A non-Archimedean Arens–Eells isometric embedding theorem on valued fields
Abstract.
In 1959, Arens and Eells proved that every metric space can be isometrically embedded into a real linear space as a closed subset. In later years, Michael pointed out that every metric space can be isometrically embedded into a real linear space as a linear independent subset and provided a short proof of Arens–Eells theorem as an application. In this paper, we prove a non-Archimedean analogue of the Arens–Eells isometric embedding theorem, which states that for every non-Archimedean valued field , every ultrametric space can be isometrically embedded into a non-Archimedean valued field that is a valued field extension of such that the image of the embedding is algebraically independent over . Using Levi-Civita fields, we also show that every Urysohn universal ultrametric sapce has a valued field structure.
Key words and phrases:
Ultrametrics, Isometric embeddings, and Non-Archimedean valued fields2020 Mathematics Subject Classification:
Primary 54E35, Secondary 54E40, 51F991. Introduction
In 1956, Arens and Eells [1] established that for every metric space there exist a real normed linear space and an isometric embedding such that is closed in . Michael [16] pointed out that for every metric space , we can take a real normed linear space and an isometric embedding such that is linearly independent over . Using this observation, Michael provided a short proof of the Arens–Eells theorem.
A metric on is said to be an ultrametric or a non-Archimedean metric if for all , where stands for the maximal operator on . A set is a range set if and . An ultrametric on is said to be R-valued if for all . Some authors try to investigate non-Archimedean analogue of theorems on metric spaces such as the Arens–Eells theorem. Megrelishvili and Shlossberg [13] proved a non-Archimedean Arens–Eells theorem, which embeds ultrametric spaces into linear spaces over . In [14, Theorem 4.3], as a improvement of their theorem, they prove a non-Archimedean Arens–Eells theorem on linear spaces over arbitrary non-Archimedean valued fields. In [7, Theorem 1.1], as a non-Archimedean analogue of the Arens–Eells theorem, the author showed that for every range set , for every integral domain with the trivial absolute value (i.e., for all ), for every -valued ultrametric space , there exist an -valued ultra-normed module over and an isometric embedding such that is closed and linearly independent over . Using this embedding theorem, the author proved a (-valued) non-Archimedean analogue of the Hausdorff extension theorem of metrics.
There are other attempts to construct an isometric embedding from an ultrametric spaces into a space with a non-Archimedean algebraic structure. Schikhof [21] established that every ultrametric spaces can be isometrically embedded into a non-Archimedean valued field using the Hahn fields, which is a generalization of a field of formal power series. In [2, Conjecture 5.34], Baroughan raise the following conjecture.
Conjecture 1.1.
Let be an odd prime and be an -valued metric space, where . Then there exist a non-Archimedean Banach algebra over and an isometry .
In this paper, as a generalization of the Schikhof theorem ([21]) and known non-Archimedean analogues ([7, Theorem 1.1] and [14, Theorem 4.3]) of the Arens–Eells theorem, we prove that for every non-Archimedean valued field , for every metric space , there exist a valud field and an isometry embedding such that is a valued field extension of , the set is closed in , and is algebraically independent over (Theorem 4.6). A key point of the proof of Theorem 4.6 is to use the notion of -adic Hahn fields (-adic Mal’cev–Neumann fields), which was first introduced by Poonen [20] as -adic analogues of ordinary Hahn fields.
The theory of -adic Hahn fields have an application to Urysohn universal ultrametric spaces, which is defined as ultrametric spaces possessing high homogeneity. We introduce the concept of -adic Levi-Civita fields as subspaces of -adic Hahn fields, which is a -adic analogue of ordinary Levi-Civita fields (see, for instance, [3]). We also show that if is or a prime, then every Urysohn universal ultrametric space has a field structure that is an extension of , where we consider that and it is equipped with the trivial valuation in the case of (see Theorem 5.5).
The paper is organized as follows. Section 2 presents notions and notations of metric spaces and valued fields. We introduce Hahn fields and -adic Hahn fields, which plays an important role in the proofs of our main results. We also prepare some basic statements on valued fields and Hahn fields. In Section 3, we show statements on algebraic independence in (-adic) Hahn fields. Section 4 is devoted to proving Theorem 4.6. We also provide an affirmative solution of Conjecture 1.1. In Section 5, we show that (-adic) Levi-Civita fields become Urysohn universal ultrametric spaces. Our arguments in this section are baed on the author’s paper [8].
Acknowledgements.
The author would like to thank to Tomoki Yuji for helpful advices on algebraic arguments.
2. Preliminaries
2.1. Generalities
In this paper, we use the set-theoretic notations of ordinals. For example, for an ordinal , we have if and only if .
2.1.1. Metric spaces
For a metric space , and for a subset of of , and for , we define . For and , we denote by the closed ball centered at with radius . We often simply represent it as when no confusion can arise. Similarly, we define the open ball .
The proofs of the next two lemmas are presented in Propositions 18.2, and 18.4, in [22], respectively.
Lemma 2.1.
Let be a set, and be a pseudo-metric on . Then satisfies the strong triangle inequality if and only if for all , the inequality implies .
Lemma 2.2.
Let be a pseudo-ultrametric space, and . Then for every , we have .
2.1.2. Valued rings
Let be a commutative ring. We say that a function is a (additive) valuation if the following conditions are satisfied:
-
(1)
for every , we have if and only if ;
-
(2)
for every pair , we have
-
(3)
for every pair , we have , where stands for the minimum operator on .
If is a field, then it is called a valued field. Note that for every valued ring , we can extend the valuation to the fractional field of . Namely, for , where and , we define . In this case, is well-defined and the pair naturally becomes a valued field. For example, for a prime , we define the -adic valuation on by declaring that is the number of the factor in the prime factorization of . The completion of with respect to is called the ring of -adic integers. The fractional field of is called the field of -adic numbers. For more discussion on -adic numbers, we refer the readers to [19] and [22].
We say that a function is a (non-Archimedean) absolute value or multiplicative valuation if the following conditions are satisfied:
-
(1)
for every , we have if and only if ;
-
(2)
for every pair , we have
-
(3)
for every pair , we have , where stands for the maximum operator on .
In what follows, for every , we consider that and . For a valuation on a ring , and for a real number , we define and .
Fix , valuations and absolute values on a ring are essentially equivalent. Namely, they are corresponding to each other as follows.
Proposition 2.3.
Let be a commutative ring. Then the following statements are true:
-
(1)
For every , and for every valuation on , the function is an absolute value on ;
-
(2)
For , and for every absolute value on , the function is a valuation on .
In this paper, we use both of valuation and absolute values on a ring due to Proposition 2.3. Since we represent a valuation (resp. an absolute value) as a symbol like , or (resp. or ), the readers will be able to distinguish them.
For a valued ring , the set becomes a ring and is a maximal ideal of . We denote by the field , and we call it the residue class field of . We also denote by the canonical projection. We simply represent it as when no confusions can arise. We say that a subset of is a complete system of representative of the residue class field if , , and is bijective.
2.2. Constructions of valued fields
In this section, we review some constructions of valued fields such as Hahn fields. For more discussion, we refer the readers to [3] and [4]. Most of the proofs in this section is refered to [20] and [23].
2.2.1. Hahn rings and fields
A non-empty subset of is said to be well-ordered if every non-empty subset of has a minimum.
We denote by the set of all subgroups of containing (equivalently, ). For the sake of convenience, we only consider the setting where in this paper.
Now we review the construction of the Hahn fields in [20]. Let and be a commutative ring. For a map , we define the support of by the set . We denote by the set of all such that is well-ordered. We often symbolically represent as , where is an indeterminate. For every pair , we define by
We also define by
Define a valuation on by . Since is well-ordered, the minimum actually exists. Note that becomes a subring of .
Proposition 2.4.
Let , and be a field. The pair becomes a valued field and it satisfies that .
Proof.
See [20, Corollary 1]. ∎
We call the Hahn ring associated with and and call it the Hahn field if is a field. Note that in general, we can define the Hahn fields even if is a linearly ordered Abelian group (see [20]).
2.2.2. The -adic Hahn fields
A -adic analogue of the Hahn fields was first introduced in [20]. Let us review a construction. A field of characteristic is said to be perfect if , or and every element of has a -th root in . The following proposition states the existence of rings of Witt vectors. The proof is presented in [23].
Proposition 2.5.
Let be a perfect field of characteristic . The there exists a unique valued ring of characteristic equipped with a valuation such that , is complete, and .
For each perfect field , we denote by the valuation field stated in Proposition 2.5 and denote by the fractional field of . We use the same symbol as the valuation on induced by in the canonical way. The ring is called the ring of Witt vectors associated with . Notice that for every prime , we have and , and the valuation coincides with the -adic valuation .
The next proposition explains the concrete representation of an elements of a ring of Witt vectors.
Proposition 2.6.
Let be a perfect field of characteristic . Then there uniquely exists a map such that is a complete system of representatives, and for all . In this case, for every , there uniquely exists a sequence in such that and for a sufficient large , we have for all .
Proof.
See Proposition 8 in the page 35 and and the argument in the page 37 in [23]. ∎
Proposition 2.7.
Let and be perfect fields of characteristic . For every homomorphism , there uniquely exists a homomorphism such that . Moreover, if , where , then . In particular, we have for all .
Proof.
See [23, Proposition 10 in the page 39]. ∎
Remark 2.1.
It is known that the construction of rings of Witt vectors is a functor (see [23, Page 39]).
Now we discuss a -adic analogue of Hahn fields, which is defined by a quotient field of a Hahn ring. For , and for a perfect field of characteristic , we define a subset of by the set of all such that in for every .
Proposition 2.8.
For every , and for every perfect field with characteristic , the set is an ideal of the ring , and is a filed.
Lemma 2.9.
Let , be a prime, and be a field of characteristic . Let be a complete system of representatives of the residue class field of . Then every element is equivalent to an element modulo , where is in . In addition, for every , the family is unique and .
Proof.
See [20, Proposition 4]. ∎
Definition 2.1.
Let denote the quotient field , and let denote the canonical projection. We define by .
Proposition 2.10.
Let , and be a perfect field of characteristic . Take a complete system of representatives of the residue class field . Then the following statements are true:
-
(1)
For every , the value is independent from the choice of .
-
(2)
The map is a valuation on .
Proof.
See [20, Proposition 5]. ∎
We call the valued field the -adic Mal’cev–Neumann field or -adic Hahn field. Notice that is nothing but the field of -adic numbers.
To consider characteristics of a valued field and its residue class field, we supplementally define by the set of all pairs such that and are or a prime satisfying the either of the following conditions:
-
(Q1)
,
-
(Q2)
and .
Note that satisfies (Q2) if and only if .
In order to discuss -adic and ordinary Hahn fields in unified manner, we make a notation as follows.
Definition 2.2.
Let , , and let be a perfect field of characteristic . We define a field by
We also define a valuation on by
A metric space is said to be spherically complete if for every sequence of (closed or open) balls with for all , we have .
Proposition 2.11.
Let , , and be a field of characteristic . Then the following statements are true:
-
(1)
;
-
(2)
;
-
(3)
is spherically complete. In particular, it is complete.
Proof.
Next we consider homeomorphic embeddings between -adic or ordinary Hahn fields. We begin with ordinary ones. Let with , and be commutative rings. We represent as the inclusion map . Let be a ring homomorphism. For , we define by , where
Then becomes a map. If , we simply write it as . Let us observe properties of .
Proposition 2.12.
Let be commutative rings, and be a ring homomorphism. Let such that , and denote by the inclusion map . Then the map is a ring homomorphism and satisfies on .
Proof.
The lemma follows from the definitions of and Hahn fields. ∎
Next we discuss -adic Hahn fields, which is defined by quotient fields of Hahn rings.
Proposition 2.13.
Let , and be perfect field with characteristic and is a homomorphism. Then the homomorphism satisfies
in particular, the map induces a homomorphism
such that on .
Proof.
Take and put . Then for every we have in . Note that for a fixed and for a sufficient large , we have for all since is well-ordered. By the strong triangle inequality, is equivalent to in as (see [3, Theorem 2.24]). Since for all (see Proposition 2.7), we also have in as . Thus the infinite sum is convergent and we have
This shows that , and hence
In particular the map induces a map .
Proposition 2.12 implies that a map satisfies . Then . ∎
Let with , , and be fields of characteristic , and be a homomorphism. Denote by the inclusion map. We define
Let and be valued fields. We say that is a valued field extension of as a valued field if and .
Proposition 2.14.
Let and be perfect field with characteristic and be a homomorphism. The map is a homomorphism such that on . In addition, if , then, the following are true:
-
(1)
is a valued field extension of ;
-
(2)
In particular, is a valued field extension of .
Proof.
Definition 2.3.
Let , , and let be a perfect field of characteristic . We make the following assumptions and definitions.
-
(1)
In the rest of the paper, whenever we take a complete system of representatives of the residue class field , in the case of , we define using the fact that . In the case of , we take such that based on Proposition 2.14.
-
(2)
We define an element as follows. If , we define by an indeterminate as in the definition of Hahn fields. If , we define . In this case, every element of can be represented as a power series of with powers in .
-
(3)
For , and , if , then we define the coefficient of in the power series representation of . If , then we fixed a complete system of representatives of , and we define the coefficient of with respect to . Of cause, in this case, the value depends on a system of representatives. Throughout this paper, we will not consider the situation where we change a system of representatives. Thus no confusion can arise even if does not explicitly appear in the notation of . Notice that we can represent .
Remark 2.2.
A group is said to be divisible if for every and for every there exists such that .
Proposition 2.15.
Let be divisible, , be an algebraically closed field with characteristic , and be a valued field such that and . Then there exists a homomorphic embedding such that fot all . Namely, the field can be regarded as a valued subfield of .
Proof.
See [20, Corollary 5]. ∎
2.2.3. Levi–Civita fields
We next discuss Levi–Civita fields and -adic Levi–Civita fields, which are will be used in Section 5.
Let , and be a field. For , in this subsection, we mainly consider the following condition.
-
(Fin)
For every , the set is finite.
We denote by the set of all satisfying the condition (Fin). For the next lemma, we refer the readers to [3, Theorem 3.18].
Lemma 2.16.
Let , be a field. Then the set is a subfield of .
We call the Levi–Civita field associated with and
Fix with and assume that has characteristic . Before defining a -adic analogue of Levi-Civita fields, we supplementally define a subset of by the set of all members satisfying the condition (Fin). We define by , where is the canonical projection.
Lemma 2.17.
Let , be a prime, and be a field of characteristic . Then the following statements are true:
-
(1)
For every complete system of representatives of , and for every , the member satisfies the condition (Fin) and for all .
-
(2)
We have is a subring of ;
-
(3)
If satisfies , then .
-
(4)
For every , there exists such that is equivalent to modulo .
Proof.
First we prove (1). By Lemma 2.9, we see that for all . Put and . Lemma 2.9 also shows that . Due to this relation, since satisfies the condition (Fin), so does . Hence the statement (1) is true.
Next we prove (2). By the definitions of and , we have (pay an attention to the difference between and appearing in and , respectively). Thus is a subring of .
As a consequence of Lemma 2.17, we obtain the next corollary.
Corollary 2.18.
Let , be a prime, and be a field of characteristic . Then the following statements are true:
-
(1)
For every complete system of representatives of , the set is equal to the set . Moreover, for every , there uniquely exists such that .
-
(2)
The set is a subfield of .
We call the -adic Levi–Civita field associated with and . We simply represent the restriction as the same symbol . In this setting, the field becomes a valued subfield of .
To use -adic and ordinary Levi-Civita fields in a unified way, we make the next definition.
Definition 2.4.
Let , , and be a field of characteristic . We define by
Proposition 2.19.
Let , , and be a field of characteristic . Then the set is a subfield of .
3. Algebraic independence over valued fields
First we remark that, for valued fields and such that is a valued field extension of , there exists a canonical injective embedding from into . Namely, we can regard as a subset of since the inclusion map satisfies and . Thus it naturally induce an homomorphic embedding .
Let and be fields with . A member of is said to be transcendental over if is not a root of any non-trivial polynomial with coefficients in . A subset of is said to be algebraically independent if any finite collection in does not satisfy any non-trivial polynomial equation with coefficients in . Note that a singleton of is algebraically independent over if and only if is transcendental over .
Lemma 3.1.
Let and be valued fields. Assume that is a valued field extension of . If such that is transcendental over , then is transcendental over .
Proof.
The lemma follows from [5, Theorem 3.4.2]. ∎
Lemma 3.2.
Let and be valued fields such that is a valued field extension of . If satisfy that:
-
(1)
the set is algebraically independent over ;
-
(2)
there exists a filed containing and satisfying for which there exist and satisfying that and is transcendental over ,
then the set is algebraically independent over .
Proof.
For the sake of contradiction, suppose that the set is not algebraically independent over . From (1) and the fact that contains , it follows that is algebraic over , and hence so is . Using Lemma 3.1 together with (2), we see that is transcendental over . This is a contradiction. Therefore, the set is algebraically independent over . ∎
Lemma 3.3.
Let , , , and let be a perfect field with characteristic . Fix a cardinal and a complete system of representatives of the residue class field . If a set of non-zero elements of satisfies that:
-
(N1)
for every pair with , and for every satisfying that , if either of and is non-zero, then ,
then for every finite subset of , there exist and such that and for all .
Proof.
Put and take a pair such that . Then either or is non-zero. We my assume that . Using the condition (N1) and the minimality of , we have for all . ∎
Let , , be a field of characteristic , and be a subfield of . Fix a complete system of representatives if . We denote by the subfield of generated by over . The definition of is “ad-hoc”, which means that it depends on not only information of , but also the inclusion map . Namely, even if are isomorphic each other as fields, it can happen that .
Proposition 3.4.
Let , , and be perfect fields with characteristic such that , Take a subfield of such that . Fix a system of representatives of . If a set of of non-zero elements of satisfies the condition (N1) in Lemma 3.3 and the following:
-
(T1)
for every pair , and for every distinct pair , if , then we have ;
-
(T2)
the set is algebraically independent over ,
then the set is algebraically independent over .
Proof.
Let be the same element of as in Definition 2.4. Take -many distinct members in . Now we prove that is algebraically independent over by induction on .
In the case of , put . Take such that . Put . Then, due to the condition (T1), we have . Note that the set is algebraically independent over . Let be the perfect subfield of generated by , and put . Notice that is a subfield of (see Proposition 2.14). The fact that implies that . By assumption (T2) and , we see that is transcendental over . Thus, by Lemma 3.1, we conclude that is transcendental over . In particular, is transcendental over .
Next, we fix and assume that the case of is true. We consider the case of . Since satisfies the condition (N1), we can take and stated in Lemma 3.3. We may assume that . Put
According to (T1) and the conclusion of Lemma 3.3, we see that . Let be a perfect subfield of generated by , and put . Similarly to the case of , we observe that . Since , we have for all . Define . Due to the condition (T1), we have . By the definition, we also have . Thus the condition (T2) shows that is transcendental over . Hence Lemma 3.2 shows that the set is algebraically independent over . This finishes the proof. ∎
Remark 3.1.
Put . The condition (T1) means that is zero or is a unique member in such that for some .
4. Isometric embeddings of ultrametric spaces
In this section, we prove our non-Archimedean analogue of the Arens–Eells theorem. As a consequence, we give an affirmative solution of Conjecture 1.1.
4.1. A non-Archimedean Arens–Eells theorem
4.1.1. Preparations
This subsection is devoted to proving the following technical theorem, which plays a central role of our first main theorem. Our proof of the next theorem can be considered as a sophisticated version of the proof of the main theorem of [21].
Theorem 4.1.
Let , , , be a field, and be a perfect field of characteristic Fix a cardinal and a complete system of representatives of the residue class field . Let be a subset of . Put . If the following condition are satisfied:
-
(A1)
is a field extension of ;
-
(A2)
the subset of is algebraically independent over ;
-
(A3)
,
then for every -valued ultrametric space with , there exists a map such that:
-
(B1)
each is non-zero;
-
(B2)
the map is an isometric embedding from into the ultrametric space ;
-
(B3)
for every pair , and for every , if either of or is non-zero, then we have ;
-
(B4)
for every pair , and for every distinct pair , if , then .
-
(B5)
for every , the set is contained in .
In this subsection, in what follows, we fix objects in the assumption of Theorem 4.1. We divide the proof of Theorem 4.1 into some lemmas.
First we prepare notations. Take , and put . Fix and , and define an -valued ultrametric on by and . Then is actually an ultrametric (see, for example, [7, Lemma 5.1]). A one-point extension of a metric space is a traditional method to prove analogues of the Arens–Eells theorem.
Put and with . For every , we also put and .
Fix . We say that a map is well-behaved if the following conditions are true:
-
(C1)
if , then is the empty map and if , then , where is the zero element of ;
-
(C2)
the map is an isometric embedding from into ;
-
(C3)
for every pair , and for every , if either of or is non-zero, then we have ;
-
(C4)
for every pair , and for every distinct pair , if , then ;
-
(C5)
for every , the set is contained in .
For an ordinal , a family is said to be coherent if the following condition is true:
-
(Coh)
for every and for every , we have .
For the sake of simplicity, we define an ultrametric on by .
We shall construct a coherent family of well-behaved maps using transfinite recursion. We begin with the following convenient criterion.
Lemma 4.2.
Proof.
First we note that the following claim is true:
-
(CL)
For every , there exists such that
On the other words, for every , there exists such that
for all
Due to (C3) for , it suffices to consider the case of . Take and . Assume that either of or is non-zero. In this case, we have . Thus . If , then the property (P1) shows that . Thus the condition (C3) is valid. If , then the property (P2) implies that . Thus, using (C5) for , the condition (C3) is satisfied. In any case, we conclude that the condition (C3) is true.
Owing to (C4) for , it is enough to confirm the case where and , or and . Take arbitrary distinct pair . First assume that , i.e., . By (P1), we have . In the case of , using (P2) and (C5) for , we have . In the case of . Then . Thus the property (CL) enables us to take such that . Thus we also have for all . Applying (C4) for to and , we obtain that . Hence, we have . Second assume that , i.e., . By (P1), it is sufficient consider that . By (P2), if , then we have . If , then the property (CL) enables us to take such that . The remaining proof is similar to that of the first case.
We next see the elementary lemma.
Lemma 4.3.
Let be the same element of as in Definition 2.4. Then, for every and for every , we can take such that , where . In this case, we have .
Proof.
We put , where and we define . By the definition of , or , we have . Thus Lemma 2.2 implies that . ∎
Next we show lemmas corresponding to steps of isolated ordinals in transfinite induction.
Lemma 4.4.
Fix with , and let be a well-behaved map. If , then we can obtain a well-behaved isometric embedding such that .
Proof.
Put . Put and . Let be the same element in as in Lemma 4.3.
Case 1. [There is no such that ]: Take a sequence in such that for all and as . Put . Since and for all (see Lemma 2.2), using the spherical completeness of ((3) in Proposition 2.11), we obtain . In this case, the set is a closed ball of radius centered at some point in . Lemma 4.3 implies that there exists in such that . We define a map by and . This definition implies the condition (C1). Next we verify the condition (C2). It suffices to show that . Take such that . Then Lemma 2.1 implies , and hence . By , we have
Thus, using Lemma 2.1 again, we have
and hence . By the construction, the mao satisfies the properties (P1)–(CL). Thus, we see that satisfies the conditions (C3)–(C5).
Case 2. [There exists such that ]: By Lemma 4.3, there exists such that . We put
Then Lemma 2.2 states that . We define a map by and . To verify the condition (C2), we show that for all . If , then we have . Similarly, we have . Since , we have
If , then . By the definition of , we have . We also see that . Then we can represent , where . Take with . Then the condition (C2) for implies that for all . From the definition of and , it follows that . Hence
Similarly to Case 1, by the construction, we see that satisfies the properties (P1)–(CL). Thus, we see that satisfies the conditions (C3)–(C5). ∎
Lemma 4.5.
Fix with , and let be a well-behaved map. If , then there exists a well-behaved map such that .
Proof.
We now define as follows. Take a sequence in such that as , and define . The value is independent of the choice of a sequence .
First we confirm that is well-behaved. Since satisfies the condition (C1), so does .
Let us prove Theorem 4.1.
Proof of Theorem 4.1.
The proof is based on [21]. Using transfinite recursion, we first construct a coherent family of well-behaved maps. Fix and assume that we have already obtained a coherent family of well-behaved maps. Now we construct as follows. If , then we define as the empty map. If , then we have and we define by . If for some with , then using Lemmas 4.4 and 4.5, we obtain a well-behaved map such that . If is a limit ordinal, then we define by , where . In this case, is well-defined since the family is coherent. Therefore, according to transfinite recursion, we obtain a well-behaved isometric embedding . In this case, note that . Now we define by . Due to (C1), we have and . Thus the condition (B1) is true. Since satisfies (C2)–(C5), the map satisfies the conditions (B2)–(B5). This finishes the proof of Theorem 4.1. ∎
4.1.2. The proof of the first main result
The following is our first main result.
Theorem 4.6.
Let , , be a cardinal, and be dividable. Put . Take an arbitrary valued field of characteristic such that and has characteristic . Then there exists a valued field such that:
-
(L1)
the field is a valued field extension of ;
-
(L2)
;
-
(L3)
for each -valued ultrametric with , there exists an isometric embedding such that the set is algebraic independent over .
Moreover, for every -valued ultrametric space , there exist a valued field and a map such that:
-
(F1)
the map is an isometric embedding;
-
(F2)
the field is a valued field extension of ;
-
(F3)
;
-
(F4)
the set is closed in ;
-
(F5)
the set is algebraically independent over ;
-
(F6)
if is complete, then can be chosen to be complete.
Proof.
Let be the algebraic closure of . Notice that is perfect. Take a perfect filed such that the transcendental degree of over is , and take a transcendental basis of over . In this case, we have . Take a system of representatives of . Notice that . Put . Applying Theorem 4.1 to and , we can take an isometric embedding satisfying the conditions (B1)–(B5). Let be the inclusion map. Using a homomorphic embedding , in what follows, we consider that is a subfield of . In this case, we see that . Put and . Then satisfies the conditions (L1) and (L2) (see Proposition 2.11). Now we prove (L3). Take an -valued ultrametric space . Applying Theorem 4.1 to , , , , and of , we can take satisfying the conditions (B1)–(B5). The condition (B2) shows that is isometric. Due to the conditions (B1), (B3), and (B4), and due to the fact that , the set satisfies the assumptions in Proposition 3.4. Then, according to Proposition 3.4, we see that is algebraic over . This means that the condition (L3).
We now prove the latter part. Let be the completion of . Then is -valued (see [3, (12) in Theorem 1.6]). Take an isometric embedding stated in the former part of Theorem 4.6. Since is complete and is isometric, the set is closed in . Let be the subfield of generated by over and put . Since is algebraically independent over , we have . Thus is closed in . The condition (F6) follows from the construction. This finishes the proof. ∎
Letting , and using Proposition 2.14, we obtain the next corollary.
Corollary 4.7.
Let , with , be a cardinal, and be dividable. Put . Then there exists a valued field such that:
-
(1)
the field is a valued field extension of ;
-
(2)
the absolute value is -valued;
-
(3)
for each -valued ultrametric with , there exists an isometric embedding such that the set is algebraic independent over .
Moreover, for every -valued ultrametric space , there exist a valued field and a map such that
-
(1)
the map is an isometric embedding;
-
(2)
the field is a valued field extension of ;
-
(3)
;
-
(4)
is closed in ;
-
(5)
is algebraically independent over ;
-
(6)
if is complete, then can be chosen to be complete.
4.2. Broughan’s conjecture
Next we give an affirmative solution of Conjecture 1.1. We begin with the definition of non-Archimedean Banach algebras. Let be a field equipped with a non-Archimedean absolute value, and be a normed linear space over a field . We say that is a non-Archimedean Banach algebra over (see [26]) if the following conditions are fullfiled:
-
(1)
is a ring (not necessarily commutative and unitary);
-
(2)
is a normed linear space over ;
-
(3)
for every pair , we have and ;
-
(4)
is complete with respect to ;
-
(5)
if has a unit , then .
Some authors assume commutativity and the existence of a unit in the definition of a Banach algebra (see, for example, [24] and [17]). For instance, a valued field extension of is a (unitary) Banach algebra over .
As an application of Corollary 4.7, we obtain the next theorem.
Theorem 4.8.
Let be a prime, be a cardinal, and be a perfect field of characteristic such that the transcendental degree of over is equal to . Then every -valued ultrametric with can be isometrically embedded into . In particular, Conjecture 1.1 is true.
Proof.
The theorem follows from Corollary 4.7. Theorem 4.8 is actually an affirmative solution of Conjecture 1.1 since for every cardinal , there exists a perfect filed of characteristic such that the transcendental degree of over is equal to , and since is a non-Archimedean (commutative and unitary) Banach algebra over (see Proposition 2.14). ∎
Remark 4.1.
An solution of Conjecture 1.1 is given as the ring of Witt vectors.
5. Algebraic structures on Urysohn universal ultrametric spaces
In this section, we show that every Urysohn universal ultrametric space has a valued field structure that are extensions of a given prime valued field. Such an algebraic structure is realized as a -adic or ordinary Levi–Civita field.
For a class of metric spaces, a metric space is said to be -injective if for every pair and in and for every pair of isometric embeddings and , there exists an isometric embedding such that . We denote by (resp. for a range set ) the class of all finite metric spaces (resp. all finite -valed ultrametric spaces). There exists a complete -injetive separable complete metric space unique up to isometry, and this space is called the Urysohn universal metric space (see [25] and [15]).
Similarly, if is countable, there exists a separable complete -valued -injective ultrametric space up to isometry, and it is called the -Urysohn universal ultrametric space, which is a non-Archimedean analogue of (see [6] and [18]). Remark that if is uncountable, then every -injective ultrametric space is not separable (see [6]). Due to this phenomenon, mathematicians often consider only the case where is countable for separability.
In [11], the author discovered the concept of petaloid spaces, which can be considered to as -Urysohn universal ultrametric spaces in the case where is uncountable.
We begin with preparation of notations. A subset of is said to be semi-sporadic if there exists a strictly decreasing sequence in such that and . A subset of is said to be tenuous if it is finite or semi-sporadic (see [10]). For a range set , we denote by the set of all tenuous range subsets of . Let us recall the definition of petaloid spaces ([11]).
Definition 5.1.
Let be an uncountable range set. We say that a metric space is -petaloid if it is an -valued ultrametric space and there exists a family of subspaces of satisfying the following properties:
-
(P1)
For every , the subspace is isometric to the -Urysohn universal ultrametric space.
-
(P2)
We have .
-
(P3)
If , then .
-
(P4)
If and , then belongs to .
We call the family an -petal of , and call the -piece of the -petal .
Notice that even if is countable, the -Urysohn universal ultrametric space has a petal structure satisfying the conditions (P1)–(P4) (see [11]). This means that a petal space is a natural generalization of Urysohn universal ultrametric spaces.
Theorem 5.1.
Let be an uncountable range set. The following statements hold:
-
(1)
There exists an -petaloid ultrametric space and it is unique up to isometry.
-
(2)
The -petaloid ultrametric space is complete, non-separable, and -injective.
-
(3)
Every separable -valued ultrametric space can be isometrically embedded into the -petaloid ultrametric space.
Based on Theorem 5.1, for a range set , we denote by the -Urysohn universal ultrametric space if is countable; otherwise, the -petaloid ultrametric space. In what follows, by abuse of notation, we call the -Urysohn universal ultrametric space even if is uncountable.
We explain an example of a petaloid space. Let be a countable set. Let be a range set. We also denote by the set of all function such that and the set is tenuous. For , we define an -ultrametric on by if ; otherwise, . For more information of this construction, we refer the readers to [6] and [18].
Lemma 5.2.
For every countable set and every range set , the ultrametric space is isometric to .
Proof.
Lemma 5.3.
Let be a subset of , and . Put . Then the following statements are equivalent:
-
(1)
for every , the set is finite;
-
(2)
the set is tenuous.
Proof.
The lemma follows from [10, Lemma 2.12] and the fact that a map defined by reverses the order on . ∎
In the next lemma, in order to treat the cases of and , even if , we define by .
Lemma 5.4.
Let , , , be a field of and characteristic . Put . Fix a complete system of representatives of the residue class field . Take . Define a subset of by the set of all such that , and define an ultrametric on by . Then is isometric to .
Proof.
According to , we see that . Then, using Lemma 2.9 or (1) in Corollary 2.18, for every , there uniquely exists such that and . Put , where . We define a map by and if . Then . Thus for all . By the definition of , we see that the map defined by is bijective. Using the definitions of and , the map becomes an isometric bijection. Thus is isometric to . Due to , Lemma 5.2 finishes the proof. ∎
Theorem 5.5.
Let , , , be a field of and characteristic . Define an ultrametric on by . Then the Levi-Civita field is isometric to .
Proof.
Lemma 5.4 implies the conditions (P1). From the definition of , the conditions (P2) is true. Next we show the condition (P3). It is sufficient to verify
(5.1) |
Take and put . We also define . Then by Lemma 5.3. Hence . Thus, the inclusion (5.1) is true.
Now we show (P4). We may assume that . Put . Since , we have . Put and . Notice that . Define a point by . Then and . Then . For the sake of contradiction, suppose that . In this setting, we can take such that . Then . In particular, . Namely, we have . This is a contradiction. Thus we have , which means that the condition (P4) holds.
Therefore, we conclude that the Levi-Civita field is isometric to . ∎
We find another application of the theory of Urysohn universal ultrametric spaces to that of valued field. For a class of metric spaces, we say that a metric space is -universal or universal for if for every there exists an isometric embedding . An ultrametric space is said to be -haloed if for every and for every , there exists a subset of such that and for all distinct (see [9]).
Theorem 5.6.
Let , , be a prime, and let be a valued field such that is an infinite set and has characteristic . Put . Then is universal for all separable -valued ultrametric spaces.
Proof.
According to , we observe that is -haloed, and hence it is -injective (see [9, Theorem 1.1]). In [9, Theorem 1.5], it is stated that every complete -injective ultrametric space contains an isometric copy of . Thus contains a metric subspace isometric to . Hence is universal for all separable -valued ultrametric spaces. ∎
For a prime , the field of -adic complex numbers is defined as the completion of the algebraic closure of . The -adic valuation can be extended on . In this case, we have (see [22]).
Corollary 5.7.
Let , and be a prime. Put . The field of -adic complex numbers is universal for all separable -valued ultrametric spaces.
6. Questions
As a sophisticated version of a non-Archimedean Arens–Eells theorem for linear spaces, we ask the next question.
Question 6.1.
Let be range sets, and be a range set such that is a multiplicative subgroup of . Take a non-Archimedean valued field and assume that
-
(1)
;
-
(2)
is -valued.
For every -valued ultrametric space , do there exist a -valued ultra-normed linear space over , and an isometric map ?
Question 6.2.
Let be or a prime, be an infinite field of characteristic , and be a rang set such that is a multiplicative subgroup of . Take an arbitrary complete valued field with the residue class field such that and is a valued field extension of . For every -valued ultrametric space with , does there exist an isometric embedding ?
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