5.4 is satisfied.
Moreover, if is a complete -sublattice of the complete lattice defined in Proposition 4.1.2, then the above three statements are true.
Proof.
By Lemma 5.3.1, is the smallest topology on such that the collection of all closed sets contains . Suppose . Then all elements of are closed in , and so . Hence . Thus .
(i)(ii): For any intersection given in (ii), from Definitions 3.2.3 and 2.6.1, we can tell . Suppose . Then . Since , let . By the definition of , every element of is closed in , and so is the intersection (of finite unions of elements of ). Again by the definition of , .
(ii)(iii):
By the definition of in Lemma 5.3.1, all elements of are closed in , and so by Definition 3.2.3 and Proposition 2.6.2,
On the other hand, we prove
as follows.
Let such that is closed in . We need to show .
Because is generated by subbasis , is the collection of all unions of finite intersections of elements of
Since is closed in , is a union of finite intersections of elements of . It follows from DeMorgan’s Laws that is an intersection of finite unions of elements of . Because , is an intersection of finite unions of elements of . Since (ii) is assumed to be true and , , as desired. Hence
Therefore,
Thus . We showed at the beginning of our proof. So by the definition of , is the coarsest topology in . Then by the definition of , is the coarsest topology on such that Equation 5.4 is satisfied.
(iii)(i): Obvious.
By Proposition 4.3.2, any intersection of elements of still lies in . Then as a result of the first distributive law of set operations, any intersection of finite unions of elements of is a union of elements of . Suppose that is a complete -sublattice of the complete lattice defined in Proposition 4.1.2. Then implies , and so (ii) is true (in this case). ∎
When , by Theorem 5.3.2, we can endow with a topology such as such that Equation 5.4 is satisfied. Then in Corollary 3.2.5 can be replaced by the set of all closed submonoids of .
Nevertheless, it is possible that . Recall that in Example 4.3.3, and , but . Thus for Example 4.3.3, statement (ii) in Theorem 5.3.2 is not true, and hence . Therefore, it is possible that we cannot endow with any topology such that the Galois correspondence depicted in Corollary 3.2.5 can be characterized in terms of closed submonoids of .
5.4. Topologies on
To replace in Corollary 3.2.6, we need
Lemma 5.4.1.
Let be a -space and let . Then the topology on generated by a subbasis
is the smallest topology on such that the collection of all closed sets contains .
Proof.
Almost the same as the proof of Lemma 5.3.1 except that , , and are replaced by , , and , respectively. ∎
For a -space with , to characterize the Galois correspondence depicted in Corollary 3.2.6 in terms of closed subgroups of , we need a topology on such that the following equation is satisfied.
(5.5) |
Remark.
See Notation 3.0.2 for .
Theorem 5.4.2.
Let be a -space and let . Let
and let
where is defined in Lemma 5.4.1. Then and the following statements are equivalent:
-
(i)
.
-
(ii)
For any intersection of finite unions of elements of , implies .
-
(iii)
; that is, is the coarsest topology on such that Equation 5.5 is satisfied.
Moreover, if is a complete -sublattice of the complete lattice defined in Proposition 4.1.3, then the above three statements are true.
Proof.
Nevertheless, we do not know whether the three equivalent statements in Theorem 5.4.2 are always true, and hence we do not know whether there always exists a topology on such that Equation 5.5 is satisfied. In other words, for any -space with , we do not know whether we can always endow with a topology such that in Corollary 3.2.6 can be replaced by the set of all closed subgroups of .
6. Generalized morphisms and isomorphisms from a -space to a -space
In this section, we shall define generalized morphisms and isomorphisms from a -space to a -space, where both and are operator semigroups.
6.1. -morphisms
In algebra, ring homomorphisms from the ring to the ring , where both and are commutative rings, are defined. However, Definition 2.3.1 does not cover this kind of morphisms because, by Example 2.1.2, and may induce different operator semigroups. So we generalize Definition 2.3.1 as follows.
Definition 6.1.1.
Let and be operator semigroups, let be a map from to , and let be a map from a -space to a -space. If and , , then is called a -morphism.
Remark.
If and is the identity map, then is also a -morphism defined by Definition 2.3.1.
Let us see two examples as follows.
Proposition 6.1.2.
Let and be differential rings, let resp. be the operator semigroup on resp. defined as in Example 2.1.5, and let a map be given by , , where and are derivations of and , respectively. Let be a ring homomorphism. Then is a differential homomorphism if and only if is a -morphism.
Proof.
Recall that a ring homomorphism is a differential homomorphism if and only if it satisfies (see e.g. [1]). Thus we only need to show that is a -morphism if and only if satisfies . Since , , the latter equivalence is obvious by Definition 6.1.1. ∎
Now Proposition 2.3.5 can be generalized to continuous functions from a topological space to another.
Proposition 6.1.3.
Let and be topological spaces. Let and denote the power sets of and , respectively. Let
and
where and denote the identity functions on and , respectively. Let a map induce a map as follows.
that is closed in , let , where denotes the set for convenience. And which is not closed in , let .
Let be the map from to given by and . Then is continuous if and only if is a -morphism.
Proof.
Our proof is analogous to that of Proposition 2.3.5.
Both and are operator semigroups. and are -space and -space, respectively.
1. Necessity Suppose that is continuous. To prove that is a morphism, it is sufficient to show that and , . The equation holds when . So it suffices to show that
, .
If , by the definitions of , and ,
,
as desired.
Since is continuous, (see e.g. [6]). Hence . On the other hand, because . So . Thus if , then
,
as desired.
2. Sufficiency Suppose that is a -morphism. To prove that is continuous, we only need to show that , with .
Assume with and . Let . Then and (because otherwise ). Hence
,
but
because . Therefore, , which is contrary to the assumption that is a -morphism. ∎
However, when we tried to generalize Proposition 2.3.2 to the case of -morphisms, we found that in Definition 6.1.1, the condition being a map is too strong for ring homomorphisms. Basically this is because in Example 2.1.2, the map which induces is not necessarily injective (and so defined in Proposition 6.1.10 is not necessarily a map). Hence we generalize Definition 6.1.1 to Definition 6.1.8, where the condition being a map is replaced by a weaker one.
First, we allow to be a binary relation, not necessarily a function.
Notation 6.1.4.
Let (resp. ) be a set, let (resp. ) be a set of functions from to (resp. a set of functions from to ), let binary relation , and let . We denote by the binary relation , where is the restriction of to .
Definition 6.1.5.
Let and be defined as in Notation 6.1.4. Let and . If , , then we say that is well-defined and let .
Proposition 6.1.6.
Let and be defined as in Notation 6.1.4. Then the following statements are equivalent:
-
(i)
is a map.
-
(ii)
is well-defined, and .
Proof.
is a map;
, ;
and , ;
and , is well-defined. ∎
Moreover, to simplify our descriptions, we need a convention as follows.
Convention 6.1.7.
In this article, unless otherwise specified, by an equation we always imply that both sides of the equation are well-defined.
Then Definition 6.1.1 is generalized to
Definition 6.1.8.
Let and be operator semigroups, let , and let be a map from a -space to a -space. If and , , then is called a -morphism.
Corollary 6.1.9.
Let be a map, not necessarily a -morphism, given as in Definition 6.1.8. If is a -morphism, then is a map. Conversely, if is a map, then and , is well-defined.
Proof.
Remark.
From Corollary 6.1.9, we can tell that the condition being a map in Definition 6.1.1 is replaced by a weaker one (i.e. being a map) in Definition 6.1.8. So Definition 6.1.1 is a special case of Definition 6.1.8.
In the remaining part of Section 6 and Section 7, a -morphism is always defined by Definition 6.1.8 unless otherwise specified.
Now Proposition 2.3.2 can be generalized to
Proposition 6.1.10.
Let resp. be a field extension, let resp. be the corresponding operator semigroup defined as in Example 2.1.2, let be a field isomorphism, and let be the map given by where
for .
Let , where and are the maps which induce and , respectively, defined as in Example 2.1.2. Then and , a map is a ring homomorphism extending if and only if is a -morphism.
Proof.
Obviously, with , and and are the rings and , respectively.
1. Sufficiency Suppose that is a -morphism. By Definition 6.1.8, , and .
Let be a constant polynomial function given by . Then by the definitions of and , we can tell that is the constant polynomial function given by . Thus, , , that is, extends .
Let . Then with and . Hence
(since is a -morphism)
(by the definition of )
(since is a field isomorphism)
(since by Corollary 6.1.9, is a function)
(since is a -morphism)
.
And
(since is a -morphism)
(by the definition of )
(since is a field isomorphism)
(since by Corollary 6.1.9, is a function)
(since is a -morphism)
.
Therefore is a ring homomorphism extending .
2.Necessity Suppose that is a ring homomorphism extending . Then , , and , and .
Let and let . Then
that is
(6.1) |
To show that is a -morphism, by Corollary 6.1.9, we need to show that is a map. Let with . By the definition of , such that and , and so .
Assume ; that is, . Then such that . However, since , , and so from Equation 6.1, we can tell that , a contradiction.
Hence must be a map. Then and ,
(by Equation 6.1)
(by the definition of and the fact that is a map).
So by Definition 6.1.8, is a -morphism. ∎
Note that up to now all polynomial rings involved are in one variable. We shall discuss ring homomorphisms involving polynomial rings in more than one variable in Sections 8 and 9.
For of -morphisms , we generalize Proposition 2.3.6 as follows.
Proposition 6.1.11.
6.2. -isomorphisms
The notion of -isomorphism is generalized as follows.
Definition 6.2.1.
Let be a -morphism from a -space to a -space . If is bijective, then we call a -isomorphism. Moreover, we denote by the set of all -isomorphisms from to .
To justify the definition, we need to show that is a -morphism from to . For this purpose, we need
Lemma 6.2.2.
Let be a -isomorphism from a -space to a -space . Then is a map, where .
Proof.
Proposition 6.2.3.
Let and be a -space and a -space, respectively, let and let be the inverse map of . Then .
7. Constructions of the generalized morphisms and isomorphisms
For a -morphism or a -morphism from a -space , Subsections 7.1, 7.2 and 7.3 discuss how to construct it by a map from a set which generates . Specifically, Subsection 7.1 is about a construction of a -morphism from , where , by a map from . And Subsection 7.2 generalizes the construction to a -morphism from , where , by a map from . Then in Subsection 7.3, the construction is further generalized to -morphisms. In Subsection 7.4, we shall introduce a way of constructing -morphisms in terms of topology.
Moreover, some notions and results in algebra are generalized in Section 7 as “byproducts” (see e.g. Propositions 7.1.2, 7.1.3 and 7.1.4).
7.1. A construction of -morphisms from
Let be a field extension and let be algebraic over . If and share the same minimal polynomial over , then
In light of this observation, we introduce the following.
Notation 7.1.1.
Let be a set, let be a set of maps from to , and let . Then by we mean that
, if , then .
Moreover, by we mean that
and ,
or equivalently, by we mean that
, .
Besides, we denote by and the set and the set , respectively.
Remark.
Apparently, defines an equivalence relation on .
The following serves as an example.
Proposition 7.1.2.
Let be a subfield of a field . Let be the operator semigroup defined in Example 2.1.2.
-
(a)
If is algebraic over , then
{all roots of },
where is the minimal polynomial of over .
-
(b)
If is transcendental over , then
and { is transcendental over }.
Proof.
For (a): Let be algebraic over .
Let . By Notation 7.1.1, (cf. Example 2.1.2), implies . So , implies . Thus is a root of . Hence
{all roots of }.
Moreover, let be a root of . Then , . It follows that , ; that is, , and so . Hence
{all roots of }.
On the other hand, by Notation 7.1.1. Therefore,
{all roots of }.
For (b): Let be transcendental over .
Then , implies . So , implies . Then , , and so .
Hence {}.
, if is also transcendental over , then as was just shown for , . So
{ is transcendental over }.
On the other hand, if is algebraic over , then by (a), , i.e. is false; That is, if is true, then is transcendental over . So
{ is transcendental over }.
Therefore, { is transcendental over }. ∎
We will talk more about the notion of transcendental elements in Section 21.
Recall that in the classical Galois theory, each element of the Galois group of a polynomial permutes its roots. From Proposition 7.1.2, we can tell that the following two results generalize this well-known fact.
Proposition 7.1.3.
Let be a -morphism from a -space (to a -space). Then , and so .
Proposition 7.1.4.
Let be a -morphism from a -space (to a -space) such that is injective. Then , and so .
Proof.
The converse of Proposition 7.1.3 may imply a potential way to construct -morphisms. However, the converse of Proposition 7.1.3 is not true. That is, for a map from a -space , the condition “, ” is not sufficient for to be a -morphism, as shown below.
Example 7.1.5.
Let be a differential ring with a derivation such that (it is easy to find such differential ring), let be the operator semigroup on given by (note that unlike the operator semigroup defined in Example 2.1.5, now ), let be the constant ring of , and let a map be given by , where and . Then and , implies . Hence , . But obviously and such that . So , a map from the -space to itself, is not a -morphism.
Thus, to make a map from a -space be a -morphism, we need to impose a condition on which is stronger than “, ”. Before we do this by Proposition 7.1.7, let’s see a fact as follows.
Proposition 7.1.6.
Let . Then implies , .
Remark.
So, if and there exists a map given by , then , (since , ). We shall see in Proposition 7.1.7 that the map is a -morphism.
Proof.
Suppose . Let such that . Then by Notation 7.1.1, it suffices to show .
Proposition 7.1.7.
Let . Then
-
(a)
if and only if there exists a map given by .
-
(b)
The map given in (a) is a -morphism.
Proof.
For (a):
;
, implies ;
given by , is a well-defined map.
For (b): Let . Then such that . Let . Then . By Definition 2.1.1, , and so by the definition of in (a), . Besides, also by the definition of . Combining the above equations, we have . Therefore, is a -morphism. ∎
For -isomorphisms (Definition 2.4.1), we have the following straightforward result of Proposition 7.1.7.
Corollary 7.1.8.
Let .
-
(a)
if and only if there exists a bijective map given by .
-
(b)
The bijective map given in (a) is a -isomorphism.
The converse of statement (b) in Proposition 7.1.7 raises a question as follows.
Question 7.1.9.
For and a -morphism from , does there exist any such that ?
The significance of this question is that, in the cases where the answer is yes, we can construct a -morphism by letting . Proposition 7.1.7 implies that there are cases where the answer to the question is positive, and we shall talk about these cases in Subsection 7.2. It seems not easy to find a case which gives negative answer to the question, but we found one as follows.
Example 7.1.10.
Define by and by . Let (Definition 2.1.6) and .
Then each can be expressed as a finite composite of alternating “powers” of or as follows, where each or denotes the composite of copies of or , respectively.
(7.1) |
(7.2) |
(7.3) |
or
(7.4) |
where is an even number for 7.1 and 7.4, is an odd number for 7.2 and 7.3, and . In particular, has the form (7.3), where , and has the form (7.2), where .
Correspondingly, , has the form
(7.5) |
(7.6) |
(7.7) |
or
(7.8) |
where is an even number for 7.5 and 7.8, is an odd number for 7.6 and 7.7, and .
Let . Since is transcendental over , the expression for in the above forms is unique. It follows that there is only one in the form of 7.1, 7.2, 7.3, or 7.4 such that . Hence we can define a map as follows.
For all , if has the form 7.1 or 7.2 such that , let , which clearly belongs to , and if has the form 7.3 or 7.4 such that , let .
Now let’s show . Let .
Suppose that has the form 7.5 or 7.6. Then has the form 7.1 or 7.2, and so also has the form 7.1 or 7.2. Hence by the definition of , , as desired.
Suppose that has the form 7.7 or 7.8. Then has the form 7.3 or 7.4, and so also has the form 7.3 or 7.4. Hence by the definition of , , as desired.
Therefore, , and . So .
For 1 and , assume that the answer to Question 7.1.9 is positive, that is, there exists such that . Recall the definitions of and at the beginning of the example. Since has the form (7.3), , and so ; but since has the form (7.2), , and so , a contradiction.
Therefore, for 1 and , the answer to Question 7.1.9 is negative.
Recall that right after Proposition 2.5.2, we claimed that for a -space and , it is possible that is not a -space. Now we show this claim as follows.
Proof.
Let , and be the same as in Example 7.1.10 and let . Let us restrict the domain of to . We can tell that the “new” , which we still denote by for convenience, is an operator semigroup on , and all the argument in Example 7.1.10 still holds. In particular, , and because 2 has the form 7.7. Suppose that is a -space. Then such that . Since , such that . It follows that must be 1, and so . Then because , and hence every element of is fixed under the action of . However, and , a contradiction. ∎
7.2. A construction of -morphisms from
-morphisms discussed in Subsection 7.1 are from -spaces generated by a single element of . To study -morphisms from other -spaces, we generalize Notation 7.1.1 as follows.
Notation 7.2.1.
Let be a set, let be a set of maps from to and let be a map. Then by we mean that
and , if , then .
Moreover, by we mean that
is bijective, and ,
where denotes the inverse of ;
or equivalently, by we mean that
is bijective, and and , .
Now Proposition 7.1.3 can be generalized to
Proposition 7.2.2.
Let be a -morphism from a -space to a -space . Then , , where . In particular, .
However, the converse of Proposition 7.2.2 is not true, as shown below.
Example 7.2.3.
To construct -morphisms, we generalize Proposition 7.1.7 as follows.
Proposition 7.2.4.
Let be a map.
-
(a)
if and only if there exists a map given by , and .
-
(b)
The map given in (a) is a -morphism.
Proof.
For (a):
;
implies , and ; (by Notation 7.2.1)
given by , and , is a well-defined map.
For (b): Let . Then and such that . Let . Then . By Definition 2.1.1, , and so by the definition of , . Besides, also by the definition of . Combining the above equations, we have
Thus is a -morphism. ∎
To construct -isomorphisms, Corollary 7.1.8 is generalized as follows.
Proposition 7.2.5.
Let be a map.
-
(a)
The following statements are equivalent:
-
(i)
.
-
(ii)
is bijective and there exists a bijective map given by , and .
-
(i)
-
(b)
The map given in (ii) is a -isomorphism.
Proof.
For (a):
;
is bijective, and and , if and only if ; (by Notation 7.2.1)
is bijective and given by , and , is a well-defined bijective map.
The converse of statement (b) in Proposition 7.2.4 raises a question as follows, which generalizes Question 7.1.9.
Question 7.2.6.
For and a -morphism from , is there any map such that , and ?
The significance of this question is that, in the cases where the answer is yes, we can construct a -morphism by letting , and . Example 7.1.10 shows a case where the answer to Question 7.2.6 is negative, but soon we shall see that there are cases where the answer is positive. Thus the property in question deserves a name.
Definition 7.2.7.
Let be an operator semigroup on , let , and let be a -morphism from . If there exists a map such that , and , then we say that is constructible by , or just say that is constructible for brevity.
Recall that it is possible that does not contain (Proposition 2.2.6). But if the identity function lies in , which is true in the cases such as Examples 2.1.2, 2.1.4 and 2.1.5, then , where we can get a positive answer to Question 7.2.6 as shown below.
Corollary 7.2.8.
Let . If , then any -morphism from is constructible by .
Proof.
Since , by Definition 2.3.1, , and . Thus is constructible by . ∎
Moreover, if the operator semigroup can be generated by a single element, then we also have a positive answer to Question 7.2.6 as shown below.
Theorem 7.2.9.
Let be a -morphism from to . If (Definition 2.1.6), where is a function from to , then there exists a map such that is constructible by .
Proof.
If is the empty map , then by Definition 7.2.7, is constructible by the empty map trivially. Hence it suffices to show the case where and are nonempty.
Let . and , set
(7.9) |
Let . Let . Suppose . Then by Equation 7.9, , and . To define a map such that is constructible by , let be any . Then , , as desired for to be constructible by .
Therefore, it suffices to show .
To show , we shall first show . Then we shall prove .
Since , , where denotes the composite of copies of .
Let . Because , by Definition 2.3.1,
Applying to the left and right sides of the above equations, we have
(7.10) |
To show , suppose .
Hence , and so
Therefore, to show , it suffices to prove .
Since , and such that . If , then by (with and ), . And if , then , and thus by (with , and the fact that ), . In both cases of , , as desired. Therefore, , as desired.
In summary, to define a map , for each , let be any element of . Then satisfies , and . Hence by Definition 7.2.7, is constructible by . ∎
7.3. A construction of -morphisms
For -morphisms, we generalize Notation 7.2.1 to
Notation 7.3.1.
Let (resp. ) be a set, let (resp. ) be a set of maps from to (resp. a set of maps from to ), let , and let be a map.
By we mean that and ,
implies .
Moreover, by we mean that
is bijective, and ,
where .
Remark.
The following generalizes Proposition 7.2.2.
Proposition 7.3.2.
Let be a -morphism from a -space to a -space . Then , , where . In particular, .
For some reasons which will be clear soon, we are more interested when has the following property.
Definition 7.3.3.
Let and be operator semigroups on and , respectively, let , and let . If , and , then we say that is distributive over .
Remark.
Then to construct -morphisms, we generalize Proposition 7.2.4 as follows.
Proposition 7.3.4.
Let and be operator semigroups on and respectively, let with , and let be a map.
-
(a)
if and only if there exists a map given by , and .
-
(b)
The map given in (a) is a -morphism if is distributive over .
Proof.
For (a):
;
and , implies that ; (by Notation 7.3.1)
given by , and , is a well-defined map.
For (b): Suppose that is distributive over .
Then Proposition 7.2.5 is generalized as follows.
Proposition 7.3.5.
Let , , and be defined as in Proposition 7.3.4. Let such that and . Then
-
(a)
The following statements are equivalent:
-
(i)
.
-
(ii)
is bijective, there exists a bijective map given by , and , and its inverse can be given by , and .
-
(i)
-
(b)
Suppose that is distributive over . Then the bijective map given in (ii) is a -isomorphism.
Proof.
For (a):
;
is bijective, and (by Notation 7.3.1);
is bijective, given by , and , is a well-defined map, and given by , and is also a well-defined map (by Proposition 7.3.4);
is bijective, given by , and , is a well-defined bijective map, and its inverse can be given by , and .
The third equivalence relation is explained as follows. The sufficiency () is obvious, so we only show the necessity (). For this purpose, we show that both and are the identity map.
Let and . Then
(by the definition of )
(by Definition 6.1.5)
(by the definition of )
(by (the version of) Definition 6.1.5).
Then is the identity map on (because ).
Analogously, we can show that is the identity map on . Therefore, both and are bijective and is the inverse of .
For (b):
Question 7.3.6.
Let and be operator semigroups on and , respectively, let , let and let be a -morphism from (to a -space). If , is there any map such that , and ?
The significance of this question is that, in the cases where the answer is yes, we can construct a -morphism by letting , and . The property in question deserves a name as follows, which generalizes Definition 7.2.7.
Definition 7.3.7.
Let and be operator semigroups on and , respectively, let , let and let be a -morphism from (to a -space). If and there exists a map such that , and , then we say that is constructible by , or just say that is constructible for brevity.
Remark.
By Convention 6.1.7, the condition “, and ” implies that is a map.
Since -morphisms are special cases of -morphisms, Example 7.1.10 shows that it is possible that a -morphism from is not constructible by any map even if is distributive over . Now let’s show some cases where -morphisms are constructible. The following generalizes Corollary 7.2.8.
Corollary 7.3.8.
Let and be operator semigroups on and , respectively, let and let . If and , then any -morphism from is constructible by .
Proof.
Since and , by Definition 6.1.8, and , . ∎
Theorem 7.2.9 is generalized for -morphisms as follows.
Theorem 7.3.9.
Let and be operator semigroups on and , respectively, let , and let be a -morphism from to . If , , and is distributive over , then there exists a map such that is constructible by .
Proof.
The proof is analogous to the one for Theorem 7.2.9.
If is the empty map , then by Definition 7.3.7, is constructible by the empty map trivially. Hence we only need to prove the case where and are nonempty.
By (the remark under) Definition 7.3.3, is a map, where , and so is because .
First, and , set
(7.12) |
It is well-defined because is a map.
Let . Let . Suppose . Then by Equation 7.12, , and . To define a map such that is constructible by , let be any . Then , , as desired for to be constructible by .
Therefore, it suffices to show .
To show , we shall first show . Then we shall prove .
Since , , where denotes the composite of copies of .
Let . Because , by Definition 6.1.8,
because is assumed to be distributive over , and so
Applying to both sides of the above equation, we have
(7.13) |
Combining the above three equations, we have . Then by 7.12 (with and ),
Since is assumed to be any element of ,
(7.14) |
Hence , and so
Therefore, to show , it suffices to prove .
Since and , and such that . If , then by 7.12 (with and ), . And if , then because is distributive over , , and hence by 7.12 (with , and the fact that ), . In both cases of , , as desired. Therefore, , as desired.
In summary, to define a map , for each , let be any element of . Then satisfies , and . Hence by Definition 7.3.7, is constructible by . ∎
7.4. Another construction of -morphisms
Let be a -morphism from a -space . If , then , . In other words, if an ordered pair , where is regarded as a binary relation, then , . This observation implies that there exists some sort of basic parts of -morphisms.
To facilitate our descriptions, we give the following notations only for this subsection.
Notation 7.4.1.
Let Id on } and let , where . Besides, both and denote -spaces.
Remark.
Clearly, is a binary relation in .
Proposition 7.4.2.
Let such that is a -subspace of . Then the following statements are equivalent:
-
(i)
is a -morphism from to .
-
(ii)
There exists a binary relation such that and is a function.
Proof.
(i)(ii): Suppose . By Definition 2.3.1, and , , and hence .
Therefore, . Hence (ii) is true.
(ii)(i): Let . Suppose .
Since and is a semigroup, . Because is a function, . Hence by Definition 2.3.1, is a -morphism from to . ∎
Proposition 7.4.2 implies that , where , may be regarded as some sort of basic part of a -morphism from a -subspace of to . This observation inspires us to define a topology on as follows.
Definition 7.4.3.
Let . Let be the topology on generated by as a subbasis; that is, let be the collection of all unions of finite intersections of elements of .
Proposition 7.4.4.
Let be an arbitrary intersection of elements of . Then is also a union of elements of .
Proof.
If , then by convention, may be regarded as the union of an empty family of elements of . So we only need to consider the case where .
Any element of is a set , where . So for any and , since is a semigroup, we can tell . Then and , .
Therefore,
∎
Thus immediately we have
Corollary 7.4.5.
is a basis of . Thus, is the collection of all unions of elements of .
Combining Proposition 7.4.2 with the above corollary, we immediately obtain the main theorem of this subsection as follows.
Theorem 7.4.6.
Let such that is a -subspace of . Then is a -morphism from to if and only if and is a function.
Part II. THEORY FOR THE CASE OF MULTIVARIABLE TOTAL OR PARTIAL FUNCTIONS
To make our theory more general, in Section 8, we shall generalize the notion of operator semigroup to incorporate functions of more than one variable. Then we shall find that some familiar concepts such as ring homomorphisms, module homomorphisms, and group homomorphisms can be characterized by -morphisms or -morphisms. Moreover, in Section 9, we shall allow elements in to be partial functions. Then we will find that some notions such as covariant functors between categories can be characterized in terms of -morphisms or -morphisms.
8. Basic notions for the case of multivariable (total) functions
Subsection 8.1 generalizes the notion of operator semigroup to incorporate multivariable functions. Accordingly, Subsections 8.2, 8.3, 8.4, 8.5 and 8.6 generalize the notions of -spaces, -morphisms, -isomorphisms, -morphisms and -isomorphisms, respectively.
8.1. Operator generalized-semigroup
Definition 8.1.1.
Let be a set, let , and let be a function from the cartesian product to . , let and let be a function from the cartesian product to . Let . Then the composite is the function from to given by
where , and .
Remark.
If , we can tell that the defined above is the usual composite of functions and .
Now Definition 2.1.1 is generalized to
Definition 8.1.2.
Let be a set and let be a subset of
{all functions from the cartesian product to }.
If , of variables, and (cartesian product), the composite , then we call an operator generalized-semigroup, abbreviated as operator gen-semigroup, on and is called the domain of .
Remark.
-
(1)
The terminology “generalized-semigroup” is informal and just for convenience because we have not defined it.
-
(2)
Obviously, any operator semigroup is an operator gen-semigroup.
The following generalizes Example 2.1.2.
Example 8.1.3.
Let be a subfield of a field . Let be the set
given by ,
where is the ring of polynomials over in variables.
For example, let . Then can be induced by (with a slight abuse of notation) . That is, is given by .
Then from Definition 8.1.2, we can tell that is an operator gen-semigroup on .
Note that the map which induces , i.e.
given by , is surjective, but it is not necessarily injective.
The following several examples will also be employed.
Example 8.1.4.
Let be a ring with identity and let be the ring of polynomials over the prime field of order 2 in noncommuting variables. Let
given by .
Then we can tell that is an operator gen-semigroup on .
The map which induces , i.e.
is given by .
Example 8.1.5.
Let be a module over a commutative ring with identity. Let
given by ,
(where ). Then is an operator gen-semigroup on .
Let be the map inducing given by .
Example 8.1.6.
Let be an abelian group. Let
given by },
(where ). Then is an operator gen-semigroup on .
Let be the map inducing given by .
For not necessarily abelian groups, we have
Example 8.1.7.
Let be a group. Recall that a word is a finite string of symbols, where repetition is allowed. Let denote the set of all words on . Let
given by .
Then we can tell that is an operator gen-semigroup on .
Let be the map inducing given by .
8.2. -spaces
Definition 8.2.1.
Let be an operator gen-semigroup on and let . Then we call
, has variables, and
the -space generated by and denote by .
Example 8.2.2.
Let be a subfield of a field . Let be the operator gen-semigroup on defined in Example 8.1.3. Then , is the ring .
Example 8.2.3.
Let be a ring with identity. Let be the operator gen-semigroup defined in Example 8.1.4. Then , is a subring of . Specifically, because .
Example 8.2.4.
Let be a module over a commutative ring with identity. Let be the operator gen-semigroup defined in Example 8.1.5. If is a nonempty subset of , then is the submodule of generated by .
Example 8.2.5.
Let be an abelian group. Let be the operator gen-semigroup defined in Example 8.1.6. If is a nonempty subset of , then is the subgroup of generated by .
Example 8.2.6.
Let be a group. Let be the operator gen-semigroup defined in Example 8.1.7. If is a nonempty subset of , then is the subgroup of generated by .
8.3. -morphisms
Definition 8.3.1.
Let be an operator gen-semigroup and let be a map from a -space to a -space. If , and of variables,
then we call a -morphism.
Remark.
Note that in Examples 8.1.3 to 8.1.7, all operator gen-semigroups are induced from polynomials, rational functions, or words on a set. Therefore, we introduce an (informal) notion as follows, which will facilitate our study on -morphisms and -morphisms (defined later).
Definition 8.3.2.
Let be an expression associated with a rule . Let .
If all of arise as symbols in , and there exist sets and such that, by the rule , yields a single element, which we denote by , in whenever every in is replaced by any , respectively, then we call the couple a (C-valued) formal function of variables and is called a domain of .
Moreover, we may call a formal function if its associated rule is clear from the context.
Remark.
- (1)
-
(2)
By the definition, a formal function may have more than one domain. For example, any field which contains is a domain of any formal function .
-
(3)
Any function of variables may be regarded as the formal function , which is an expression with its associated rule being the mapping defined by the function . The converse is not true, however, because a formal function may have more than one domain, and different domains of a formal function may induce different functions. Therefore, the notion of formal function generalizes the notion of function.
When operator gen-semigroups are induced from formal functions, as is the case in Examples 8.1.3 to 8.1.7, the following lemma will be useful.
Lemma 8.3.3.
Let be a set of D-valued formal functions on a domain and let be an operator gen-semigroup on induced by ; that is, there is a surjective map given by , where is given by for of variables, . Let be a map from a -space to a -space. Then the following statements are equivalent:
-
(i)
is a -morphism.
-
(ii)
, and of variables,
(8.1)
Proof.
(ii)(i): Since is surjective, , such that . Then from statement (ii) and the definition of , we can tell that , and of variables, . Hence by Definition 8.3.1, must be a -morphism. ∎
Lemma 8.3.3 characterizes a -morphism by the formal functions which induce . When an operator gen-semigroup is induced from formal functions, the criterion for a -morphism by Lemma 8.3.3 is normally more convenient than that by Definition 8.3.1. Thus Lemma 8.3.3 simplifies our proofs of some of our results as follows.
The following several results serve as examples of -morphisms.
Firstly, Proposition 2.3.2 is generalized to
Proposition 8.3.4.
Let be a subfield of a field , let be the operator gen-semigroup on defined in Example 8.1.3, and let nonempty . Then a map is a ring homomorphism with fixed pointwisely if and only if it is a -morphism.
Proof.
We may give a proof which is analogous to that of Proposition 2.3.2. However, now we employ Lemma 8.3.3 instead. By Example 8.1.3,
given by .
Thus and are the rings and , respectively.
Let . Then Lemma 8.3.3 applies. Hence we only need to show that is a ring homomorphism with fixed pointwisely if and only if statement (ii) in Lemma 8.3.3 is true in this case. Specifically, it suffices to show that the following statements are equivalent:
-
(1)
, , and , and .
-
(2)
, and ,
(8.2)
(1)(2): Since is a polynomial, it is not hard to see that statement (2) is true.
(2)(1): Considering the case where is a constant polynomial, we can tell from statement (2) that , .
Let . Let and . Then
(by Equation 8.2)
.
And
(by Equation 8.2)
. ∎
However, ring homomorphism in Proposition 8.3.4 is between extensions over the same field. We will deal with ring homomorphisms between extensions over different fields in Subsection 8.5.
Proposition 8.3.5.
Let be a module over a commutative ring with identity. Let be the operator gen-semigroup defined in Example 8.1.5. If and are nonempty subsets of , then a map is an -homomorphism of -modules if and only if it is a -morphism.
Proof.
Let . Then Lemma 8.3.3 applies. Thus we only need to show that is an -homomorphism of -modules if and only if statement (ii) in Lemma 8.3.3 is true in this case. Specifically, it suffices to show that the following statements are equivalent:
-
(1)
and , and .
-
(2)
, and ,
(8.3)
(1)(2): Because is a linear polynomial whose constant term is zero, it is not hard to see that statement (2) is true.
(2)(1): Let and . Let and . Then
(by Equation 8.3)
.
And
(by Equation 8.3)
. ∎
-homomorphism in Proposition 8.3.5 is between submodules of the same -module. We will deal with general -homomorphisms of -modules in Subsection 8.5.
Proposition 8.3.6.
Let be an abelian group and let be the operator gen-semigroup on defined in Example 8.1.6. If and are nonempty subsets of , then a map is a group homomorphism if and only if it is a -morphism.
Proof.
By Example 8.1.6,
given by }.
Obviously, and are the subgroups of generated by and , respectively.
We may apply Lemma 8.3.3 as well in this proof. However, our argument as follows seems more straightforward.
Suppose that is a -morphism. Then by Definition 8.3.1, , and of variables,
In words, commutes with the (multiplicative) operation equipped by . Hence is a group homomorphism.
Apparently the converse of the above argument is also true. That is, if is a group homomorphism, then commutes with the operation equipped by , and hence it is a -morphism. ∎
Group homomorphism in Proposition 8.3.6 is between subgroups of the same abelian group. We will deal with group homomorphisms between different abelian groups in Subsection 8.5.
For not necessarily abelian groups, we have
Proposition 8.3.7.
Let be a group and let be the operator gen-semigroup on defined in Example 8.1.7. If and are nonempty subsets of , then a map is a group homomorphism if and only if it is a -morphism.
Proof.
The proof is comparable with that of Proposition 8.3.6.
By Example 8.1.7,
given by ,
Obviously, and are the subgroups of generated by and , respectively.
Suppose that is a -morphism. Then by Definition 8.3.1, , and of variables,
In words, commutes with the multiplication equipped by . Hence is a group homomorphism.
It is not hard to see that the converse of the above argument is also true. Specifically, if is a group homomorphism, then commutes with the multiplication equipped by , and hence it is a -morphism. ∎
8.4. -isomorphisms
Definition 2.4.1 still applies: if a -morphism is bijective, then we call it a -isomorphism. To justify the definition, we need to show Proposition 2.4.2 with being an operator gen-semigroup as follows.
Proposition 8.4.1.
(Proposition 2.4.2) Let be a -isomorphism from a -space to a -space and let be the inverse map of . Then .
8.5. -morphisms
Notation 8.5.1.
Let (resp. ) be a set, let (resp. ) be a subset of {all functions from the cartesian product to } (resp. a subset of {all functions from to }), let , and let . We denote by the binary relation , where denotes the function obtained by restricting every variable of to .
Definition 6.1.5 is generalized to
Definition 8.5.2.
Let and be defined as in Notation 8.5.1. Let , and . If , , then we say that is well-defined and let
Remark.
Hence Proposition 6.1.6 is generalized to
Proposition 8.5.3.
Let and be defined as in Notation 8.5.1. Then the following statements are equivalent:
-
(i)
is a map.
-
(ii)
, such that , is well-defined.
Proof.
is a map;
, ;
, and have the same number of variables, assumed , and , ;
, such that , is well-defined. ∎
Then Definition 6.1.8 is generalized to
Definition 8.5.4.
Let and be operator gen-semigroups and let . Let be a map from a -space to a -space. If , and of variables,
then is called a -morphism.
Then the first half of Corollary 6.1.9 is generalized to
Corollary 8.5.5.
Let be a -morphism from a -space . Then , and have the same number of variables, and is a map.
Proof.
When operator gen-semigroups are induced from formal functions, the following, which generalizes Lemma 8.3.3, will facilitate study of -morphisms.
Lemma 8.5.6.
Let resp. be a set of D-valued resp. -valued formal functions on resp. . Let resp. be an operator gen-semigroup on resp. such that there is a map being given by , where is given by for of variables, resp. such that there is a map being given analogously. Let be a map. Let and let be a map from a -space to a -space.
Then the following statements are equivalent:
-
(i)
is a -morphism.
-
(ii)
, and of n variables,
(8.4)
Proof.
Obviously, .
(i) (ii): Assume -variable and . Then
(because and by Definition 8.5.4)
(because and by Corollary 8.5.5, is a map)
, as desired.
(ii) (i): To show that is a -morphism, we first need to show that is a map.
Let with . By the definition of , such that and , and hence . Thus and have the same number of variables, which we assume to be . Then by Equation (8.4) and Convention 6.1.7, both and have variables.
Assume ; that is, . Then
such that
However, because , and hence from Equation 8.4, we can tell
a contradiction.
Hence must be a map. Then , and of variables,
(by the definition of )
(by Equation (8.4))
(by the definition of )
(because and is a map).
Since , by Definition 8.5.4, is a -morphism. ∎
Lemma 8.5.6 characterizes a -morphism by formal functions. When operator gen-semigroups are induced from formal functions, the criterion for a -morphism by Lemma 8.5.6 is usually more convenient than that by Definition 8.5.4. Thus Lemma 8.5.6 simplifies our proofs of the following several results.
Proposition 8.5.7.
Let resp. be a field extension, let an operator gen-semigroup resp. be defined as in Example 8.1.3 on resp. over resp. , let be a field isomorphism, and let
be the map given by where
and
Let , where and are the maps inducing and , respectively, defined as in Example 8.1.3.
Then and , a map is a ring homomorphism extending if and only if is a -morphism.
Proof.
We may prove Proposition 8.5.7 in a way analogous to the one for Proposition 6.1.10. However, now we employ Lemma 8.5.6 instead.
Thus and are the rings and , respectively.
Let and . Then Lemma 8.5.6 applies. Hence it suffices to show that is a ring homomorphism extending if and only if statement (ii) in Lemma 8.5.6 is true in this case. Specifically, we only need to show that the following statements are equivalent:
-
(1)
, , and , and .
-
(2)
, and ,
(8.5)
(1) (2): , and ,
(because statement (1) is true)
(because , )
(by the definition of ).
(2) (1): Considering the case where is a constant polynomial, we can tell from Equation 8.5 and the definition of that , .
Let . Let and let . Then
(by Equation 8.5)
( by the definition of )
.
And
(by Equation 8.5)
( by the definition of )
. ∎
The following is for ring homomorphisms between not necessarily commutative rings.
Proposition 8.5.8.
Proof.
Obviously and . Hence is .
Let and let . Then Lemma 8.5.6 applies. By Lemma 8.5.6, we only need to show that is a ring homomorphism if and only if statement (ii) in Lemma 8.5.6 is true in this case. Specifically, it suffices to show that the following statements are equivalent:
-
(1)
and , and .
-
(2)
, and ,
(8.6)
(1) (2): Since is a polynomial, it is not hard to see that statement (2) is true.
(2) (1): Considering the case where is the constant polynomial 1, we can tell from Equation 8.6 that .
Let . Let and . Then
(by Equation 8.6)
.
And
(by Equation 8.6)
. ∎
Proposition 8.3.5 is generalized to
Proposition 8.5.9.
Let and be modules over a commutative ring with identity, let resp. be the operator gen-semigroup on resp. defined as in Example 8.1.5, and let
where and are the maps inducing and , respectively, defined as in Example 8.1.5.
If and are nonempty subsets of and , respectively, then a map is an R-homomorphism of R-modules if and only if is a -morphism.
Proof.
Our proof is comparable with that of Proposition 8.3.5.
Clearly, and are the submodules of and generated by and , respectively.
Let and let . Then Lemma 8.5.6 applies. By Lemma 8.5.6, it suffices to show that is an -homomorphism of -modules if and only if statement (ii) in Lemma 8.5.6 is true in this case. Specifically, we only need to show that the following statements are equivalent:
-
(1)
and , and .
-
(2)
, and ,
(8.7)
(1) (2): Because is a linear polynomial whose constant term is zero, it is not hard to see that statement (2) is true.
(2) (1): Let and let . Let and let . Then
(by Equation 8.7)
.
And
(by Equation 8.7)
. ∎
Proposition 8.3.6 is generalized as follows.
Proposition 8.5.10.
Let and be abelian groups, let resp. be the operator gen-semigroup on resp. defined as in Example 8.1.6, and let , where and are the maps inducing and , respectively, defined as in Example 8.1.6.
If and are nonempty subsets of and respectively, then a map is a group homomorphism if and only if is a -morphism.
Proof.
Besides, and are the subgroups of and generated by and , respectively.
Let and let . Then Lemma 8.5.6 applies. By Lemma 8.5.6, it suffices to show that is a group homomorphism if and only if statement (ii) in Lemma 8.5.6 is true in this case. Specifically, we only need to show that the following statements are equivalent:
-
(1)
, .
-
(2)
, and ,
It is not hard to see (1)(2). ∎
Proposition 8.3.7 is generalized to
Proposition 8.5.11.
Let and be groups, let resp. be the operator gen-semigroup on resp. defined as in Example 8.1.7, and let , where and are the maps inducing and , respectively, defined as in Example 8.1.7.
If and are nonempty subsets of and , respectively, then a map is a group homomorphism if and only if is a -morphism.
Proof.
Our proof is comparable with that of Proposition 8.5.10.
Clearly, and are the subgroups of and generated by and , respectively.
Let and let . Then Lemma 8.5.6 applies. By Lemma 8.5.6, we only need to show that is a group homomorphism if and only if statement (ii) in Lemma 8.5.6 is true in this case. Specifically, it suffices to show that the following statements are equivalent:
-
(1)
, .
-
(2)
, and ,
It is not hard to see (1)(2). ∎
8.6. -isomorphisms
For and being operator gen-semigroups, Definition 6.2.1 still applies: a -morphism from a -space to a -space is a -isomorphism if it is bijective. To justify the definition, we need to show Lemma 6.2.2 and Proposition 6.2.3 for the case of operator gen-semigroups as follows.
Lemma 8.6.1.
(Lemma 6.2.2) Let be a -isomorphism from a -space to a -space . Then is a map, where .
Proof.
Let with of variables. Then by Corollary 8.5.5, both and have variables.
Let .
Then to show that is a map, by Proposition 8.5.3 and Definition 8.5.2, it suffices to show that
. Moreover, since is injective, we only need to show :
(by Definition 8.5.4)
(because and is a map by Corollary 8.5.5)
(because and is a map)
(by Definition 8.5.4).
∎
Proposition 8.6.2.
(Proposition 6.2.3) Let and be a -space and a -space, respectively, let and let be the inverse map of . Then .
Proof.
By Lemma 8.6.1, is a map. Then by Proposition 8.5.3, , such that , is well-defined, and from (8.8), we can tell that the number of variables of is also .
Thus , of variables and , it follows from (8.8) that
Hence by Definition 8.5.4, is a -morphism from to .
Moreover, since is bijective, by Definition 6.2.1, . ∎
9. Basic notions for the case of (multivariable) partial functions
In this section, we shall allow elements in to be partial functions. Then we will find that some more familiar concepts such as ring homomorphisms between fields and covariant functors between categories can be characterized in terms of -morphisms or -morphisms.
Subsection 9.1 generalizes the notion of operator gen-semigroup. Then correspondingly, the notions of -spaces, -morphisms, -isomorphisms, -morphisms and -isomorphisms are generalized in Sections 9.2, 9.3, 9.4, 9.5 and 9.6, respectively.
9.1. Partial-operator generalized-semigroup
Recall the concept of partial functions as follows.
Definition 9.1.1.
Let and be sets. A partial function of variables from the cartesian product to is a (total) function from a subset of to , and , which may be empty, is called the domain of definition of .
Moreover, let , where . If lies in the domain of definition of , then is said to be well-defined.
Remark.
By the definition, any function is a partial function.
We are going to extend our study to partial functions. First, we generalize Definition 8.1.1 to
Definition 9.1.2.
Let be a set, let , and let be a partial function from the cartesian product to . , let and let be a partial function from the cartesian product to . Let . Then the composite is the partial function from to given by
where , and .
Remark.
-
(1)
Note that is in the domain of definition of if and only if each is well-defined and is well-defined.
-
(2)
We may denote the -tuple by (in e.g. Section 14) for brevity.
Before we generalize Definition 8.1.2, we need a terminology as follows.
Terminology 9.1.3.
Let and be partial functions from to . If the domain of definition of contains the domain of definition of , which is denoted by , and , then we say that is a restriction of .
Remark.
Trivially, if , then is a restriction of any .
Definition 9.1.4.
Let be a set and let be a subset of
{all partial functions from the cartesian product to }.
If , of variables, and (cartesian product), the composite is a restriction of some element of , then we call a partial-operator generalized-semigroup, abbreviated as par-operator gen-semigroup, on and is called the domain of .
Remark.
-
(1)
Trivially, is a par-operator gen-semigroup.
-
(2)
Obviously, any operator gen-semigroup is a par-operator gen-semigroup.
Definition 2.1.6 is generalized to
Definition 9.1.5.
Let be a set and be a subset of
{all partial functions from the cartesian product to }.
Then we call the intersection of all par-operator gen-semigroups on containing the par-operator gen-semigroup on generated by , and denote it by .
Remark.
From Definition 9.1.4, we can tell that is the smallest par-operator gen-semigroup on which contains .
Example 9.1.6.
Let be a subfield of a field . Let
the partial function given by .
Then it is not hard to see that is a par-operator gen-semigroup on . The map which induces , i.e. given by , is surjective, but it is not necessarily injective.
Example 9.1.7.
Let be a differential subfield of a differential field ; that is, field is endowed with a derivation and the derivation of restricts to the derivation of . Let (Definition 9.1.5), where is the par-operator gen-semigroup on defined as in Example 9.1.6 and is the operator semigroup defined in Example 2.1.5. Then by Definition 9.1.5, is a par-operator gen-semigroup on .
Example 9.1.8.
For this example, in Definitions 9.1.2 and 9.1.4 is generalized to be a class. Let be a category and let be the collection of all morphisms in . , let
word on each of arises in }.
Let the partial function given by
,
where is defined to be the composite for (where ).
In particular, for , is the identity function on .
Thus, for , and , the following four statements are equivalent:
-
(i)
is in the domain of definition of the induced by .
-
(ii)
for , is well-defined.
-
(iii)
is well-defined.
-
(iv)
, the domain (object) of is the target (object) of .
Then it is not hard to see that is a par-operator gen-semigroup on . Let be the map inducing given by .
9.2. -spaces
Definition 9.2.1.
Let be a par-operator gen-semigroup on and let .
Then we call
well-defined , ,
the -space generated by and denote by .
Moreover, if is also a -space, then we say that is a -subspace of and write or .
Example 9.2.2.
Let be a subfield of a field . Let be the par-operator gen-semigroup on defined in Example 9.1.6. Then , we can tell that the -space is the field .
Example 9.2.3.
Let be a differential subfield of a differential field . Let be the par-operator gen-semigroup on defined in Example 9.1.7. Then we claim that , the -space is the differential subfield of generated by over .
Example 9.2.4.
Let be a category. Let and be defined as in Example 9.1.8. Then the -space since . Because there is a bijection between the objects in and their identity morphisms , we may view as .
9.3. -morphisms
Now we generalize Definition 8.3.1 to
Definition 9.3.1.
Let be a par-operator gen-semigroup. Let be a map from a -space to a -space. If , and , neither nor is well-defined or
then we call a -morphism.
Remark.
Note that in Examples 9.1.6 and 9.1.8, par-operator gen-semigroups are induced from rational functions or words on a set. Hence we generalize Definition 8.3.2 as follows.
Definition 9.3.2.
Let be an expression associated with a rule . Let .
If all of arise as symbols in , and there are sets and such that, by the rule , yields nothing or a single element in , which we denote by , whenever every in is replaced by any , respectively, then we call the couple a (-valued) formal partial function of variables and is called a domain of .
Moreover, in the case where and yields , we say that is well-defined or is in the domain of definition of .
Besides, we may call a formal partial function if its associated rule is clear from the context.
Remark.
Analogous to a formal function, a formal partial function may have more than one domain. And the notion of formal partial function generalizes the notion of partial function.
When par-operator gen-semigroups are induced from formal partial functions, the following lemma, which generalizes Lemma 8.3.3, will be useful.
Lemma 9.3.3.
Let be a set of -valued formal partial functions on a domain and let be a par-operator gen-semigroup on induced by ; that is, there is a surjective map being given by , where is given by for of variables, . Let be a map from a -space to a -space. Then the following statements are equivalent:
-
(i)
is a -morphism.
-
(ii)
, and of variables, neither nor is well-defined or
(9.1)
Proof.
(i) (ii): By Definition 9.3.1, , and of variables, neither nor is well-defined or .
Then by the definition of , (ii) must be true.
(ii) (i): Since is surjective, , such that . Then from statement (ii) and the definition of , we can tell that , and of variables, neither nor is well-defined or
.
So by Definition 9.3.1, must be a -morphism. ∎
Lemma 9.3.3 characterizes a -morphism by formal partial functions which induce . When a par-operator gen-semigroup is induced from formal partial functions, the criterion for a -morphism by Lemma 9.3.3 is normally more convenient than that by Definition 9.3.1.
The following is comparable with Proposition 8.3.4, except that now is defined by Example 9.1.6, and hence and must be fields.
Proposition 9.3.4.
Let be a subfield of a field , let be the par-operator gen-semigroup on defined in Example 9.1.6, and let nonempty . Then a map is a ring homomorphism with fixed pointwisely if and only if it is a -morphism.
Proof.
By the definition of in Example 9.1.6,
the partial function given by .
Then and are the fields and , respectively. Let . Then Lemma 9.3.3 applies. Hence it suffices to show that is a ring homomorphism with fixed pointwisely if and only if statement (ii) in Lemma 9.3.3 is true in this case. Specifically, we only need to show that the following statements are equivalent:
-
(1)
, , and , and .
-
(2)
, and , neither nor is well-defined or
(9.2)
(1) (2): Let , where and . Let . Then
is well-defined;
;
(by statement (1));
(because and statement (1) is true)
is well-defined.
Moreover, in the case where any of the above equivalent conditions is true,
.
Combining the above four equations, we have
(2) (1): Considering the case where is a constant polynomial, we can tell from statement (2) that , .
Let . Let and .
Then
(by Equation 9.2)
.
And
(by Equation 9.2)
. ∎
However, in Proposition 9.3.4 is between field extensions over the same field . We will deal with ring homomorphisms between field extensions over different fields in Subsection 9.5.
To characterize differential ring homomorphisms between differential fields by -morphisms, we need a lemma as follows.
Lemma 9.3.5.
Let be a par-operator gen-semigroup, let be a map from a -space to a -space, let have variables, and , let have variables. Suppose that both of the following two conditions are satisfied,
-
(i)
, neither nor is well-defined or ;
-
(ii)
and ,
neither nor is well-defined or
Then , where ,
neither nor is well-defined or
Proof.
Let , where . Then
is well-defined;
is well-defined, where , , , and ; (by Definition 9.1.2)
is well-defined, where , ; (by condition (ii))
is well-defined. (by Definition 9.1.2)
And in the case where any of the above five equivalent conditions is true,
(by Proposition 10.1.1)
and
(by Definition 9.1.2)
(by (i) and the fact that each (by Proposition 10.1.1))
(by condition (ii))
(by Definition 9.1.2) ∎
Proposition 9.3.6.
Let be a differential subfield of a differential field , let be the par-operator gen-semigroup on defined in Example 9.1.7, and let nonempty . Then a map is a differential ring homomorphism with fixed pointwisely if and only if it is a -morphism.
Proof.
Let and . By Example 9.2.3, and are the differential subfields of generated by and over , respectively.
Recall that a ring homomorphism is a differential homomorphism if and only if commutes with . Then by Definition 9.3.1, we only need to show that the following statements are equivalent:
-
(1)
is a ring homomorphism with fixed pointwisely and , .
-
(2)
, and of variables, neither nor is well-defined or
(1) (2): Let be the par-operator gen-semigroup on defined as in Example 9.1.6 and let be the operator semigroup defined in Example 2.1.5.
Since is a ring homomorphism with fixed pointwisely, by Proposition 9.3.4, is a -morphism. By Definition 9.3.1, , , and of variables, neither nor is well-defined or . Moreover, since , we can tell that and , .
Therefore, , , and of variables, neither nor is well-defined or .
Let have variables, let have variables, , and let , where . By the statement in the preceding paragraph, Lemma 9.3.5 applies (with ). It follows that neither nor is well-defined or
Moreover, since , from Definitions 9.1.4 and 9.1.5, we can tell that the above is a restriction of some element of , and , can be obtained by a finite number of compositions (defined by Definition 9.1.2) of elements of (or elements which “originally come from” ). Thus, by induction and application of Lemma 9.3.5 as in the preceding paragraph, we can show that , of variables and , neither nor is well-defined or
9.4. -isomorphisms
To par-operator gen-semigroups, Definition 2.4.1 still applies: if a -morphism is bijective, then we call it a -isomorphism. To justify the definition, we need to show Proposition 2.4.2 with a different proof as follows.
Proposition 9.4.1.
(Proposition 2.4.2) Let be a -isomorphism from a -space to a -space and let be the inverse map of . Then .
9.5. -morphisms
We first generalize Notation 8.5.1 to
Notation 9.5.1.
Let (resp. ) be a set, let (resp. ) be a subset of {all partial functions from the cartesian product to } (resp. a subset of {all partial functions from to }), let , and let . We denote by the set , where denotes the partial function obtained by restricting every variable of to .
Then for the case of partial functions, Proposition 8.5.3 needs to be modified as follows.
Proposition 9.5.2.
Let and be defined as in Notation 9.5.1. Then the following statements are equivalent:
-
(i)
is a map.
-
(ii)
, .
-
(iii)
, and , or neither nor is well-defined.
Proof.
Obvious. ∎
Definition 9.5.3.
Let and be defined as in Notation 9.5.1. Let , and . If , , then we say that is well-defined and let .
Remark.
Then by Proposition 9.5.2, we immediately get
Corollary 9.5.4.
Let and be defined as in Notation 9.5.1. Suppose that is a map. Let . If and such that is well-defined, then is well-defined and
Definition 8.5.4 is generalized to
Definition 9.5.5.
Let and be par-operator gen-semigroups and let . Let be a map from a -space to a -space. If
-
(i)
is a map and
-
(ii)
, and , neither nor is well-defined or
then is called a -morphism.
Remark.
The following explains why, unlike Definition 8.5.4, we put condition (i) in Definition 9.5.5 explicitly.
Proposition 9.5.6.
Proof.
By Proposition 9.5.2, is a map if and only if , and , or neither nor is well-defined. We can tell that this condition for being a map is not guaranteed by condition (ii) in Definition 9.5.5. For example, condition (ii) in Definition 9.5.5 cannot rule out the possibility where there exist , and such that only one of and is well-defined (because in this case, it is possible that neither nor is well-defined). ∎
When par-operator gen-semigroups are induced from formal partial functions, the following lemma, which generalizes both Lemmas 8.5.6 and 9.3.3, will be useful.
Lemma 9.5.7.
Let resp. be a set of D-valued resp. -valued formal partial functions on resp. . Let resp. be a par-operator gen-semigroup on resp. such that there is a map being given by , where is given by for of variables, resp. such that there is a map being given analogously. Let be a map. Let and let be a map from a -space to a -space. Then the following statements are equivalent:
-
(i)
is a -morphism.
-
(ii)
, and of variables, neither nor is well-defined or
(9.3)
Proof.
Clearly, .
(i) (ii): By Definition 9.5.5, is a map and , and of variables, neither nor is well-defined or
Let , let and let have variables. Then
is well-defined;
is well-defined;
is well-defined;
is well-defined (because and is a map);
is well-defined.
And in the case where the above equivalent conditions are true,
(because and is a map)
, as desired.
(ii) (i): By Definition 9.5.5, to show that is a -morphism, we first need to show that is a map.
Let with . By the definition of , such that and , and thus . Hence and have the same number of variables, assumed . Then by Equation (9.3) and Convention 6.1.7, also has variables if is well-defined somewhere in . Analogously, also has variables if is well-defined somewhere in .
Assume that ; that is, . Then and such that only one of and is well-defined or both of them are well-defined but
By statement (ii), if (resp. ) is well-defined, then (resp. ) is also well-defined. Hence .
However, , and hence neither nor is well-defined or both of them are well-defined and . It follows from (ii) that neither nor is well-defined or both of them are well-defined and
a contradiction.
Hence must be a map.
Moreover, by statement (ii), , and of variables, there are only two cases as follows.
Case (1): Neither nor is well-defined, and hence neither nor is well-defined. Because , by Definition 9.5.3, neither nor is well-defined.
Case (2):
(by Equation (9.3))
(because and is a map).
Therefore, , and of variables, neither nor is well-defined or
Since , by Definition 9.5.5, is a -morphism. ∎
Lemma 9.5.7 characterizes a -morphism by formal partial functions. When par-operator gen-semigroups are induced from formal partial functions, the criterion for a -morphism by Lemma 9.5.7 is more convenient than that by Definition 9.5.5.
The following, which generalizes Proposition 9.3.4, is analogous to Proposition 8.5.7, except that now is defined by Example 9.1.6, and hence and must be fields.
Proposition 9.5.8.
Let (resp. ) be a field extension, let a par-operator gen-semigroup resp. be defined as in Example 9.1.6 on resp. over resp. , let be a field isomorphism, and let
be the map given by where
and
.
Let , where and are the maps inducing and , respectively, defined as in Example 9.1.6.
Then and , a map is a ring homomorphism extending if and only if is a -morphism.
Proof.
By the definition of in Example 9.1.6,
the partial function given by .
and
the partial function given by .
Thus and are the fields and , respectively.
Let and . Then Lemma 9.5.7 applies. Hence it suffices to show that is a ring homomorphism extending if and only if statement (ii) in Lemma 9.5.7 is true in this case. Specifically, it is sufficient to show that the following statements are equivalent:
-
(1)
, , and , and .
-
(2)
, and , neither nor is well-defined or
(9.4)
(1) (2): Let , where and . Let . Then
is well-defined;
;
(by the definition of );
(by the definition of )
is well-defined.
Moreover, in the case where the above equivalent conditions are true,
(because statement (1) is true)
(because , )
(by the definition of ),
as desired.
(2) (1): Considering the case where is a constant polynomial, we can tell from Equation 9.4 and the definition of that , .
Let . Let and .
Then
(by Equation 9.4)
( by the definition of )
.
And
(by Equation 9.4)
( by the definition of )
. ∎
We are going to show that a covariant functor can be characterized by a -morphism. Because in every category, there is a bijection between objects and their identity morphisms , we view as and as in the following.
Proposition 9.5.9.
Let (resp. ) be a category, let resp. be the collection of all morphisms in resp. , and let , , and the par-operator gen-semigroup resp. on resp. be defined as in Example 9.1.8. Let , where and are the maps inducing and , respectively, defined as in Example 9.1.8.
Suppose that a function maps each identity morphism in to an identity morphism in . Then is a covariant functor if and only if it is a -morphism.
Proof.
By Example 9.2.4, and are -space and -space, respectively. Let and let be the identity function. Then Lemma 9.5.7 applies. Hence it suffices to show that is a covariant functor if and only if statement (ii) in Lemma 9.5.7 is true in this case. Specifically, we only need to show that the following statements are equivalent:
-
(1)
, ; if in , then in ; and if in , then in and .
-
(2)
, and of variables, neither nor is well-defined or
(9.5)
(1) (2): Let and let (cf. Example 9.1.8). Then
is well-defined;
is well-defined (by the definition of in Example 9.1.8);
is well-defined (explained below);
is well-defined (by the definition of in Example 9.1.8).
The second equivalence is explained as follows. (resp. ) is the collection of all morphisms in category (resp. ), and hence , (resp. ) is a morphism in (resp. ). Then is well-defined if and only if , the domain of is the target of . And this occurs if and only if is well-defined because is a covariant functor.
Therefore, is well-defined if and only if is well-defined.
Moreover, in the case where the above equivalent conditions are satisfied, since is a covariant functor,
that is (by the definition of in Example 9.1.8),
(2) (1): Firstly, we view (resp. ) as (resp. ). That is, , may be regarded as , and vice versa, and so do and . It is supposed that maps each identity morphism in to an identity morphism in , and hence with a slight abuse of notation, , , more precisely, .
Secondly, let be a morphism in . Let be . Then
(by the definition of in Example 9.1.8)
(by Equation 9.5)
(by the definition of in Example 9.1.8).
Hence in since and .
9.6. -isomorphisms
To and as par-operator gen-semigroups, Definition 6.2.1 still applies: a -morphism from a -space to a -space is a -isomorphism if it is bijective. To justify the definition, we need to show Lemma 6.2.2 and Proposition 6.2.3 for the case of par-operator gen-semigroups as follows.
Lemma 9.6.1.
(Lemma 6.2.2) Let be a -isomorphism from a -space to a -space . Then is a map, where .
Proof.
Let , and . Then to show that is a map, by Proposition 9.5.2, it suffices to show that neither nor is well-defined or .
Proposition 9.6.2.
(Proposition 6.2.3) Let and be a -space and a -space, respectively, let and let be the inverse map of . Then .
Proof.
By Definition 9.5.5, is a map and , and , neither nor is well-defined or
Hence we have
Fact (A): and ,
neither nor is well-defined or
(9.6) |
Let , and . Then there are two cases as follows.
Case (1): is well-defined.
We are showing that is also well-defined and
Since is a map, by Corollary 9.5.4, , is well-defined and
Then by Fact (A) and Equation (9.6), ,
Thus by (the version of) Definition 9.5.3, is well-defined and
Case (2): is not well-defined.
We are showing that is not well-defined, either.
Let . Assume that is well-defined. Then by Definition 9.5.3, is also well-defined, a contradiction. Hence is not well-defined. Thus by Fact (A), is not well-defined, either, and hence is not well-defined by (the version of) Definition 9.5.3.
Combining Cases (1) and (2), we can tell that , and ,
or neither nor is well-defined.
Furthermore, since is bijective, by Definition 6.2.1, . ∎
10. Basic properties and more notions
In Sections 10 to 14 as follows, our study will focus on par-operator gen-semigroups. The following statement is obvious by Definitions 2.1.1, 8.1.2, and 9.1.4: Each operator semigroup is an operator gen-semigroup, and each operator gen-semigroup is a par-operator gen-semigroup.
Thus our research on par-operator gen-semigroups also applies to operator gen-semigroups and operator semigroups. From now on, unless otherwise specified, always denotes a set and always denotes a par-operator gen-semigroup on defined by Definition 9.1.4. In Sections 10 to 14, unless otherwise specified, -morphisms and -morphisms are always defined by Definitions 9.3.1 and 9.5.5, respectively.
For par-operator gen-semigroups, Sections 10, 11, 12, 13 and 14 generalize results obtained in Sections 2, 3, 4, 5 and 7, respectively. Throughout Sections 10 to 14, by statements such as “Proposition (or Definition, etc.) x.x.x (which is the index) still holds (or still applies)” we mean that the proposition (or definition, etc.) still holds for (or still applies to) the case of par-operator gen-semigroups given specifically in the preceding paragraph.
Roughly speaking, a result for an operator semigroup still holds for a par-operator gen-semigroup if the correctness of the proof does not depend on the number of the variables or the domain of definition of any element of . However, if the correctness of the proof of a result does depend on the number of the variables or the domain of definition of an element of , the result may still hold but need a different proof.
10.1. Basic properties of -spaces and quasi--spaces
In this subsection, we shall generalize properties of -spaces or quasi--spaces obtained in Subsection 2.2.
Proposition 2.2.7 is generalized to
Proposition 10.1.1.
Let be a -space. Then , and of variables, if is well-defined. Hence , . In particular, .
Proof.
Let , let and let have variables. By Definition 9.2.1, such that . And , of variables and (cartesian product) such that By Definition 9.1.4, the partial function defined by Definition 9.1.2 is a restriction of some element of . So if is well-defined.
Therefore, , , and of variables, if is well-defined. Hence , and thus (by Definition 9.2.1). ∎
Definition 2.2.8 still applies. Then by Proposition 10.1.1, Proposition 2.2.9 still holds. Propositions 2.2.10 and 2.2.12 still hold because their proofs still apply. However, we could not generalize Propositions 2.2.13 and 2.2.14 because now the elements of may have more than one variable. Proposition 2.2.16 and its proof still apply.
10.2. Basic properties of -morphisms and -isomorphisms
In this subsection, we shall generalize properties of -morphisms and -isomorphisms obtained in Subsections 2.3 and 2.4.
Proposition 2.3.6 still holds, but now it requires a different proof as follows.
Proposition 10.2.1.
(Proposition 2.3.6) Let be a -morphism from to . Then . Moreover, if or more generally, , then .
Remark.
We keep the original index of the proposition (i.e. 2.3.6) because we shall use it later (explicitly or implicitly).
Proof.
By Definition 9.3.1, , and of variables, neither nor is well-defined or
Hence , and thus by Definition 2.2.8, is a quasi--subspace of .
Moreover, if or , then by Proposition 2.2.10, is a -space, and hence . ∎
By the way, we generalize Proposition 10.2.1 to -morphisms as follows, which generalizes Proposition 6.1.11 to the case of par-operator gen-semigroups.
Proposition 10.2.2.
Let be a -morphism from a -space to a -space . Suppose . Then . Moreover, if or more generally, , then .
Proof.
By Definition 9.5.5, , and , neither nor is well-defined or
Since , we can tell , and so by Definition 2.2.8, is a quasi--subspace of .
Moreover, if or , then by Proposition 2.2.10, is a -space and hence . ∎
Proposition 10.2.3.
Proof.
The identity map on lies in , and hence we only need to show that is a semigroup with composition of functions as the binary operation.
Let , , and . Then by Definition 9.3.1,
is well-defined;
is well-defined;
is well-defined.
Moreover, in the case where the above equivalent conditions are satisfied,
.
Then again by Definition 9.3.1, .
Composition of functions is associative. Therefore, is a monoid with composition of functions as the binary operation. ∎
Proposition 2.3.9 is still obvious.
10.3. Galois -extensions
Definition 2.5.1 and the remark right after it still apply. Proposition 2.5.2 still holds, but now it requires a different proof as follows.
Proposition 10.3.1.
(Proposition 2.5.2) Let be a -space and let be a subset of . Then . Moreover, if or , then .
Proof.
Let , and of variables. By Definition 9.3.1, , neither nor is well-defined, or
Therefore, if is well-defined. Thus . Hence is a quasi--space, and so .
Moreover, if or , then by Proposition 2.2.10, is a -space, and hence . ∎
Proposition 2.5.3 is still obvious.
10.4. Galois -monoids and Galois -groups
10.5. Generated submonoids of and subgroups of
11. Galois correspondences
In this section, we shall find that all results obtained in Subsection 3.2 still apply to the case where is a par-operator gen-semigroup.
Definition 3.2.1 and the remark right after it still apply. Lemma 3.2.2 still holds because its proof still applies.
Definition 3.2.3 and the remark right after it still apply. Lemma 3.2.4 still holds because its proof still applies.
Then by Propositions 8.3.5, 8.3.6, 8.3.7, 8.5.8, 9.3.4, 9.3.6 and 9.5.9, we may apply Corollaries 3.2.5 and 3.2.6 to modules, abelian groups, non-abelian groups, rings, field, differential field, and categories, respectively. We can tell that Galois correspondences exist not only for Galois -groups (by Corollary 3.2.6), but also for Galois -monoids (by Corollary 3.2.5).
12. Lattice structures of objects arising in Galois correspondences on a -space
In this section, results obtained in Section 4 are generalized for par-operator gen-semigroups.
12.1. , and
As said in Subsection 10.1, we could not generalize Proposition 2.2.13 because now the elements of may have more than one variable. So to generalize Proposition 4.1.1, we need a notion as follows.
Definition 12.1.1.
Let be a -space. If , then the intersection of all the quasi--subspaces of containing , denoted by , is called the quasi--subspace of generated by .
Remark.
By Proposition 2.2.12, is the smallest quasi--subspace of which contains .
Then Proposition 4.1.1 is generalized to
Proposition 12.1.2.
Let be a -space. Then is a lattice with inclusion as the binary relation if , and . Moreover, is a complete lattice if let and let , .
12.2. and
Lemma 4.2.1 still holds because its proof still applies. Proposition 4.2.2 would still hold if Proposition 4.1.1 in it were replaced by Proposition 12.1.2 as follows.
Proposition 12.2.1.
is a complete -sublattice of the complete lattice defined in Proposition 12.1.2.
Proof.
Analogously, Lemma 4.2.4 still holds, and Proposition 4.2.5 would still hold if Proposition 4.1.1 in it were replaced by Proposition 12.1.2 as follows.
Proposition 12.2.2.
is a complete -sublattice of the complete lattice defined in Proposition 12.1.2.
12.3. and
13. Topologies employed to construct Galois correspondences
In this section, results obtained in Section 5 are generalized for par-operator gen-semigroups.
13.1. Topologies on a -space for
Lemma 5.1.1 and its proof still apply.
However, we could not generalize a part of Theorem 5.1.2 straightforwardly because we could not generalize Proposition 2.2.13. Instead, we have the following, which in a sense is analogous to Theorem 5.3.2.
Theorem 13.1.1.
Let be a -space and let . Let
and let
where is defined in Lemma 5.1.1. Then and the following statements are equivalent:
-
(i)
.
-
(ii)
For any intersection of finite unions of elements of , implies .
-
(iii)
; that is, is the coarsest topology on such that Equation 5.1 is satisfied.
Moreover, if is a complete -sublattice of the complete lattice defined in Proposition 12.1.2, then the above three statements are true.
Proof.
By Lemma 5.1.1, is the smallest topology on such that the collection of all closed sets contains . Suppose . Then all elements of are closed in , and thus . Hence . Therefore .
(i) (ii): Let . Let be any intersection of finite unions of elements of . By the definition of , each element of is closed in , and so is . Suppose . Then . Again by the definition of , . Hence (ii) is true.
(ii) (iii): By the definition of in Lemma 5.1.1, all elements of are closed in . So by Definition 3.2.1 and Proposition 10.3.1,
{ is nonempty and closed in }.
On the other hand, we prove
{ is nonempty and closed in }
as follows.
Let is nonempty and closed in . We need to show .
Because is generated by subbasis , is the collection of all unions of finite intersections of elements of . Thus is a union of finite intersections of elements of . It follows from DeMorgan’s Laws that is an intersection of finite unions of elements of . Because and , by (ii), , as desired. Hence
{ is nonempty and closed in }
Therefore, Equation 5.1 is satisfied for .
Thus . We showed at the beginning of our proof, and thus by the definition of , is the coarsest topology in . Hence by the definition of , is the coarsest topology on such that Equation 5.1 is satisfied.
(iii) (i): Obvious.
Let be any intersection of finite unions of elements of . By Proposition 12.2.1, any intersection of elements of lies in . Then as a result of the first distributive law of set operations, is a union of elements of . Suppose that is a complete -sublattice of the complete lattice defined in Proposition 12.1.2. Then implies , and hence (ii) is true (in this case). ∎
13.2. Topologies on a -space for
Lemma 5.2.1 and its proof still apply.
We could not generalize a part of Theorem 5.2.2 straightforwardly because we could not generalize Proposition 2.2.13. Instead, we have the following, which is analogous to Theorem 13.1.1.
Theorem 13.2.1.
Let be a -space and let . Let
and let
where is defined in Lemma 5.2.1. Then and the following statements are equivalent:
-
(i)
.
-
(ii)
For any intersection of finite unions of elements of , implies .
-
(iii)
; that is, is the coarsest topology on such that Equation 5.3 is satisfied.
Moreover, if is a complete -sublattice of the complete lattice defined in Proposition 12.1.2, then the above three statements are true.
Proof.
13.3. Topologies on
13.4. Topologies on
14. Constructions of the generalized morphisms and isomorphisms
In this section, results obtained in Section 7 are generalized for par-operator gen-semigroups.
14.1. A construction of -morphisms from
Although the contents of this subsection will be further generalized in the next subsection, we still keep this subsection because we will need it in Sections 19 and 20.
Notation 7.1.1 is generalized for partial functions of multiple variables as follows.
Notation 14.1.1.
Let all partial functions from to }, where is a set. Let . By we mean that
, of variables and of variables,
implies ,
where denotes the -tuple .
Besides, let denote the set .
Moreover, by we mean that and .
Besides, let denote the set .
Remark.
-
(1)
Suppose . Then and of variables, implies , and hence by Convention 6.1.7, that is well-defined implies that is well-defined.
-
(2)
Apparently, defines an equivalence relation on .
Then Proposition 7.1.3 still holds but its proof changes as follows.
Proposition 14.1.2.
(Proposition 7.1.3) Let be a -morphism from a -space . Then , .
Proof.
Let have variables, let have variables and let .
If , where denotes the -tuple , then by Definition 9.3.1, both and are well-defined and
Therefore, . ∎
Analogously, it is not hard to tell that Proposition 7.1.4 still holds.
Proposition 7.1.6 can be generalized for par-operator gen-semigroups, but we omit it for brevity.
Proposition 7.1.7 is generalized to
Proposition 14.1.3.
Let .
-
(a)
The following statements are equivalent:
-
(i)
.
-
(ii)
There exists a map given by , and of variables such that is well-defined.
-
(i)
-
(b)
The map given in (ii) is a -morphism.
Proof.
For (a):
;
, of variables and of variables, implies ;
given by , and of variables such that is well-defined, is a well-defined map (cf. Remark (1) right after Notation 14.1.1).
For (b):
Let . Then , of variables such that .
Let have variables. By Definition 9.1.2, is a partial function from to where . By Definition 9.1.4, is a restriction of some element of . Hence
is well-defined;
(by Definition 9.1.2)
(by the definition of in (ii))
(by Definition 9.1.2)
(by the definition of in (ii))
is well-defined;
as desired for to be a -morphism (by Definition 9.3.1). ∎
Corollary 7.1.8 is generalized to the following, which is a straightforward result of Proposition 14.1.3.
Corollary 14.1.4.
Let .
-
(a)
The following statements are equivalent:
-
(i)
.
-
(ii)
There exists a bijective map given by , and of variables such that is well-defined.
-
(i)
-
(b)
The bijective map given in (ii) is a -isomorphism.
14.2. A construction of -morphisms from
For partial functions of multiple variables, we generalize Notation 7.2.1 as follows.
Notation 14.2.1.
Let be a set, let be a map and let be a subset of all partial functions from to }.
By we mean that
, , , and ,
.
Moreover, by we mean that is bijective, and ; or equivalently, by we mean that is bijective and , , , and ,
.
Remark.
Suppose . Then of variables and ,
.
Hence by Convention 6.1.7, is well-defined if is well-defined.
Then Proposition 7.2.2 still holds but its proof changes as follows.
Proposition 14.2.2.
(Proposition 7.2.2) Let be a -morphism from a -space to a -space . Then , , where . In particular, .
Proof.
Let , let , let , and let . If , by Definition 9.3.1, both and are well-defined and
(since )
(by Definition 9.3.1)
(because )
(by Definition 9.3.1)
(because ).
Hence by Notation 14.2.1, . ∎
Proposition 14.2.3.
Let be a map.
-
(a)
The following statements are equivalent:
-
(i)
.
-
(ii)
There exists a map given by , , and such that is well-defined.
-
(i)
-
(b)
The map given in (ii) is a -morphism.
Proof.
For (a):
;
, of variables, of variables, , and ,
implies
given by
, and such that is well-defined, is a well-defined map.
For (b):
Let . Then , of variables and such that . For , we denote by for brevity.
Let have variables. By Definition 9.1.2, is a partial function from to given by , where and . By Definition 9.1.4, is a restriction of some element of . Hence
is well-defined;
(by Definition 9.1.2)
(by the definition of in (ii))
(by Definition 9.1.2)
(by the definition of in (ii))
is well-defined;
as desired for to be a -morphism (by Definition 9.3.1).
∎
Then Proposition 7.2.5 is generalized to the following.
Proposition 14.2.4.
Let be a map.
-
(a)
The following statements are equivalent:
-
(i)
.
-
(ii)
is bijective and there exists a bijective map given by , , and such that is well-defined.
-
(i)
-
(b)
The map given in (ii) is a -isomorphism.
Proof.
For (a):
;
is bijective and , , and
;
is bijective and given by
, and such that is well-defined, is a well-defined bijective map.
For (b):
Definition 7.2.7 is generalized to
Definition 14.2.5.
Let be a par-operator gen-semigroup on , let , and let be a -morphism from . If there exists a map such that , , and such that is well-defined, then we say that is constructible by , or just say that is constructible for brevity.
Corollary 7.2.8 still holds, but its proof changes a little as follows.
Corollary 14.2.6.
(Corollary 7.2.8) Let . If , then any -morphism from is constructible by .
Proof.
However, we could not generalize Theorem 7.2.9. Indeed, in Theorem 7.2.9, if were replaced by a par-operator gen-semigroup (generated by Definition 9.1.5, where is a partial function from some to ) and if we defined in a way as in the proof of Theorem 7.2.9, then because any may have more than one variable, we could not show that is a well-defined map.
14.3. A construction of -morphisms
In this subsection, unless otherwise specified, -morphisms are defined by Definition 9.5.5, and and are par-operator gen-semigroups.
Notation 14.3.1.
Let (resp. ) be a set, let (resp. ) be a subset of all partial functions from to } (resp. a subset of all partial functions from to }), let , and let be a map.
By we mean that , , , and , implies
Moreover, by we mean that is bijective, and , where .
Remark.
If and is the identity map on , then is equivalent to (defined by Notation 14.2.1).
Proposition 7.3.2 would still hold if and in it were generalized to be par-operator gen-semigroups, but the proof would change as follows.
Proposition 14.3.2.
Let be a -morphism from a -space to a -space . Then , , where . In particular, .
Proof.
By Definition 9.5.5, , and , if , then
(by Definition 9.5.5)
(since )
(by Definition 9.5.5),
and hence . By Notation 14.3.1, . ∎
Definition 7.3.3 is generalized as follows.
Definition 14.3.3.
Let and be par-operator gen-semigroups on and , respectively, let , and let . If and such that is well-defined,
then we say that is distributive over .
Proposition 14.3.4.
Let and be par-operator gen-semigroups on and , respectively, let with , and let be a map.
-
(a)
The following statements are equivalent:
-
(i)
.
-
(ii)
There exists a map given by
, and such that is well-defined.
-
(i)
-
(b)
Suppose that is a map and is distributive over . Then the map in (ii) is a -morphism.
Proof.
For (a):
;
, and ,
implies
given by
, and such that is well-defined, is a well-defined map.
For (b):
Since condition (i) in Definition 9.5.5 is satisfied, it suffices show that condition (ii) in Definition 9.5.5 is satisfied.
Let have variables and let . Then , of variables and such that . For , we denote by for brevity.
By Definition 9.1.2, is a partial function from to given by , where and . By Definition 9.1.4, is a restriction of some element of . Hence
is well-defined;
(by Definition 9.1.2)
(by the definition of in (ii))
(since is distributive over )
(by Definition 9.1.2)
(by the definition of in (ii))
is well-defined;
as desired for to be a -morphism. ∎
Proposition 14.3.5.
Let , , and be defined as in Proposition 14.3.4. Let such that and .
-
(a)
The following statements are equivalent:
-
(i)
.
-
(ii)
is bijective, there exists a bijective map given by
, and such that is well-defined, and its inverse can be given by
, and such that is well-defined.
-
(i)
-
(b)
Suppose that is distributive over and is a map. Then the bijective map given in (ii) is a -isomorphism.
Proof.
For (a):
;
is bijective, and (by Notation 14.3.1);
is bijective, given by
, and such that is well-defined, is a well-defined map, and given by
, and such that is well-defined, is also a well-defined map (by Proposition 14.3.4);
is bijective, there is a bijective map given by
, and such that is well-defined, and its inverse can be given by
, and such that is well-defined.
The third equivalence relation is explained as follows. The sufficiency () is obvious, so we only show the necessity (). For this purpose, we show that both and are the identity map.
Let and such that is well-defined. Then
(by the definition of )
(by Definition 9.5.3)
(by the definition of )
(by (the version of) Definition 9.5.3)
Then is the identity map on (because ).
Analogously, we can show that is the identity map on . Therefore, both and are bijective and is the inverse of .
For (b):
By (b) in Proposition 14.3.4, is a -morphism. Since is bijective, is a -isomorphism. ∎
Definition 14.3.6.
Let and be par-operator gen-semigroups on and , respectively, let , let and let be a -morphism from to a -space. If and there exists a map such that , , and such that is well-defined, then we say that is constructible by , or just say that is constructible for brevity.
Corollary 7.3.8 would still hold if and in it were generalized to be par-operator gen-semigroups, as shown below. The following also generalizes Corollary 14.2.6.
Corollary 14.3.7.
Let and be par-operator gen-semigroups on and , respectively, let and let . If and , then any -morphism from to a -space is constructible by .
Proof.
As we said at the end of Subsection 14.2, we could not generalize Theorem 7.2.9. For the same reason, we could not generalize Theorem 7.3.9, either.
For a par-operator gen-semigroup , we could not generalize results in Subsection 7.4 because now any may have more than one variable.
Part III. SOLVABILITY OF EQUATIONS
In Part III, we shall deviate from the topics of Parts I and II and study solvability of equations. A main goal of the classical Galois theory is to study the solvability by radicals of polynomial equations. In Section 15, we shall introduce a new understanding of solvability of polynomial equations, which involves a composition series of the Galois group of the polynomial. Analogously, for homogeneous linear differential equations, in Section 16, we shall introduce a solvability which involves a normal series of the differential Galois group of the differential equation. In Section 17, we shall generalize our results to “general” equations in terms of the theory developed in Parts I and II.
15. A solvability of polynomial equations
In this section, we shall show that for any separable polynomial over a field with a splitting field , there are polynomials such that the Galois group of each is simple and for any root of in , there is a formula which only involves rational functions over on the roots of in .
15.1. Canonical extensions and canonical solvability
Lemma 15.1.1.
Let be an algebraic field extension. Suppose , where denotes the fixed field of the Galois group , and has a composition series . Then there are intermediate fields such that , is Galois, and is a simple group or the trivial group , .
Proof.
, let . Then and , , and hence
Since the normal series is a composition series, by the fundamental theorem of the classical Galois theory or infinite Galois theory, , is Galois and
is a simple group or the trivial group. ∎
Lemma 15.1.1 implies that we may define a notion of solvability other than the solvability by radicals. In Subsection 15.2, it will be clear why we introduce the notions as follows.
Definition 15.1.2.
A field extension is called an irreducible Galois extension if is Galois and is a simple group or the trivial group. A field extension is called a canonical field extension if there is a tower of fields , where and , is an irreducible Galois extension.
Definition 15.1.3.
A polynomial is said to be canonically solvable if there is a canonical field extension such that is over , is not over any proper subfield of , and is a splitting field of .
Proposition 15.1.4.
Every separable polynomial is canonically solvable, and so is every polynomial over a field of characteristic 0.
15.2. Formulas for roots of separable polynomials
Now let’s explain why we introduced the notion of canonical solvability.
As is well-known, when we say that a polynomial over a field is solvable by radicals, we imply that there are formulas for the roots of the polynomial which may involve extraction of roots (), rational functions, and no operation else. Thus actually we assume that solving equations in the form of is a basic operation. Analogous to this assumption, now we are assuming that solving a separable polynomial whose Galois group is simple is a basic operation. Said differently, we assume that the roots of any separable polynomial with a simple Galois group can be determined (by symbols, formulas, or values). Then theoretically, there exists a formula for any root of any separable polynomial, explained below by a process described in steps.
Step 15.2.1.
Given a separable polynomial over a field with a splitting field , by Proposition 15.1.4 and Lemma 15.1.1, there exist intermediate fields such that , is Galois, and is a simple group or the trivial group, . Obviously we only need to deal with the case where each is a simple group, or equivalently, where each is an irreducible Galois extension (Definition 15.1.2) and .
Step 15.2.2.
Since is an irreducible Galois extension and , there is a separable polynomial such that is the splitting field of in and the Galois group of is simple. Now we assume that every root of can be determined (by a symbol, a formula, or a value). Then , which is a splitting field of , can be generated over by the roots of in . Hence every element of can be determined. Said precisely, , there is a formula for which only involves rational functions over on the roots of in .
Step 15.2.3.
Similarly, because is an irreducible Galois extension and , there is a separable polynomial over such that is the splitting field of in and the Galois group of is simple. Again we assume that every root of in can be determined. Then , which is a splitting field of , can be generated over by the roots of in . And as was shown in Step 15.2.2, can be generated over by the roots of in . Hence by substitutions, can be generated over by the roots of and in . Thus, , there is a formula for which only involves rational functions over on the roots of and in .
Step 15.2.4.
By induction, eventually for each element of , there is a formula which only involves rational functions over on the roots of in , where each is a separable polynomial over whose splitting field in is and whose Galois group is simple. Since is a splitting field of , for any root of in , there is a formula which only involves rational functions over on the roots of in .
In fact, in the above process, the operation of solving equation is “reduced” to the operations of solving , where the Galois group of each is simple, which implies that the operation of solving is “irreducible”. This is why we use the terminology “canonical”.
The above process applies to any polynomial which is canonically solvable. Therefore, the following is obvious, which also explains why we introduced the notion of canonical solvability.
Corollary 15.2.5.
Let be a polynomial over a field with a splitting field such that . If is separable, or more generally, is canonically solvable, then there exist polynomials such that the Galois group of each is simple and for any root of in , there is a formula which only involves rational functions over on the roots of in .
16. A solvability of homogeneous linear differential equations
For homogeneous linear differential equations, we can do a similar thing as we just did for polynomial equations, although the process will be relatively complicated.
16.1. Normal Zariski-closed series and composition Zariski-closed series
In differential Galois theory, the Picard-Vessiot extension of a homogeneous linear differential equation is the analogue of a splitting field of a polynomial (see e.g. [1, 4, 9]). However, by the fundamental theorem of differential Galois theory, only Zariski-closed subgroups (not necessarily all subgroups) of a differential Galois group have a bijective correspondence with intermediate differential fields of a Picard-Vessiot extension. Hence we first give the following definition.
Definition 16.1.1.
Let be a normal series where every is a Zariski-closed subgroup of a general linear group over a field. Then we call it a normal Zariski-closed series of . Moreover, if any possible proper refinement of this normal series is not a normal Zariski-closed one, then we call it a composition Zariski-closed series of .
Remark.
We do not require that a composition Zariski-closed series to be a composition series.
The following is comparable with Lemma 15.1.1.
Lemma 16.1.2.
Let a differential field extension be a Picard-Vessiot extension such that the constant field of is algebraically closed. Suppose that the differential Galois group has a normal Zariski-closed series
Then there are intermediate differential fields such that
, is a Picard-Vessiot extension and , and , .
Proof.
Let be the constant field of , which is algebraically closed. Hence we can apply the fundamental theorem of differential Galois theory (see e.g. [1, 4, 9]) for any Picard-Vessiot extension of .
, let . Then
and, by the fundamental theorem, each because each is a Zariski-closed subgroup of . Hence
Moreover, , is a differential field with . And since the constant field of is algebraically closed, by the fundamental theorem, is a Picard-Vessiot extension because . Hence by the definition of Picard-Vessiot extensions (see e.g. [1]), is also the constant field of . Again by the fundamental theorem, is also a Picard-Vessiot extension because . By induction, eventually each is a Picard-Vessiot extension and each by the fundamental theorem. ∎
Because , always has a normal Zariski-closed series. Thus we can employ Lemma 16.1.2 to any Picard-Vessiot extension whenever the constant field is algebraically closed.
16.2. Canonical extensions and canonical solvability
Let be a homogeneous linear differential equation over for which is a Picard-Vessiot extension with an algebraically closed field of constants. Then Lemma 16.1.2 implies that can be “decomposed” into homogeneous linear differential equations such that , is a Picard-Vessiot extension for , and hence we can solve by solving . We shall explain this process in Subsection 16.3. In light of this idea, we define a notion of solvability other than solvability by quadratures.
Definition 16.2.1.
A Picard-Vessiot extension is said to be irreducible if for any with , is not a Picard-Vessiot extension of . Otherwise it is said to be reducible.
Remark.
Trivially, is an irreducible Picard-Vessiot extension.
Then by the fundamental theorem of differential Galois theory, the following, which characterizes reducible Picard-Vessiot extensions, is obvious.
Corollary 16.2.2.
Let be a Picard-Vessiot extension with the constant field of being algebraically closed. Then is reducible if and only if has a proper nontrivial normal Zariski-closed subgroup.
Definition 16.2.3.
A differential field extension is called a canonical extension if there is a tower of differential fields , where , such that , is an irreducible Picard-Vessiot extension.
Definition 16.2.4.
Let be a homogeneous linear differential equation. If there is a Picard-Vessiot extension for such that
-
(i)
the constant field of is algebraically closed;
-
(ii)
there does not exist any proper differential subfield of such that is over and the constant field of is algebraically closed; and
-
(iii)
is a canonical extension;
then is said to be canonically solvable.
The following characterizes canonical solvability in terms of a composition Zariski-closed series of the Galois group.
Proposition 16.2.5.
Let be a homogeneous linear differential equation. Then is canonically solvable if and only if there is a Picard-Vessiot extension for such that conditions (i) and (ii) in Definition 16.2.4 are satisfied and has a composition Zariski-closed series.
Proof.
Assume that there is a Picard-Vessiot extension for such that conditions (i) and (ii) in Definition 16.2.4 are satisfied. Then by Definition 16.2.4, it is sufficient to show that the following two statements are equivalent.
(a) is a canonical extension.
(b) has a composition Zariski-closed series.
1. (a) (b):
Suppose that is a canonical extension. By Definition 16.2.3, there is a tower of differential fields where each is an irreducible Picard-Vessiot extension. Hence by the definition of Picard-Vessiot extensions, the constant field of every is the same, which we assume to be . Because condition (i) in Definition 16.2.4 is assumed to be satisfied, is algebraically closed, and hence we can apply the fundamental theorem of differential Galois theory to any Picard-Vessiot extension of any as follows.
because both and are Picard-Vessiot extensions. Analogously, since can be viewed as defined over , is a Picard-Vessiot extension, and hence because is also Picard-Vessiot. Applying this argument for each , eventually we obtain
(16.1) |
Because is an algebraically closed field, each is a Zariski-closed subgroup of the general linear group , where is the order of . Thus by Definition 16.1.1, (16.1) is a normal Zariski-closed series of .
Suppose that (16.1) has a proper refinement which is still normal Zariski-closed. Then and a Zariski-closed subgroup of such that with . Hence by the fundamental theorem of differential Galois theory, , , and is a Picard-Vessiot extension of . But is an irreducible Picard-Vessiot extension, a contradiction.
Therefore, any possible proper refinement of (16.1) is no longer a normal Zariski-closed one. By Definition 16.1.1, (16.1) is a composition Zariski-closed series of .
2. (b) (a):
Suppose that the differential Galois group has a composition Zariski-closed series: .
By Lemma 16.1.2, there are intermediate differential fields such that , where , , is a Picard-Vessiot extension and , . Thus,
(16.2) |
is a composition Zariski-closed series.
Besides, by the definition of Picard-Vessiot extensions, all share the same constant field, which we assume to be . Because condition (i) in Definition 16.2.4 is assumed to be satisfied, is algebraically closed. Hence we can apply the fundamental theorem of differential Galois theory to any Picard-Vessiot extension of , .
Suppose that there exists a reducible Picard-Vessiot extension , where . Then by Definition 16.2.1, with such that is a Picard-Vessiot extension of . Hence is also the constant field of . By the fundamental theorem, , where is also Zariski-closed. Moreover, since is a Picard-Vessiot extension, so is . Then again by the fundamental theorem, . Therefore, . And because . But (16.2) is a composition Zariski-closed series, a contradiction.
Thus, every (Picard-Vessiot extension) of is irreducible. By Definition 16.2.3, is a canonical extension. ∎
16.3. Formulas for solutions of homogeneous linear differential equations
Now we show that in a sense we may develop formulas for the solutions of any homogeneous linear differential equation by the following process described in steps. To simplify our descriptions, we introduce
Terminology 16.3.1.
Let be a differential field extension generated by . Then by differential field operations over on we mean rational functions over on and the derivation function of .
Step 16.3.2.
Let be a homogeneous linear differential equation for which is a Picard-Vessiot extension such that conditions (i) and (ii) in Definition 16.2.4 are satisfied. Let be the algebraically closed field of constants of .
Because , there must be a normal Zariski-closed series of in the form . By Lemma 16.1.2, there are intermediate differential fields such that with every being a Picard-Vessiot extension. Thus there exist homogeneous linear differential equations such that , is a Picard-Vessiot extension for .
Step 16.3.3.
Now is a Picard-Vessiot extension for . Just as the solutions of are assumed to be , now the solutions of in are assumed to be determined (by symbols, formulas, or values). Then because can be generated over by the solutions of in , every element of can be determined. Said precisely, , there is a formula for which only involves differential field operations over on the solutions of in (cf. Terminology 16.3.1).
Step 16.3.4.
Analogously, because is a Picard-Vessiot extension for , can be generated over by the solutions of in . Hence , there is a formula for which only involves differential field operations over on the solutions of in . As was shown in Step 16.3.3, every element of can be obtained by a formula which only involves differential field operations over on the solutions of in . Hence by substitutions, every element of can be determined by a formula which only involves differential field operations over on the solutions of and in .
Step 16.3.5.
By induction, eventually every element of can be determined by a formula which only involves differential field operations over on the solutions of equations in . Therefore, for each solution of in , there is a formula which only involves differential field operations over on the solutions of equations in .
In fact, in the above process, the operation of solving equation is “decomposed” into the operations of solving . Moreover, Proposition 16.2.5 tells us that in the case where has a composition Zariski-closed series, , is an irreducible Picard-Vessiot extension, which implies that any further “decomposition” of any , , would be trivial. This explains why in this case we say that is canonically solvable.
The following summarizes this subsection.
Corollary 16.3.6.
Let be a homogeneous linear differential equation for which is a Picard-Vessiot extension such that the constant field is algebraically closed.
Suppose that , where . Then there exist intermediate differential fields such that , , and homogeneous linear differential equations such that , is a Picard-Vessiot extension for . And for any solution of in E, there is a formula which only involves differential field operations over on the solutions of equations in cf. Terminology 16.3.1.
17. A possible strategy for equation solving
Now for “general” equations (with unknowns), we shall generalize our results (in the two preceding sections) in terms of the theory developed in Parts I and II. Because this section is only a sketch of some ideas, it is not developed in a rigorous way.
17.1. Splitting -spaces, Galois -groups and Galois -monoids of equations
To simplify our descriptions, we introduce
Terminology 17.1.1.
Let be a par-operator gen-semigroup on a set and let be a set. If and , there is a formula for which only involves some given operations on elements of and , then we say that T is defined on over .
Remark.
Note that we do not require .
For instance, the par-operator gen-semigroups defined in Examples 2.1.2, 8.1.3, 9.1.6 and 9.1.7 are all defined over .
Notation 17.1.2.
Let be an equation (with unknowns) and let be a set. We denote by the set of all solutions of in .
Example 17.1.3.
Let be a field extension and let be the par-operator gen-semigroup on defined in Example 9.1.6. Let such that splits over and let be the set of roots of in . Then is a splitting field of over , where denotes the equation .
Example 17.1.4.
Let be a differential field extension such that every constant of lies in . Let be the par-operator gen-semigroup on defined in Example 9.1.7. Let be a homogeneous linear differential equation over with being a fundamental set of solutions of it in . Then is a Picard-Vessiot extension (for ), where denotes the equation .
Inspired by the above two examples, we generalize the concepts of splitting fields and Picard-Vessiot extensions as follows.
Definition 17.1.5.
Remark.
-
(1)
If has no solution in , then of course .
-
(2)
We use the terminology “splitting” to be consistent with the classical Galois theory although maybe nothing of “split” like a polynomial.
-
(3)
It is possible that and/or .
-
(4)
if .
Moreover, we generalize the notions of Galois groups of polynomials and differential Galois groups of homogeneous linear differential equations as follows.
Definition 17.1.6.
17.2. Formulas for elements of splitting -spaces of “general” equations
Generalizing the strategies used in Subsections 15.2 and 16.3, the following shows that we can determine the splitting -space of an equation provided that some conditions are satisfied.
Proposition 17.2.1.
Given an equation , we are required to determine , where is a par-operator gen-semigroup on a set over a set .
If there exist equations , where every is nonempty and is assumed to be determined (by a symbol, a formula, or a value), and par-operator gen-semigroups on such that the following two conditions are satisfied,
-
(i)
, is defined over , where and , and all given operations on are also given as operations on (cf. Terminology 17.1.1);
-
(ii)
;
then for each element of , there is a formula which only involves the given operations on and .
Proof.
Since and is defined over (condition (i)), by Terminology 17.1.1, , there is a formula for which only involves the given operations on and .
Analogously, because and is defined over , , there is a formula for which only involves the given operations on and . And as was shown above, for each element of , there is a formula which only involves the given operations on and . Hence by substitutions and the assumption that all given operations on are also given operations on , any can be determined by a formula which only involves the given operations on , , and .
By induction, eventually every element of can be determined by a formula which only involves the given operations on and . Thus, (condition (ii)), there is a formula for which only involves the given operations on and . ∎
The strategy described by Proposition 17.2.1 is actually “decomposing” into each of which the solutions are assumed to be known. This is just the strategy used in Sections 15 and 16.
Note that in Sections 15 and 16, the “decomposition” of an equation into depends on a normal series of the Galois -group of (cf. Lemmas 15.1.1 and 16.1.2). However, for any other equation , we do not know whether a similar situation arises, i.e. whether there is a connection between the “decomposition” of described in Proposition 17.2.1 and a normal series of the Galois -group of (defined by Definition 17.1.6).
Moreover, we do not know whether there exists any case where the “decomposition” of is related to the Galois -monoid of (defined by Definition 17.1.6) rather than the Galois -group of .
Of course, the possible strategy for equation solving described in this section is just a very rough sketch and much more work is needed.
Part IV. OTHER TOPICS AND FUTURE RESEARCH
In Part IV, we shall briefly introduce some more results which we think to be important or deserve deeper study in the future.
18. Dualities of operator semigroups
Let be an operator semigroup and let be a -space. Then , and , . The symmetry in this equation implies that and may in a way exchange their roles with each other. Indeed, there exists a duality between and the maximum operator semigroup on which “accommodates” , as explained below.
In the following two results, for any set of functions from to and , we may denote by the set .
Theorem 18.0.1.
Let be an operator semigroup on a set , let be a -space, let , and let
and .
-
(a)
is an operator semigroup on , is a -space, and is an operator semigroup on .
-
(b)
, where the equation holds if and only if .
Remark.
-
(1)
Obviously , and intuitively, can be interpreted as the maximum operator semigroup on which “accommodates” (i.e. , would still be a -endomorphism of if were extended to ).
-
(2)
Since , if , then (b) implies . That is, when reaches its “maximum” . This is why we said that there exists a duality between and .
Proof.
For (a):
It follows from Proposition 2.3.8 and Definition 2.1.1 that is an operator semigroup on , and hence . Since Id on lies in , . Hence the -space is also a -space.
To show that is an operator semigroup on , we only need to show that it is closed under composition. Let , and . Then from the definition of , we can tell that and
Hence , and thus is an operator semigroup on .
For (b):
By Definition 2.3.1, and , commutes with on , i.e. , and hence is a -endomorphism (cf. Definition 2.3.7) of the -space (since by Proposition 2.2.7). Therefore, .
We are going to prove that if and only if .
Suppose . because . To show , let . Then by the definition of , and , and , and thus . Hence . Thus, .
Conversely, suppose . We already showed . To show , let such that . Then and and . It follows that , and thus . Hence . Therefore, . ∎
Analogously, there exists a duality between and the maximum operator semigroup which “accommodates” , though an additional condition is required.
Theorem 18.0.2.
Let be an operator semigroup on a set , let be a -space, let , and let
, and .
-
(a)
is an operator semigroup on , is a space and is an operator semigroup on .
-
(b)
If , is bijective onto , then
-
(c)
Suppose that , is bijective onto . Then and the equality holds if and only if .
Remark.
-
(1)
Apparently , and intuitively, can be interpreted as the maximum operator semigroup on which “accommodates” (i.e. , would still be a -automorphism of if were extended to ).
-
(2)
Since , if and , is bijective onto , then (c) implies . This is why we said that there exists a duality between and .
Proof.
For (a):
It follows from Proposition 2.4.3 and Definition 2.1.1 that is an operator semigroup on , and hence . Because Id on lies in , . Hence the -space is also a space.
To show that is an operator semigroup on , it suffices to show that it is closed under composition. Let , and . Then from the definition of , we can tell that and
Hence . Thus, is an operator semigroup on .
For (b):
Assume that , is bijective onto . By Definition 2.3.1, and , commutes with on , i.e. , , and hence is a automorphism of the space (since is bijective onto S). Therefore, .
For (c):
Assume that , is bijective onto . Since , , is bijective onto . Then by (b), .
We are going to prove that if and only if .
Suppose . because . To show , let . Then by the definition of , , and , and hence (because is bijective onto S). Hence . Thus, .
Conversely, suppose . We already showed . To show , let such that . Then and and . It follows that , and hence . Thus, . Therefore, . ∎
In Theorem 18.0.1, if were generalized to be a (par-)operator gen-semigroup, then statement (a) would still hold (though generally it would be false that because is always an operator semigroup). However, since the elements of End are all unary functions but the elements of a (par-)operator gen-semigroup are possibly not, statement (b) in Theorem 18.0.1 would no longer hold.
Analogously, in Theorem 18.0.2, if were generalized to be a (par-)operator gen-semigroup, then statement (a) would still hold (though generally it would be false that because is always an operator semigroup), but statements (b) and (c) would no longer hold.
19. Fixed sets and transitive actions of and
In this section we develop, among other results, two analogues of the well-known fact that the Galois group of an irreducible polynomial acts transitively on its roots.
In this section, is always a par-operator gen-semigroup unless otherwise specified.
In the classical Galois theory, the fixed field of , where is a field, is denoted by . To study and , which generalize the notion of , we first give
Notation 19.0.1.
Let , where is the domain of . Let
i.e. is the set of where is the only such that . And let
i.e. is the set of where is the only such that .
Remark.
For example, let , and be defined as in Proposition 7.1.2, then it is not hard to tell that .
19.1. Fixed sets and transitive actions of
The following explains why we have “End” in notation .
Lemma 19.1.1.
Let be a -space. Then , and hence .
Proof.
Recall a well-known fact in the classical Galois theory: the Galois group of an irreducible polynomial acts transitively on its roots. By Proposition 7.1.2, the following is an analogue of this fact.
Theorem 19.1.2.
Let be a -space. Suppose that the identity function lies in and , . Then with , such that .
Proof.
Let with . There are two cases as follows.
Suppose . It is assumed that , . Hence . By Proposition 14.1.3, the map given by , and of variables such that is well-defined, is a -morphism from to itself (because by Proposition 10.1.1). Then by Lemma 19.1.1, . Moreover, if let , then by the definition of , , as desired.
Suppose . Then by Notation 19.0.1, because . If let be the identity function on , then apparently and , as desired. ∎
Combining the above two results, we have
Corollary 19.1.3.
Let be a -space. Suppose that the identity function lies in and , . Then .
19.2. Fixed sets and transitive actions of
Lemma 19.2.1.
Let be a -space. Then , and hence .
Proof.
And for , the analogue of Theorem 19.1.2 is as follows, which, by Proposition 7.1.2, is also an analogue of the well-known fact that the Galois group of an irreducible polynomial acts transitively on its roots.
Theorem 19.2.2.
Let be a -space. Suppose that the identity function lies in and , . Then with , such that .
Proof.
Let with . There are two cases as follows.
Suppose . It is assumed that , . Hence . By Corollary 14.1.4, the map given by , and of variables such that is well-defined, is a -isomorphism from to itself (because by Proposition 10.1.1). Then by Lemma 19.2.1, . Moreover, if let , then by the definition of , , as desired.
Suppose . Then by Notation 19.0.1, because . If let be the identity function on , then and , as desired. ∎
Combining the above two results, we have
Corollary 19.2.3.
Let be a -space. Suppose that the identity function lies in and , . Then .
20. Two questions on Galois -extensions and normal subgroups of Galois -groups
In this section, is always a par-operator gen-semigroup unless otherwise specified.
As was promised at the end of Section 3, this section focuses on the following.
Problem 20.0.1.
Let be a -space and let . If and , then
-
(1)
when does ?
-
(2)
when do and
In particular, if let be fields such that is a Galois extension of , then, as is well-known, is Galois over (i.e. ) if and only if is normal in (that is, (1) is true if and only if (2) is true). However, for a general -space, the answer to Problem 20.0.1 seems to be not so simple. We can only give some sufficient conditions for the two questions in Problem 20.0.1. Nevertheless, one still can see that some, if not all, results and notions introduced in this section have analogues in well-known Galois theories.
20.1. -stable subsets of a -space
Definition 20.1.1.
Let be a -space with and . If and , then we say that is stable under or is H-stable.
Remark.
Another choice of this terminology is -invariant.
Lemma 20.1.2.
Let be a -space with . Let . Then is -stable.
Proof.
Suppose that and such that . Then and (otherwise and ). Assume that , , then , a contradiction. It follows that there is such that . However, since , . Then because , and hence , a contradiction again. ∎
Lemma 20.1.3.
Let be a -space with . Suppose that , is -stable, and every element of can be extended to an element of . Then .
Proof.
implies that no element of is fixed under the action of . Then no element of is fixed under the action of , and hence (cf. Definition 2.5.1 for ).
is -stable, and hence . Hence , and therefore .
On the other hand, since every element of can be extended to an element of , . Thus .
Therefore, . ∎
Now we obtain our first answer to the first question in Problem 20.0.1 as follows, where an “extra” condition is that “every element of can be extended to an element of ”.
Corollary 20.1.4.
Let be a -space with . Suppose that , , , and every element of can be extended to an element of . Then .
A converse of Lemma 20.1.2 is as follows, which is our first answer to the second question in Problem 20.0.1.
Proposition 20.1.5.
Let be a -space with . Suppose that is -stable. Then there exists a group homomorphism such that , and so and .
Proof.
Let be given by .
Since is stable, , . And because . Hence , which implies that is a well-defined map.
Moreover, if , then , and so , which implies that is a group homomorphism.
Finally, because is defined as the restriction to , : ∎
Combining Proposition 20.1.5 with Corollary 20.1.4, we immediately obtain our second answer to the first question in Problem 20.0.1.
Corollary 20.1.6.
Let be a -space with . If , , is -stable, and each element of can be extended to an element of , then .
20.2. Normal -subsets of a -space
Definition 20.2.1.
Let be a subset of a -space . If , (cf. Notation 14.1.1 for ), then we call a normal -subset of and write or .
Remark.
Any normal -subset of a -space is -stable, as shown below.
Lemma 20.2.2.
Let be a -space. Suppose that is a normal -subset of . Then is stable under .
Proof.
Recall that Proposition 7.1.4 still holds for being a par-operator gen-semigroup (cf. Subsection 14.1). Hence by Proposition 7.1.4, , and .
Let and . Then , and thus because . Hence by Definition 20.1.1, is stable under . ∎
The combination of Lemma 20.2.2 and Proposition 20.1.5 yields the following, which is obvious because . It is our second answer to the second question in Problem 20.0.1.
Corollary 20.2.3.
Let be a -space with . Suppose that is a normal -subset of . Then there exists a group homomorphism such that , and hence and .
A converse of Corollary 20.2.3 is as follows, which also explains the terminology “normal -subsets”.
Proposition 20.2.4.
Let be a -space. Suppose that the identity function lies in , , , and . Then .
Proof.
Finally, combining Corollary 20.2.3 with Corollary 20.1.4, we immediately obtain our third answer to the first question in Problem 20.0.1:
Corollary 20.2.5.
Let be a -space with . If , , , and each element of can be extended to an element of , then .
21. -transcendental elements and -transcendental subsets
In this section, the algebraic notions of transcendental elements over a field and purely transcendental field extensions are generalized for formal partial functions (cf. Definition 9.3.2). To simplify our descriptions, we introduce
Definition 21.0.1.
Let and be formal partial functions of variables. If for each set ,
-
(i)
is a domain of if and only if is a domain of ; and
-
(ii)
, neither nor is well-defined or ;
then we say that is equivalent to and write .
Moreover, throughout this section, unless otherwise specified, always stands for a set and denotes a set of formal partial functions.
21.1. -transcendental elements
To generalize the concept of transcendental elements in field theory, we give
Definition 21.1.1.
Let and let . If implies , , then we say that is -transcendental.
The above definition of transcendental elements coincides with that in field theory, as shown below.
Proposition 21.1.2.
Let be a subfield of a field and let be the polynomial ring . Then , is transcendental over if and only if is -transcendental.
Proof.
Let . Then
is algebraic over ;
nonzero such that ;
such that and ;
is not -transcendental.
∎
21.2. -transcendental subsets
Definition 21.1.1 may be generalized in more than one way. One of them is as follows.
Definition 21.2.1.
Let . If and , is -transcendental, i.e. implies , , then we say that is -transcendental.
Remark.
By we imply that , .
The following shows that the above definition of -transcendental subsets of coincides with the concept of purely transcendental field extension in field theory.
Proposition 21.2.2.
Let be a subfield of a field , let
let be the par-operator gen-semigroup on defined in Example 9.1.6, and let nonempty . Then is a purely transcendental field over if and only if or is -transcendental.
Proof.
Since , the field is purely transcendental over if and only if or is algebraically independent over . Hence we only need to show that is -transcendental if and only if is algebraically independent over .
is -transcendental ;
, and ,
implies ;
, and ,
implies that is a zero polynomial;
, and ,
implies that is a zero polynomial;
is algebraically independent over . ∎
22. A generalized first isomorphism theorem
We shall derive a theorem which generalizes the first isomorphism theorems for groups, rings, and modules.
Notation 22.0.1.
In this section, a -morphism is always defined by Definition 9.5.5. Moreover, is always a -morphism from a -space to a -space .
22.1. “Quotient space”
Notation 22.1.1.
For , we denote by the set .
Remark.
To define a par-operator gen-semigroup (corresponding to ) on , which we shall denote by (Notation 22.1.6), we define the elements (or for brevity) of as follows.
Notation 22.1.2.
Let -variable . Then by or () we denote the relation
and is well-defined}.
Remark.
If is well-defined, by Proposition 10.1.1, , and hence is well-defined.
To better understand the notation, we give
Example 22.1.3.
Let be a group homomorphism from a group to a group (cf. Proposition 8.5.11). Let be the identity element of . Then is the kernel of , , is a coset of the kernel, and is the set of all cosets of the kernel. Moreover, we shall see that is a group operation among elements of .
Proposition 22.1.4.
Let -variable . Then is a partial function from to , where denotes the cartesian product of copies of .
Proof.
It suffices to show that , the set has at most one element.
For this purpose, suppose that such that both and are well-defined and . By Notation 22.1.2, it suffices to show .
Then the following is obvious.
Corollary 22.1.5.
Let -variable . Then given by
such that is well-defined, is a well-defined partial function.
Notation 22.1.6.
We denote by the set , where is given in Corollary 22.1.5.
Proposition 22.1.7.
is a par-operator gen-semigroup on .
Remark.
If the proposition holds, then , and thus by Definition 2.2.8, is a quasi--space.
Proof.
By Corollary 22.1.5, , is a partial function from some to .
Let have variables, and , let have variables. To show that is a par-operator gen-semigroup on , by Definition 9.1.4, we only need to show that the composite is a restriction of some element of . For this purpose, since and hence , it suffices to show
(22.1) |
Let be the domain of . Then by Definition 9.1.2, is the partial function from to given by
where , , and .
To prove Equation (22.1), now we are showing that can be given exactly the same as above.
By Corollary 22.1.5, , can be given by
such that is well-defined. Combining this with the preceding paragraph, we can tell that can be given by
such that is well-defined.
By Corollary 22.1.5, a well-defined is given the value with being well-defined. Combining this with the preceding paragraph, we see that can be given by
such that is well-defined.
Since we showed that can be given exactly the same as above, Equation (22.1) holds, as desired. ∎
Notation 22.1.8.
By we denote the set Id on .
Corollary 22.1.9.
is a par-operator gen-semigroup on and is a -space. Specifically, .
Proof.
By Proposition 22.1.7, is a par-operator gen-semigroup on , so is Id on . Hence . Since Id , . Thus, . ∎
in Corollary 22.1.9 is actually a “quotient space” of corresponding to . To see this, we give
Example 22.1.10.
Let be the same as in Example 22.1.3. Let be the kernel of . Then the -space is the quotient group , where is the set of all group operations among (a finite number of) elements of . Moreover, defined as follows is the group isomorphism from to induced by .
22.2. A generalized first isomorphism theorem
Notation 22.2.1.
Let be given by and let .
Then . We are going to show that is a -isomorphism from the -space to if is a -space. But first, we need to show that is a -morphism, which requires a lemma as follows.
Lemma 22.2.2.
Suppose that is a map. Then is also a map. Moreover, .
Proof.
Apparently . Thus, to show that is a map, it suffices to show that is a map.
For this purpose, let such that . Let , and hence . Let and such that .
Then to show that is a map, by Proposition 9.5.2, it suffices to show that
neither nor is well-defined or :
is well-defined;
is well-defined (since and is a map);
is well-defined (by (ii) in Definition 9.5.5);
is well-defined (by Corollary 22.1.5);
is well-defined (since and );
is well-defined (by Corollary 22.1.5);
is well-defined (by (ii) in Definition 9.5.5);
is well-defined (since and is a map).
And in the case where the above equivalent conditions are satisfied,
as desired.
Therefore, is a map, so is .
Moreover, let . Then . Since both and are maps,
∎
Proposition 22.2.3.
is a -morphism from the -space to the -space .
Proof.
Since is a -morphism from to a -space, by Definition 9.5.5,
-
(i)
is a map and
-
(ii)
, and , neither nor is well-defined or
Then by Lemma 22.2.2, is a map and . Hence to show that is a -morphism, again by Definition 9.5.5, we only need to show that , and , neither nor is well-defined or
For this purpose, let , and .
Then we have the main theorem of this section as follows.
Theorem 22.2.4.
Suppose that is a -space. Then is a -isomorphism from the -space to the -space .
Proof.
Since is a -space, is also a -morphism to the -space . Then by Proposition 22.2.3, is a -morphism from the -space to the -space . Moreover, because is bijective, is a -isomorphism from the -space to the -space . ∎
Then by Proposition 10.2.2, the following is immediate.
Corollary 22.2.5.
If and , then is a -space and is a -isomorphism from the -space to the -space .
23. On topological spaces
In Subsection 23.1, we introduce some notions and properties of topological spaces which are related to our theory. To show an application, in Subsection 23.2, we employ Corollaries 3.2.5 and 3.2.6 and obtain the Galois correspondences on topological spaces.
23.1. Basic notions and properties
Most results in this subsection will not be employed in Subsection 23.2, but they may be useful for future research on topological spaces.
To simplify our descriptions, we define some notations for Section 23 as follows.
Notation 23.1.1.
Unless otherwise specified, and always denote topological spaces. Let and denote the power sets of and , respectively. Moreover,
, given by
and
, given by ,
where and denote the identity functions on and , respectively.
Remark.
By Definition 2.1.1, it is easy to see that and are operator semigroups on and , respectively.
Then we have
Proposition 23.1.2.
is a -space and
.
Proof.
By Definition 2.2.1, is a -space because .
For brevity, we take out a part from Proposition 6.1.3 as follows.
Definition 23.1.3.
Let be a map. Then we define its induced map by
-
(1)
that is closed in , let , and
-
(2)
that is not closed in , let ,
where denotes the set for convenience.
Remark.
Proposition 23.1.4.
A map is continuous if and only if its induced map is a -morphism.
To facilitate our following discussion, we give
Notation 23.1.5.
Let denote the set of all functions from to and let denote the set . Let and let . Moreover, let denote the set of all continuous functions from to , and let denote the set .
Then by Proposition 23.1.4, the following is obvious.
Corollary 23.1.6.
.
It would be desirable if . However, it is possible that (and so ). That is, there may exist a -morphism from to such that is not the induced map of any map from to , as shown below.
Example 23.1.7.
Before going further, we define
Notation 23.1.8.
We denote by the map from to given by . Besides, let .
Obviously is surjective. It would be desirable if were always injective. However, there are cases where is not injective, as shown below.
Example 23.1.9.
Let and . Let and be the (trivial) topologies on and , respectively. Let two maps be given by and , respectively. Then . And by Definition 23.1.3, and . Thus . Hence the map from to is not injective.
Surprisingly, there are two important cases where is injective, as shown by the following two results.
Proposition 23.1.10.
Suppose that every finite point set in is closed, i.e. satisfies the axiom. Then is injective.
Proof.
Assume that is not injective. Then there are two different maps such that . Hence such that .
Proposition 23.1.11.
is injective.
Proof.
Assume that is not injective. Then there are two different maps such that . Hence such that . Then by the same argument as in the proof of Proposition 23.1.10, is closed in and .
It follows that . Hence and (because ), and thus (otherwise , a contradiction). Analogously, .
Then because and . Hence ; that is, is not closed in . Hence by Definition 23.1.3,
and .
Because it is assumed that , .
Since , we can tell from the above equation that . Then because it was shown that , by Definition 23.1.3, , which is contrary to the assumption that . ∎
Moreover, there is an important case where commutes with composition of functions, as shown below.
Lemma 23.1.12.
Let and be continuous maps from to . Suppose that is injective. Then .
Proof.
It suffices to show that . There are two cases as follows.
(1) :
Then it suffices to show .
Since is continuous, by e.g. [6], . Hence . On the other hand, , and hence . Therefore, , as desired.
(2) :
Assume that is closed. Since is continuous, is also closed. However, since is injective, . Hence is not closed (because ), a contradiction. Thus is not closed. Then by Definition 23.1.3,
as desired. ∎
Notation 23.1.13.
We denote by the set of all homeomorphisms from to . Let .
Then we immediately have
Corollary 23.1.14.
constitutes a group, which we still denote by , with composition of functions as the binary operation.
also constitutes a group, as shown below.
Proposition 23.1.15.
With composition of functions as the binary operation, constitutes a subgroup, which we still denote by , of defined in Proposition 2.4.3.
Remark.
Recall that was defined in Notation 23.1.1.
Proof.
It follows from Lemma 23.1.12 that is closed under composition of functions as the binary operation; that is, , . Composition of functions is associative. Moreover, by Definition 23.1.3, the identity map on induces the identity map on , which we denote by .
By Definition 23.1.3, it is not hard to see that , is bijective. , let . Then , by Lemma 23.1.12, , and analogously, .
Therefore, constitutes a group with composition of functions as the binary operation.
Moreover, is isomorphic to , as shown below.
Proposition 23.1.16.
, which is the restriction of to , is a group isomorphism from to .
Remark.
was defined by Notation 23.1.8.
23.2. Galois correspondences on topological spaces
Firstly, we need to recall some notations. By Definitions 2.5.1, 2.6.1, 3.2.1 and 3.2.3, we immediately have
Proposition 23.2.1.
Let and let . Then
and
Remark.
was given in Proposition 23.1.2.
Then by applying Corollary 3.2.5 to a topological space , we immediately get the following, which, roughly speaking, shows the Galois correspondence between the Galois -monoids of and the fixed subsets of under the actions of the -endomorphisms of .
Corollary 23.2.2.
Let be a topological space, let be the power set of , and let . Then the correspondences
and
define inclusion-inverting mutually inverse bijective maps between and .
Moreover, we may employ Theorem 5.1.2 (resp. Theorem 5.3.2) to investigate whether we can endow with a topology such that
{ is closed and nonempty}
(resp. whether we can define a topology on End such that is the set of all closed submonoids of ).
We will not go further for this subject.
Analogously, by Proposition 23.1.2 and Definition 2.4.1, we have
Autbijective .
And by Definitions 2.6.1, 3.2.1 and 3.2.3, we immediately have
Proposition 23.2.3.
Let . Then
and
Then by applying Corollary 3.2.6 to a topological space , we immediately have the following, which shows the Galois correspondence between the Galois -groups of and the fixed subsets of under the actions of the -automorphisms of .
Corollary 23.2.4.
Let be a topological space, let be the power set of , and let . Then the correspondences
and
define inclusion-inverting mutually inverse bijective maps between and .