Problems in Modern Galois Theory¹¹1The abstract of this report was given as a lecture at the 1932 International Congress of Mathematics in Zurich.

One can separate the work which deals with the foundations of Galois theory into two types. The first of these includes the tasks which seek new paths to justify the classical Galois theory, while those of the second kind deepen the definition of the Galois group, facilitating its application to far more areas than is possible using just the classical theory.

Belonging to the first kind, the work of F. Mertens [50], S. Schatunowski [62], and A. Loewy [46, 47] is of particular note. Mertens develops the definition of the Galois group and proves the associated fundamental theorems without using the notion of the normal field [extension], the Galois resolvent, etc. He builds upon the definition of irreducibility in extended domains. Given an equation $f(x)=0$ with $x_{1}$ as a root, he finds a factor $f_{1}(x;x_{1})$ of the polynomial $\frac{f(x)}{x-x_{1}}$ that is irreducible over $K[x_{1}]$. He proceeds to find a $K[x_{1},x_{2}]$-irreducible factor of the polynomial $\frac{f_{1}(x;x_{1})}{x-x_{2}}$, where $x_{2}$ is a root of $f_{1}(x;x_{1})$. Continuing this process, he arrives at the family

of polynomials, which forms a family of fundamental relations. A polynomial $\varphi(x_{1},x_{2},...,x_{n})$ is then equal to zero (in the variables $x_{i}$) if and only if it can be represented in the form

where the $P_{i}$’s are polynomials. The Galois group is thus defined to consist of those permutations that transform each of the $Z_{i}(x_{i+1})$ into a polynomial of the form (1.2). The order of this Galois group is equal to the product of degrees of the polynomials in (1.1).

Schatunowski establishes a theory formally covering the same ideas as Mertens, yet at the same time assumes a much more general point of view. His work builds upon Kronecker’s idea of functional relations, which, given a root $x_{i}$ of the equation

consists of considering any quantity depending on $x_{i}$ as a function of an undetermined variable, while taking congruence modulo $f(x)$ as the equals sign. Since one considers functions of several roots of equation (1.3) in Galois theory, Schatunowski aims to construct a family of functional relations in the variables $x_{1},...,x_{n}$, such that the quotient according to these relations is isomorphic to the algebraic number field generated by the roots of equation (1.3). The ideal generated by

is not suited for this, since the $f(x_{i})$ are not $\equiv 0$ modulo this ideal. The ideal generated by

also fails to serve this purpose, since the statement

fails to hold, whereas $\psi_{i}V\equiv 0(\text{ mod }f(x_{1}),f(x_{2}),...,f(x_{n}))$ does hold, where $V$ denotes the Vandermonde determinant of $x_{1},...,x_{n}$. In accordance with modern convention we say that (1.4) and (1.5) are not prime ideals.
Schatunowski constructs the desired system of relations by starting with what he calls Cauchy relations. One obtains these relations in the following manner: As the first relation one takes $f(x_{1})$; as the second relation the quotient of the division of $f(x_{1})$ by $x_{1}-x_{2}$; as the third relation the quotient of the second relation, ordered by powers of $x_{2}$, by $x_{2}-x_{3}$, etc. The Cauchy relations then give rise to residue classes which in turn are isomorphic to the corresponding number field if and only if equation (1.3) is without affect. [German: affektlos] In this case each of these ideals is irreducible over the ideal preceding it. If this is not the case Schatunowski considers the general Mertensian relations. The degrees of these relations provide information regarding the transivity and primitivity relations of equation (1.3). For instance, the [Galois] group is $k$-times transitive whenever the first $k$ relations coincide with those of Cauchy.
Schatonowski’s work contains the introductory foundations of what is known as the theory of polynomial ideals nowadays. It deals with the rings whose relations are reducible. Such a ring is called a “semifield” if one can adjoin a (finite) number of new relations such that the ring becomes a field.
Paritcularly notable among terms introduced by Schatunowski is the notion of an extension of the second kind. He uses this to refer to rings that are obtained by adjoining a finite number of moduli to a base ring. In other words, one sets certain nonvanishing elements of the base ring equal to zero. It may be more natural to refer to these extensions as quotient rings, in accordance with the term quotient group. It is known that a field does not admit any extensions of the second kind. In this, it is assumed that all fields can only have characteristic zero. However, if we consider the prime characteristic $p$, this fails upon adding the new modulus $p$, and a field remains a field.
Assuming the Mertens-Schatunowskian definition of the Galois group, one can easily prove the following theorem, originally due to I. Schur ([70], see also [80]):
The [Galois] group of a factor field is a subgroup of the group of the original field.

The Loewyian justification of Galois theory has some overlap with the Mertens-Schatunowskian theory, in particular the avoiding the assumption of normal fields. Rather than taking a single primitive element, Loewy takes several algebraic elements as possible generators for a field $P$, which he refers to as conductors²²2Translated from the German “Dirigenten”.-YS. The first conductor $\rho_{1}$ is a root of an equation with coefficients in the domain of rationality; the second conductor is a root of an equation of the type $f(\rho_{1};z)=0$; the third of the type $f(\rho_{1},\rho_{2};z)=0$, etc. Provided all these equations are irreducible, Loewy uses “transformations of the field $P$” to refer to the replacement of a conductor $\rho_{i}$ with a conjugate root belonging to all equations in which $\rho_{i}$ occurs (which also necessitates the replacement of all subsequent conductors). He proves that such a transformation disturbs none of the relations between the conductors. However, it is in fact possible that such a transformation maps some elements of the field $P$ outside of $P$. The transformations that map elements $P$ to $P$ form a group, which is called the group of automorphic transformations. From a group theoretic standpoint this means the following: A non-normal field $P$ is not at all determined by its [Galois] group; it is determined by its Galois group $\mathfrak{G}$ and the subgroup $\mathfrak{H}$ corresponding to a primitive element of $P$. The automorphic group of $P$ is isomorphic to $\mathfrak{K}/\mathfrak{H}$, where $\mathfrak{K}$ denotes the normalizer of $\mathfrak{H}$ in $\mathfrak{G}$. The collection of all transformations of $P$, using the above definition, do not form a group in the familiar sense. Loewy calls these algebraic structures mixed groups and investigates their structural properties [47]. Each mixed group $\mathfrak{T}$ contains a kernel $\mathfrak{G}$ that is defined to be the largest ordinary group contained inside $\mathfrak{T}$ and consists of some of the cosets (residue classes) of $\mathfrak{G}$. It is important, that a factor group of $\mathfrak{T}$ by any (not neccesarily normal) subgroup $\mathfrak{H}$ of $\mathfrak{G}$ remains a mixed group, whose kernel is given by $\mathfrak{N}/\mathfrak{G}$, where $\mathfrak{N}$ denotes the normalizer of $\mathfrak{H}$ in $\mathfrak{G}$. These facts permit viewing a mixed group as a more adequate picture of a field.
A similar group construction was introduced by H. Brandt [7], which is called the Brandt groupoid.

Before we transition to the work of the second kind, we must set out the modern notion of the term “Galois group”, which deviates from older notions. It is difficult for me to say who this new notion originates from. The older Galois theory views the elements of the Galois group, the transformations (or permutations), as interchanges among the roots of a generating equation (which may well be reducible), which do not disturb any relations between the roots. On the other hand, the modern Galois theory considers assignments that are simultaneously undergone by all elements of a field $K$, without disturbing the established relations between them. In other words, each element of a Galois group is a map of a (normal) field $K$ to itself, or, as one says in group theory, an automorphism. That is, an assignment of all elements of a field to elements of the same field, such that sums and products and taken to sums and products respectively.

Of utmost importance in Galois theory is the mutual correspondance between subfields of $K$ and subgroups of its Galois group. More specifically one can formulate this in the following manner ([43]; [74], appendix; [3]): One assigns to each subfield $U$ of $K$ the largest subgroup $\mathfrak{H}(U)$ of $\mathfrak{G}$ which fixes all elements of $U$. On the other hand one can assign to each subgroup $\mathfrak{H}$ of $\mathfrak{G}$ the largest subfield $U(\mathfrak{H})$ of $K$, whose elements remain unchanged by $\mathfrak{H}$. If $U_{1}>U_{2}$, then $\mathfrak{H}(U_{1})<\mathfrak{H}(U_{2})$, and vice-versa. Moreover it holds that

However, one can only readily develop Galois theory, whenever

So that (1.8) holds, $K$ is required to be finite over its domain of rationality (Krull, [43]).
So that (1.7) holds, $K$ must be of the first kind over its domain of rationality (Baer-Hasse, [74], appendix).
If $K$ possesses the field of rational numbers as a subfield (one says that $K$ has characteristic zero), then $K$ is of the first kind in any case. On the other hand, if $K$ has characteristic $p$ (that is, there exists a prime number $p$, that is equal to zero inside $K$), then $K$ is of the first kind if and only if a generating element of $K$ satisfies an irreducible equation with entirely distinct roots³³3This coincides precisely with the modern notion of separability.-YS.
K is finite over its domain of rationality, if there exists a finite number of basis elements, so that each element of $K$ has a linear representation with in terms of the basis elements and coefficients from the domain of rationality.

For the case of $K$ being infinite, Krull [43] broadened the fundamental theorem of Galois theory by considering only the closed subgroups of $\mathfrak{H}$ in lieu of all subgroups. By this he means the following. Given $\gamma$ an element of $\mathfrak{G}$ and $x$ a finite normal subfield of $K$, a mapping of $x$ to itself is induced by $\gamma$. If both $\gamma$ and $\gamma^{*}$ induce the same mapping of $x$, we say that $\gamma^{*}$ is contained in an $x$-neighborhood of $\gamma$. This definition of neighborhood satisfies all the Hausdorff neighborhood axioms:

1. a)

   Each element $\gamma$ is contained in each of its own neighborhoods.

2. b)

   The intersection of two neighborhoods of $\gamma$ contains a new neighborhood of $\gamma$. Moreover, it itself is a neighborhood of $\gamma$, since the intersection of the $x_{1}-$ and $x_{2}$-neighborhoods is the $x_{3}$-neighborhood, where $x_{3}$ is both finite and normal as it is the union⁴⁴4The author uses the German word “Vereinigung”, or union, but in this case the notion of a compositum of fields seems more appropriate.-YS of $x_{1}$ and $x_{2}$.

3. c)

   Given $\delta$ an element of a $x$-neighborhood of $\gamma$, there exists a neighborhood of $\delta$ which is entirely contained inside the $x$-neighborhood of $\gamma$. Moreover, the $x$-neighborhoods of $\gamma$ and $\delta$ coincide, since they contain the largest number of elements of $\mathfrak{G}$ that act on the field $x$ in precisely the same way.

If $\gamma$ and $\delta$ are distinct elements of $\mathfrak{G}$, there exists a neighborhood of $\gamma$ not containing the element $\delta$. This is because $\gamma$ and $\delta$ being distinct implies the existence of elements in $K$ that behave differently under $\gamma$ and $\delta$. As each of these elements generates a finite field [extension], it follows that each of the fields corresponds to a different neighborhood of $\gamma$ and $\delta$.
This definition permits the definition of accumulation elements. Given $\mathfrak{H}$ a subgroup of $\mathfrak{G}$, an accumulation element of $\mathfrak{H}$ should contain elements of $\mathfrak{H}$ in each of its neighborhoods. An accumulation element of $\mathfrak{H}$ is not necessarily contained in $\mathfrak{H}$ itself. However, in the case where a subgroup $\mathfrak{H}$ of $\mathfrak{G}$ contains all of its accumulation elements, it is called closed. Each group which corresponds to a subfield of $K$ is closed. Conversely, to each closed group $\mathfrak{H}$ there is a corresponding subfield $U$ of $K$, so that $\mathfrak{H}[U(\mathfrak{H})]=\mathfrak{H}$ holds true. Thus, in order to be able to broaden the fundamental theorems of Galois theory to infinite fields [extensions], one must only consider those subgroups of $\mathfrak{H}$ which are closed.

The conditions for the presence of the relation (1.7) were investigated thoroughly by R. Baer [3]. He found that, in any case, there exists an intermediate field [extension] $S$, which he called the rigid field between $K$ and the domain of rationality. The rigid field is characterized by $K$ being orderly (that is (1.8) always holds) when one takes $S$ to be the domain of rationality, while all elements of $S$ remain invariant under automorphisms of $K$. I cannot elaborate on the additional interesting statements of this work here.

It is very difficult to define the Galois group in the case where the field in question $K$ has a higher degree of transcendence than its domain of rationality. The reason for this lies in the fact that the universal norm of such a field (that is the field, which contains all fields conjugate to subfields of $K$) is an infinite field [extension] whose definition is difficult to summarize analytically. To theoretically construct a group which possesses the desired main properties of the Galois group, one can sketch the following scheme. Let $x_{1},x_{2},...,x_{n}$ be the generating elements of a field $K$, among which certain algebraic relations may be established, which we will denote by (I). One can determine each subfield $U$ of $K$ in an analogous sense through generating elements $\xi_{1},\xi_{2},...,\xi_{m}$, where the $\xi_{i}$ are expressed rationally through the $x_{i}$:

The equations

determined a new field, whose generators $[x_{1},x_{2},...,x_{n};y_{1},y_{2},...,y_{n};y_{1}^{\prime},y_{2}^{\prime},...,y_{n}^{\prime};...]$ are related by the relations (I) and (II). This fields can be called the relative norm of $K$ (with respect to $U$). The assignment of $(x_{1},...,x_{n})$ to $(y_{1},...,y_{n})$ is designated a transmutation of $K$ or a permutation of its relative norm. If $U$ runs through all subfields of $K$, then the sought after universal norm is generated by the relative norms. The compositum of all previously constructed permutations is a group possessing all main features of the Galois group.

One can construct the Galois group of a field of algebraic functions in a slightly different manner. Instead of considering the functions of the field, one envisions the totality of their coordinate systems, which form a so-called absolute Riemann surface (see [86]). In this case each transformation of the Galois group assigns to each value of a function of $K$ a new value, such that the relations between values of different functions are preserved. As each function is determined by the totality of its values, such transformations also determine the functions to which the given functions are assigned. The various monodromy groups which do not disturb certain domains of rationality are contained in this group. It can very well occur that a transformation leads several functions of $K$ outside of $K$. In the case of an independent variable this is reflected when a function determined up to a multiplicative constant by its zeroes and poles [of the form]

is assigned to a product

where the numerator and denominator are contained in different ideal classes. Each transformation assigning divisors to equivalent divisors belongs to the so-called group of intrinsic transformations [38], which play an analogous role to the group of automorphic transformations introduced by A. Loewy [47].

Of importance is the theory of algebraic functions of a group, which is closely related to the Galois group which has just been defined.
Let $\mathfrak{u}_{1}^{\mathfrak{p},\mathfrak{p}^{\prime}},\mathfrak{u}_{2}^{\mathfrak{p},\mathfrak{p}^{\prime}},...,\mathfrak{u}_{p}^{\mathfrak{p},\mathfrak{p}^{\prime}}$ be the linearly independent Abelian integrals of the first kind defined on the Riemann surface of $K$. The Jacobi inversion problem consists in the solution of the system of equations

where the lower limits $\mathfrak{p}_{i}$ are given and the upper limits $\mathfrak{p}_{i}^{\prime}$ are sought for and the congruences are taking from the period systems as relations. When none of the points $\mathfrak{p}_{i}$ are collinear, this problem is clearly solvable ([54];[4]). If we take the $\mathfrak{p}_{i}$, $\mathfrak{p}_{i}^{\prime}$ to be coordinates of the points $P$, $P^{\prime}$ in a $p$-dimensional space, then each coordinate system of the parameters $\mathfrak{v}_{i}$ determines a transformation assigning each point $P$ to a certain point $P^{\prime}$ (in this case one must not regard points differing by the order of their coordinate values as different, so that these points may be defined unambiguously using the values of the symmetric functions of $\mathfrak{p}_{i}$). The group of these transformations is locally isomorphic⁵⁵5The author uses the term im Kleinen isomorph, where the adjective im Kleinen seems to be a weaker statement that local when used to qualify other definitions. Whether this is the same as locally isomorphic seems uncertain and as such any clarification would be greatly appreciated.-YS to the $p$-parameter group of translations and admits an analytical expression in terms of the addition formulas of the Abelian functions. It is a subgroup of an extended Galois group which transforms the coordinates of the point $P$ independently of each other. We will refer to this group as the Jacobian group in the sequel.

The problem of finding equations with prescribed [Galois] groups belongs to the most important problems of modern Galois theory and to this day has not been solved. It can be summarized in the following three ways:

1. I.

   One finds any equations, whose [Galois] groups are isomorphic to a given group $\mathfrak{G}$.

2. II.

   One finds the most general parametric form of the coefficients of an equation whose [Galois] group is isomorphic to $\mathfrak{G}$ or a subgroup of $\mathfrak{G}$. Being able to present the coefficients in this form should provide a necessary and sufficient condition for the [Galois] group of the equation to be isomorphic with $\mathfrak{G}$ or one of its subgroups.

3. III.

   One outlines a procedure for the determination of equations, whose [Galois] group is isomorphic to $\mathfrak{G}$. This procedure should produce all equations of this kind if continued sufficiently.

Problem II always admits a solution whenever the generalized Lüroth theorem (also called the rational minimal basis theorem) holds for a given group $\mathfrak{G}$. This theorem may be formulated in the following manner:
For $K_{n}(x_{1},x_{2},...,x_{n})$ the field of rational functions of the variables $x_{1},x_{2},...,x_{n}$, every subfield of $K_{n}(x_{1},x_{2},...,x_{n})$ is isomorphic to $K_{m}(x_{1},x_{2},...,x_{m})$ $(m\leq n)$. (One can also say: this field is purely transcendental)
This theorem was proven in the case $n=1$ by Lüroth [48]. The proof for $n=2$ was found by Castelnuovo [15]. Both G. Fano [21] and F. Enriques [20] found a counterexample to the case $n=3$. However, the full scope of this theorem is not required for the solving Problem II. Restricting ourselves to the case where the subfield in question contains the field of elementary symmetric functions in $x_{1},x_{2},...,x_{n}$, the validity of the theorem becomes fully equivalent to the solvability of task II. Until now the validity of this theorem “in a narrower setting” remains open (see [74], remark of B.L. Van der Waerden). If this also does not hold true in general, one can determine whether any abstractly given finite group is “Lürothian” or not. That is, whether the field $K(a_{1},a_{2},...,a_{n};\varphi)$ is purely transcendental, where $\mathfrak{G}$ is represented as the permutation group of $x_{1},x_{2},...,x_{n}$, $a_{1},a_{2},...,a_{n}$ denote the elementary symmetric functions of $x_{1},x_{2},...,x_{n}$, and $\varphi$ is a function of $x_{1},x_{2},...,x_{n}$ belonging to $\mathfrak{G}$.

E. Noether [55] deduced Problem I from Problem II by using Hilbert’s irreducibility theorem, which states for any irreducible polynomial one can choose numerical values for some of the variables so that the polynomial is irreducible in the remaining variables.
One can tighten this result by not just deducing Problem I, but also Problem III from Problem II. For that one uses the following procedure specified by M. Bauer [5],[6]. It is known that the Galois group of an algebraic equation

is transformed into one of its subgroups when one regards the values of its domain of rationality not absolutely, but modulo a prime number (or a prime ideal). That is, one replaces the domain of rationality with one of its factor rings [16][70][80]. Then again, it also known that the Galois group of a field of characteristic $p$ [German: Primzahlmodulkongruenz, or prime number modular congruence] is cyclic, and that a generating permutation of the latter group consists of cycles, whose orders are equal to the irreducible components of our congruence. As a result, if

holds, where $X_{n_{i}}^{(p)}$ denotes a modulo $p$ irreducible polynomial of degree $n_{i}$ ($n_{1}+n_{2}+...+n_{k}=n$), then the [Galois] group of (2.1) contains a permutation whose cycles have order $n_{1},n_{2},...,n_{k}$.

We assume that the field $K(a_{1},a_{2},...,a_{n};\varphi)$ is purely transcendental (see the definitions in no. 2). There thus exist rational functions $\pi_{1},\pi_{2},...,\pi_{n}$ of $a_{1},a_{2},...,a_{n};\varphi$ so that $a_{1},a_{2},...,a_{n};\varphi$ can themselves be expressed in terms of the $\pi_{i}$. Next let $\mathfrak{G}_{1},\mathfrak{G}_{2},...,\mathfrak{G}_{s}$ be a system of subgroups of $\mathfrak{G}$, so that each proper subgroup of $\mathfrak{G}$ is a subgroup of at least one of the $\mathfrak{G}_{i}$. Such a system can be certainly be constructed, for instance by taking all proper subgroups of $\mathfrak{G}$ as the $\mathfrak{G}_{i}$.
Let $\varphi_{i}$ be a function of $x_{1},x_{2},...,x_{n}$ belonging to $\mathfrak{G}_{i}$ $(i=1,2,..,s)$, and let $F_{i}(x_{i})$ be the polynomial of smallest degree, whose coefficients are rational functions of $\pi_{1},\pi_{2},...,\pi_{n}$ and for which $\varphi_{i}$ is a root ($i=1,2,...,s$). One can see that the degree of $F_{i}(z_{i})$ is equal to the index $(\mathfrak{G}:\mathfrak{G}_{i})$ ($i=1,2,...,s)$.
The [Galois] group of the equation $F_{i}(z_{i})=0$, as a transitive permutation group, contains a permutation $\overline{S_{i}}$, which fixes no indices. This permutation corresponds to at least one permutation in $\mathfrak{G}$. Let $S_{i}$ be such a permutation, and let $n_{1},n_{2},...,n_{k}$ be the orders of its cycles. Take any prime number $p_{i}>n-2$ and let

where $X_{n_{j}}^{(p_{i})}$ denotes a modulo $p_{i}$ irreducible polynomial of degree $n_{j}$. This determines the congruence classes modulo $p_{i}$ for the $a_{i}$. Using the expressions for the coefficients of the equation $F(z)=0$, and we obtain the congruence

which surely has one or more rational roots. Let $\varphi_{i}$ be one of these roots. Then $\varphi_{i}$ is transformed into the other rational roots by means of some permutations $\Sigma_{1},\Sigma_{2},...,\Sigma_{\nu}$ of the symmetric permutation group $\mathfrak{S}$ of $x_{1},...,x_{n}$.
By fixing (2.3), the “broad” class of permutations containing $p_{i}$ is determined, that is the totality of all permutations similar to $S_{i}$ within $\mathfrak{S}$. However, by fixing the congruence class of $\varphi_{i}$, we determine a division ⁶⁶6From the German Abteilung. According to Pioneers of Representation Theory: Frobenius, Burnside, Schur, and Brauer by Charles W. Curtis this is a term a term introduced by Frobenius. A division of a [Galois] group is the union of conjugacy classes of a given group element and all of the powers of this element coprime to its order.-YS. If $\mathfrak{U}$ is a division of $S_{i}$, then

are precisely those divisions of $\mathfrak{G}$ possessing the cycle type of $S_{i}$. One of these division must correspond to the division of a permutation $\overline{S}_{i}$ of our choosing. Plugging the values of $\varphi_{i}^{\Sigma_{j}}(j=1,2,...,\nu)$ into the expressions for the coefficients of $F_{i}(z_{i})\equiv 0$, then at least one of the resulting congruences $F_{i}(z_{i})\equiv 0(\text{ mod }p_{i})$ corresponds to $\overline{S}_{i}$ and thus has no rational roots. In other words $p_{i}$ belongs to a permutation class of $K(x_{1},x_{2},...,x_{n})$ which is not contained in $\mathfrak{G}_{i}$.
Taking $i=1,2,...,s$, we obtain the congruence classes modulo $P=p_{1}p_{2}...p_{s}$ for $a_{1},a_{2},...,a_{n};\varphi$ and thus for $\pi_{1},\pi_{2},...,\pi_{n}$. Fixing the $\pi_{i}$ within the congruence classes we have just determined and plugging them into the equation $f(x)=0$, we obtain precisely $\mathfrak{G}$ as the group of this equation. On the one hand, it is contained in $\mathfrak{G}$ by virtue of the parametric expressions of the $a_{i}$, but on the hand is not contained in any of the groups $\mathfrak{G}_{1},\mathfrak{G}_{2},...,\mathfrak{G}_{s}$ due to the established congruence conditions.

In particular, if one wants to construct equations without affect, one can follow a process due to M. Bauer. Take three arbitrary prime numbers $p,q,r$ ($r\geq n-2$) and impose the following three congruence conditions on the polynomial $f(x)$:

The [Galois] group of the equation $f(x)=0$ thus contains an $n$-cycle, an $(n-1)$-cycle and a transposition and thus the symmetric group [5][6][80].

The procedure outlined in No. 4 allows the construction of all possible equations with [Galois] group $\mathfrak{G}$ provided one continues the process sufficiently. This follows from the following result of Frobenius [23]:
If the [Galois] group of the equation $f(x)=0$ contains a permutation with the cycles of length $n_{1},n_{2},...,n_{k}$ ($\Sigma n_{i}=n)$, then there exist infinitely many prime numbers $p$ such that the congruence $f(x)\equiv 0\text{ (mod $p$)}$ splits into irreducible polynomials of degree $n_{1},n_{2},...,n_{k}$.
The somewhat vague term “all equations” can be made precise by giving ourselves the problem of finding all equations with [Galois] group $\mathfrak{G}$, such that their coefficients don’t exceed a certain bound. To this end one must tighten the result of Frobenius in the following sense:
One finds bounds under which a prescribed number of prime numbers with the desired properties are located.
Such bounds have been provided by L. Kronecker [42] and F. Mertens [49] in the case of arithmetic progressions. I have carried out this estimate for the problem of Frobenius [81]. The result is as follows. If

for all $d\mid f$, then the interval $(1,x)$ surely contains $V$ prime numbers belonging to the division of $S$. The constants $A_{d},g_{d},h_{d},W$ depend on certain subfields of $K$ and on the number $V$. In order to express this bound explicitly in terms of the coefficients of equation (2.1), it is neccesary to estimate certain constants of K. R. Remak [61] very recently constructed an upper and a lower bound for the regulator of a field, which is particularly important for estimating formula (2.4).

If we seek only a solution to Problem I the solvability of Lüroth’s problem is unneccesary. If $\varphi$ is a function belonging to $\mathfrak{G}$ which satisfies the equation $F(\varphi)=0$, then the coefficients of the polynomial $F(\varphi)$ are given by rational functions of the coefficients $a_{1},a_{2},...,a_{n}$ of the polynomial $f(x).$ Taking $f(x)\equiv X_{n_{1}}^{(p_{j})}X_{n_{2}}^{(p_{j})}...X_{n_{k}}^{(p_{j})}\quad\text{(mod $p_{j}$)}$, we can determine the $a_{i}$ modulo $p_{j}$. Plugging these values into the congruence $F(z)\equiv 0\quad\text{(mod $p_{j}$)}$, we obtain at least one rational root. If the congruence has multiple roots, we choose the one that corresponds to the desired division. Letting $j$ run through $1,2,...,s$, we can determine $a_{1},a_{2},...,a_{n};\varphi$ moduli $p_{1},p_{2},...,p_{s}$ and thus modulo $P=p_{1}p_{2}...p_{s}$, and can thus be represented in the form $a_{i}=a_{i}^{(0)}+Pt_{i}$, $\varphi=\varpi_{0}+Pu$, where $a_{1}^{(0)},a_{2}^{(0)},...,a_{n}^{(0)};\varphi_{0}$ denote constants. Substituting this into the equation $F(\varphi)=0$, we obtain the Diophantine equation $\Phi(t_{1},t_{2},...,t_{n};u)=0$. The Problem now follows from the solution of this equation. Note that this equation possesses the following properties:

1. i)

   It can always be solved using fractions. That is, one can replace the $a_{i}$ with the elementary symmetric functions of the $n$ arbitrary rational numbers. This means the solution has $n$ “degrees of freedom.”

2. ii)

   It can always be solved using $p$-adic numbers, where the prime number $p$ can be chosen arbitrarily.

The solvability of the equation $\Phi=0$ does not depend on the choice of function belonging to $\mathfrak{G}$.

Much work has been devoted to the solution of the problems in question in the cases of a few special groups. In the first place we have the work of D. Hilbert [35], in which Problem I is solved for symmetric and alternating groups of any order using the irreducibility theorem. The genuine construction of equations with alternating groups was recently carried out to near completion by I. Schur [72]. That is, given $n\equiv 0\text{mod $4$}$, Schur shows that the equation

possesses the alternating group as its Galois group. On the other hand he shows (Crelle: 165, 1931) that when $n\equiv 1\text{(mod) 2}$, the equation

also has the alternating group as its Galois group.

In the solution of Problem II it is important to solve Lüroth’s problem using rational functions with rational coefficients. This question in the case of solvable groups of prime degree is addressed in the work of S. Breuer [10][11] and Ph. Furtwängler [27]. Furtwängler proposed the following sufficient condition in the case of cyclic groups of prime degree $p$:

The problem permits a solution whenever it is possible to construct an integer number system $e_{0},e_{1},...,e_{p-2}$, such that $1)$ the Hankel determinant is

and $2)$ the congruences $\sum\limits_{i=0}^{p-2}e_{i}g^{i}\equiv 0\text{(mod $p$)}$ hold, where $g$ is a primitive root of $p$.
This criterion is not always satisfied, for instance in the case of $p=47$. Later, Furtwängler gives some general instructions for the construction of rational minimal bases of metacyclic groups.
Breuer derived several similar criteria by using his theorem about the decomposition of a field of rational functions of $n$ variables into two subfields, where one depends on the full metacyclic functions of the generating variables.

Problems I and III have been extended to the setting of relative fields. First of all, the solution of problem III can be regarded as solved for relative fields when a known rational basis contains certain roots of unity in the coefficients rather than being rational-valued (see for instance E. Fischer, [22]). This is however not an extension of the problem. A solution to problems I and III can only be considered a satisfactory extension to relative fields if we are capable of also saying something about the absolute Galois group.
One can call the work of O. Ore [57] and H. Hasse [31][30] as the first in this direction, even though they are not directly concerned with this problem. Hasse starts with a number field $k$ and a number of its prime ideals $\mathfrak{p}_{i}$. He then finds infinitely many extension fields $K$, in which the $\mathfrak{p}_{i}$ decompose into prime ideals of prescribed order and multiplicity. He then significantly tightens this result by taking the prescribed decompositions to be regular and requires $K/k$ to be relatively abelian. He provides the existence proof under certain restrictive conditions.

The general form of problem I can be formulated for relative fields in the following sense (see [83]):
Problem A. Let an algebraic number field $k$ be given, whose [Galois] group [over $\mathbb{Q}$] is given by $\mathfrak{g}$. Furthermore, let an abstract group $\mathfrak{G}$ and a normal subgroup $\mathfrak{H}$ be given, such that the factor group $\mathfrak{G}/\mathfrak{H}$ is isomorphic to $\mathfrak{g}$. One must find neccesary and sufficient conditions such that there exists an extension field $K$ of $k$ whose [Galois] group [over $\mathbb{Q}$] is isomorphic to $\mathfrak{G}$.
The following example shows that this problem is not solvable in a few cases. Let $k$ be cyclic of prime degree $l$. Let $\mathfrak{G}$ be cyclic of order $l^{2}$. The critical prime numbers of $k$ are clearly $\equiv 1(\text{mod }l)$, but not $\equiv 1(\text{mod }l^{2})$. It follows from the number theoretic monodromy theorem that $K$ contains at least one inertia group of order $l^{2}$ (see [82]). The prime number $p$ corresponding to this inertia group must also be critical in $k$, which is impossible since it is known that it satisfies the congruence $p\equiv 1\quad(\text{mod }l^{2})$.
This example also shows that the solvability of problem A is not just determined by the structure of the group $\mathfrak{G}$ but also via certain arithmetic properties of the field $k$.

Especially important results regarding solving problem A concerning abelian groups were obtained by A. Scholz [64][63]. His investigations are mostly concerned with two-step solvable groups (that is, groups whose commutator subgroups are abelian) and are far-reaching in two directions. First of all he established a very utilitarian classification of two-tiered groups. The easiest of his classes, which he called disposition groups, allows for a solution to problem A independently of the arithmetic properties of the field $k$ [64]. One can define the disposition group as a group $\mathfrak{G}$, whose normal abelian subgroups $\mathfrak{H}$ are the direct product of all cyclic groups that that are conjugated by them. In addition each of these cyclic groups is required to have exactly $\mathfrak{H}$ as its normalizer.
Scholz proved the following two theorems on disposition groups:

1. i)

   A disposition group $\mathfrak{G}$ is fully determined as soon as one knows the group $\mathfrak{G}/\mathfrak{H}$ and the order of a generating element of $\mathfrak{H}$.

2. ii)

   Given an algebraic number field with [Galois] group $\mathfrak{G}/\mathfrak{H}$, one can always find a extension field $K$, whose [Galois] group is isomorphic to $\mathfrak{G}$.

Scholz later showed that the first of these theorems can be expanded to the case where neither $\mathfrak{G}/\mathfrak{H}$ nor $\mathfrak{H}$ are required to be abelian, but are completely arbitrary.
Secondly, Scholz investigated other types of two-tiered groups extensively [63]. He namely found two maximal types among all two-tiered groups (that is those types, such that every two-tiered group can be represented as a factor group or a product of groups of this type): ring groups and branch groups. Ring groups are always factor groups of certain powers of disposition groups. Branch groups on the other hand possess a feature that does not allow such a representation: their commutator subgroup is not properly contained in any abelian subgroup. There the question of whether relative fields with [Galois] branch groups exist remains open.

I have made a contribution towards problem A [83]. In this I somewhat broadened the definition of the disposition group by throwing out the requirement that each generating cyclic subgroup of $\mathfrak{H}$ permits no normalizer besides $\mathfrak{G}$. From this arise the so-called Scholz groups, corresponding to the purely branched fields (that is those containing no unbranched extension fields of $k$) with relative discriminants, whose prime ideal divisors within $k$ are not critical. The question of the uniqueness of a Scholz group given the factor group $\mathfrak{G}/\mathfrak{H}$, the order of a generating element of $\mathfrak{H}$ and its normalizer, remains open. The question of the existence of a relative field with a given Scholz group can be reduced to to the existence of principal prime ideals $\mathfrak{p}$ with prescribed values of a Hasse norm remainder symbol$\left(\frac{p,K}{\mathfrak{p}}\right)$ (see Hasse [32] III). However this question exceeds the scope of the currently known analytical ideal theory.

If the group $\mathfrak{G}$ does not belong to the type of Scholz group, then one can only expect the existence of a corresponding extension field when it contains an absolute Teilklassenkörper or its relatively critical prime ideals are also critical within $k$. In this case we are subject to the full control of the individual oddities of the field $k$.
The [Galois] groups the of absolute class fields are also not allowed to be completely arbitrary. On the one hand they are restricted by the aforementioned monodromy theorem, due to which all interia groups generate the full Galois group of the field, while the interia groups of a relatively unbranched field induce an unambiguous mapping on the inertia groups of the base field $k$. Operating on this principle, I carried out a classification of the possible types of [Galois] groups of absolute class fields [82]. On the other hand, F. Pollaczek [60] and Scholz [66] discovered and developed a restriction of the absolute class fields based on the properties of the Grundeinheitensystem.

It is known that an algebraic number field is not fully determined by its Galois group. The known invariants that fully determine a field are the so-called Artin symbols $\left(\frac{K}{\mathfrak{p}}\right)$ ([32] III, p. 6), that is the permutation classes containing individual prime numbers. There are infinitely many such invariants, which implies that they cannot be independent from one another. The existing relations between them can be derived provided we know the analytical form in which we can represent the prime numbers belonging to certain permutation classes, that is corresponding to the same value $\left(\frac{K}{\mathfrak{p}}\right)$. It can be proven, that such analytical forms depend on the structural properties of the corresponding [Galois] groups $\mathfrak{G}$. This connection yields a deep insight into the arithmetic structure of a group with known Galois group.

First I will recall the classical case of an abelian field. For a prime number to belong to a given permutation (in this case each permutation class consists of only one permutation), it must be representable in the form of one of the arithmetic progressions $ax+b$ unambiguously corresponding to the element (permutation) of the Galois group in question. The number $a$ is one and the same for all permutations and consists of the prime numbers that come up in the driscriminant of the group, while the $b$ are assigned to the different elements (permutations) of the Galois group.

The other, now classic, case is that of complex multiplication of elliptic functions. Given a number field that is relatively abelian over an imaginary quadratic field, then each prime number belonging to a given permutation class can be represented by one or more quadratic forms, whose discriminants depend on the field (more precisely: equal to the discriminant of the imaginary quadratic field) and whose classes are assigned to the permutation classes.

This fact was suspected by Kronecker (“Jugendtraum”, or “youthhood dream”) and proven by R. Fueter [24][25]. The principles that it is based on follow from the general class field theory. Consider only those numbers of a field $k$ which satisfy certain congruences modulo an ideal $\mathfrak{f}$ (which one calls leader⁷⁷7Führer in the original German.-YS. Taking these numbers as principal ideals, and letting $h$ equal the number of classes in the newly defined sense, it follows that there exists a relatively abelian field $K$ of relative degree $h$. This field, a so-called class field, has the property a prime ideal of $k$ decomposes fully in $K$ if and only if it lies in the principal class (Ph. Furtwängler, [26]; T. Takagi, [76]). Conversely we can regard any relatively abelian field over $k$ as a class field with a suitably chosen leader (Fueter, [24]; Takagi, [76]; Hasse, [32]).

We can ask ourselves the general question concerning the analytic form of prime numbers belonging to classes generating by powers of a permutation $S$. Let a normal algebraic number field $K$ be given, and denote its [Galois] group by $\mathfrak{G}$. Furthermore let $\mathcal{S}$ be a permutation in $\mathfrak{G}$. For a prime number $p$ to belong to the classes generated by powers of $\mathcal{S}$, it is necessary and sufficient for the subfield $K_{\mathcal{S}}$ of $K$ belonging to $\mathfrak{Z}_{\mathcal{S}}$ to contain a prime ideal divisor $\mathfrak{p}$ of $p$ of the first degree, where $\mathfrak{Z}_{\mathcal{S}}$ denotes the cyclic subgroup of $\mathfrak{G}$ generating by the powers of $\mathcal{S}$.
Let $\mathfrak{a}_{1},\mathfrak{a}_{2},...,\mathfrak{a}_{k}$ be a system of representatives of all distinct ideal classes of $K_{\mathcal{S}}$, and let $(\mu_{1}^{(i)},\mu_{2}^{(i)},...,\mu_{n}^{(i)})$ be respective bases of the ideals $\mathfrak{a}_{i}(i=1,2,...,h)$. Then we have that $N(\mu_{1}^{(i)}x_{1}+\mu_{2}^{(i)z}x_{2}+...+\mu_{n}^{(i)}x_{n})$ is the form (admitting a decomposition) of $n$-th degree in the variables $x_{1},x_{2},...,x_{n}$, whose coefficients have the number $N(\mathfrak{a}_{i})$ as their greatest common divisor. The quotient

is thus a primitive form of $n$-th degree. If we allow the $x_{i}$ to take on any integer values, then the norm $f_{i}(x_{1},x_{2},...,x_{n})$ takes on the values of the norms of all ideals, whose classes are opposite to the class of $\mathfrak{a}_{i}$. Since if $b$ lies in the opposite class to that of $\mathfrak{a}_{i}$, it follows that $b\mathfrak{a}_{i}$ is a principal ideal, which may be associated with a number $\mu_{1}^{(i)}x_{1}+\mu_{2}^{(i)}x_{2}+...+\mu_{n}^{(i)}x_{n}$ from the ideal $\mathfrak{a}_{i}$. Thus, it holds that

Given $\mathfrak{p}$ a prime ideal of $K_{s}$ of first degree, it holds thus that $N(\mathfrak{p})=p$, one wishes to find the representative $\mathfrak{a}_{i}$, whose class is opposite to the class of $\mathfrak{p}$. Thus $\mathfrak{p}$ can be represented in the form $f_{i}(x_{1},x_{2},...,x_{n})$. Conversely, if $\mathfrak{p}$ can be represented in the form $f_{i}(x_{1},x_{2},...,x_{n})$, then $N(\mathfrak{a}_{i})p$ can be represented in the form $N(\mu_{1}^{(i)}x_{1}+\mu_{2}^{(i)}x_{2}+...+\mu_{n}^{(i)}x_{n})$. The number $\mu_{1}^{(i)}x_{1}+\mu_{2}^{(i)}x_{2}+...+\mu_{n}^{(i)}x_{n}$ is divisible by $\mathfrak{a}_{i}$, and the norm of the quotient ideal $\frac{\mu_{1}^{(i)}x_{1}+\mu_{2}^{(i)}x_{2}+...+\mu_{n}^{(i)}x_{n}}{\mathfrak{a}_{i}}$ is equal to $p$. It follows from this that there exists an ideal with norm $p$. This ideal must be a prime ideal of degree 1.

To set up the conditions for a prime number $p$ to be a member of the division of $\mathcal{S}$, we must exclude its membership to the powers $\mathcal{S}^{k}$ of $\mathcal{S}$, whose exponents $k$ are not relatively prime to the order $f$ of $\mathcal{S}$. For this to occur $p$ in $K_{\mathcal{S}}$ must contain at least one prime ideal of first degree, while this does not happen for any $K_{\mathcal{S}^{k}}$, $(k,f)\neq 1$. Given

the corresponding forms for the fields $K_{\mathcal{S}^{k}}$, constructed in the same way as $f_{i}(x_{1},x_{2},...,x_{n})$, we have that $p$ belongs to the division of $\mathcal{S}$ if and only if it admits a representation in terms of one of the forms (3.1), but none in terms of the forms (3.2).

How can one characterize the membership of a prime number $p$ to the class of $\mathcal{S}$? I can do this only if I know a number $a$ such that $p^{f}\equiv 1\text{ (mod $a$)}$, but that this congruence fails for any power of $p$ whose exponent is less than $f$. Using this to construct the subfield $K(\eta)$ of the $a$-th roots of unity, it follows that $p$ remains irreducible in $K(\eta)$. If $p\equiv b\text{ (mod $a$)}$, then $\eta^{p}\equiv\eta^{b}\text{ (mod $p$)}$ holds. Now we form the element

where $\omega$ denotes an element of $K$, and $\Phi(\xi)=0$ is an equation satisfied by $\xi$. If one takes as established that $p$ belongs to the division of $\mathcal{S}$, then $p$ belongs to the class of $\mathcal{S}$ if and only if the congruence $\Phi(\xi)\equiv 0\text{ (mod $p$)}$ has rational roots. That is, if $p$ has at least one prime ideal divisor of first degree in $K(\xi)$. One can hence set up a system of forms such that $p$ can be represented by it if and only if $p$ belongs to the class of $\mathcal{S}$. One can certainly find the corresponding number $a$ for each $p$; there are however different forms corresponding to different $p$. I can therefore not set up a finite number of forms that are valid for all prime numbers.

The classically regarded criteria in No. 2 and 3 do not follow from this criterion. To set up a more general criterion we regard the case where $\mathfrak{G}$ possesses an abelian normal subgroup $\mathfrak{H}$. Then one can understand $K$ as a relatively abelian field with respect to the field $k$ corresponding to $\mathfrak{H}$. $K$ is thus a class field of $k$. By the general reciprocity law of E. Artin [2] there is a one-to-one correspondence between the permutations in $\mathfrak{H}$ and the ideal classes (more specifically: the residue classes of a certain ideal class subgroup) of $k$, which has the character of an isomorphism. The forms (3.1) corresponding to $k$ decompose thusly in the system of forms $\mathfrak{B}_{i}$, of which each represents one of the aforementioned residue classes. The number of these systems of forms is equal to the order of $\mathfrak{H}$. The Artin reciprocity law says that a prime number $p$ can be represented by one of the forms in the system $\mathfrak{B}_{i}$ if and only if it belongs to the class of $S_{i}$, where $S_{i}$ is one of the permutations in $\mathfrak{H}$ corresponding to the system $\mathfrak{B}_{i}$. This system of forms has the advantage that the degree of its forms in generally smaller. For instance, if $K$ is absolutely abelian, then $k$ is the rational field, so that the norms coincide with the numbers themselves. The partitioning of numbers of the rational field in the “narrower” sense is nothing else but their distribution among the congruence classes by a certain relation, which one calls leader. If $k$ is quadratic, we arrive at the quadratic forms, in complete concurrence with the general theory.

We still note that the forms corresponding to the field $k$ allow for the so-called form composition. For instance, given

then it follows that

On the other hand, if $\mathfrak{a}\mathfrak{b}$ lies in the opposing ideal class to $\mathfrak{a}_{3}$, then $N(\mathfrak{a}\mathfrak{b})=f_{3}(z_{1},z_{2},...,z_{n})$, where $z_{1},z_{2},...,z_{n}$ are certain integer-valued bilinear expressions in $x_{1},x_{2},...,x_{n}$ and $y_{1},y_{2},...,y_{n}$. One can obtain these by considering the bilinear expression $\sum\limits_{i,j}\mu_{i}^{(1)}\mu_{j}^{(2)}x_{i}x_{j}$, where $(\mu_{1}^{(1)},\mu_{2}^{(1)},...,\mu_{n}^{(1)})$, $(\mu_{1}^{(2)},\mu_{2}^{(2)},...,\mu_{n}^{(2)})$ denote the bases of the ideals $\mathfrak{a}_{1},\mathfrak{b_{2}}$ respectively. By expressing the $\mu_{i}^{(1)}\mu_{j}^{(2)}$ using a basis $(\mu_{1}^{(3)},\mu_{2}^{(3)},...,\mu_{n}^{(3)})$ of the ideal $\mathfrak{a}_{1}\mathfrak{b}_{2}$: $\mu_{i}^{(1)}\mu_{j}^{(2)}=\sum\limits_{s}c^{s}_{ij}\mu_{s}^{(3)}$, i.e. $\mathfrak{a}_{1}\mathfrak{a}_{2}\mathfrak{a}\mathfrak{b}=\sum\limits_{i,j,s}c_{ij}^{s}x_{i}x_{j}\mu_{s}^{(3)}$, one then sets the $z_{s}$ equal to the coefficients of $\mu_{s}^{(3)}$: $z_{s}=\sum\limits_{i,j}c_{ij}^{s}x_{i}x_{j}$. It is easy to understand that this composition of forms corresponds to the multiplication of their corresponding forms.

To obtain a simplest analytic expression of prime numbers belonging to different permutation classes of a field we must find maximal abelian normal subgroups of its [Galois] group $\mathfrak{G}$. We note that a permutation $\mathcal{S}$ in $\mathfrak{G}$ can be contained in a abelian normal subgroup of $\mathfrak{G}$ if and only if its class is abelian, i.e. all permutations in its class commute with each other. If $\mathfrak{C}_{1},\mathfrak{C}_{2},...,\mathfrak{C}_{k}$ are all the abelian classes in $\mathfrak{G}$, we have that each abelian normal subgroup consists of those permutations in the classes which commute with one another. For instance, given $\mathfrak{C}_{1}$ and $\mathfrak{C}_{2}$ commuting, then the join, that is the smallest group containing $\mathfrak{C}_{1}$ and $\mathfrak{C}_{2}$, is an abelian normal subgroup of $\mathfrak{G}$. The uniqueness of a maximal abelian normal subgroup cannot be guaranteed since the commutativity of classes is not a transitive property.
The following example shows that cases exist where $\mathfrak{G}$ contains multiple different maximal abelian subgroups. Define $\mathfrak{G}$ using 3 generating elements $s_{1},s_{2},s_{3}$, where are subject to the following relations:

($p$ is a prime number). Both subgroups $(s_{1},s_{2})$ and $(s_{1},s_{3})$ are abelian normal subgroups of $\mathfrak{G}$. Both are maximal since they are only properly contained in $\mathfrak{G}$, which is not abelian. On the other hand, they are distinct from one another.

As an example we will consider the general cubic number field $K$. Its alternating group possesses a quadratic subfield $k$, and by using Fueter-Takagi theory one can regard $K$ as a ring class field of $k$. Corresponding to the ring classes of $k$ in consideration we have a system of binary quadratic forms, which decompose into the 3 subsystems $\mathfrak{B}_{1},\mathfrak{B}_{2},\mathfrak{B}_{3}$, to which one can associate each of the three permutations in $\mathfrak{H}$. Given $D$ the discriminant of the system of forms, we have that $\left(\frac{D}{p}\right)=\pm 1$ is the necessary and sufficient condition for $p$ to admit a representation through one of these forms. The principal system is the system from $\mathfrak{B}_{1},\mathfrak{B}_{2},\mathfrak{B}_{3}$, say $\mathfrak{B}_{1}$, which satisfies the property $\mathfrak{B}_{1}\mathfrak{B}_{1}=\mathfrak{B}_{1}$. Then all prime numbers relatively prime to the discriminant of $K$ belong to the following three types:

1. i)

   $\left(\frac{D}{p}\right)=-1$, $p$ does not belong to $\mathfrak{H}$, hence belonging to one of the transpositions.

2. ii)

   $\left(\frac{D}{p}\right)=+1$, $p$ can be represented through one of the forms $\mathfrak{B}_{2},\mathfrak{B}_{3}$. $p$ belongs to one of the 3-cycles.

3. iii)

   $\left(\frac{D}{p}\right)=+1$, $p$ can be represented through one of the forms $\mathfrak{B}_{1}$. $p$ belongs to one of the identity permutation.

The cubic field was investigated in this respect by Dedekind [17], Voronoi [85] and Takagi. More recently Hasse [33] investigated the cubic field from a class field theoretic perspective, by also considering the critical prime ideals. This work was kindly brought to my attention by B. Delaunay.

I want to mention a very elegant procedure due to A. Speiser [73] which serves the purpose of determining the order $\mathfrak{f}$ of a permutation whose class contains a prime number $p$. I allow myself to modify the proof somewhat. Given

the equation to be examined, we consider the difference equation

We know its general solution to be given by

where $\alpha_{1},\alpha_{2},...,\alpha_{n}$ are the roots of equation (3.3) and $C_{1},C_{2},...,C_{n}$ are arbitrary constants. Setting $y(1)=y(2)=y(n-1)=0,y(n)=1$, we have $C_{i}=\frac{1}{f^{\prime}(\alpha_{i})}$. Now take $GF[p^{\mathfrak{f}}]$ as a base field. If $u$ is the smallest (integer-valued) period of the function $y(m)$ modulo $p$, it follows that $\alpha_{i}^{u}\equiv 1\text{ (mod $p$) ($i=1,2,...,n$)}$, and vice-versa. This is true since if $y(u)\equiv y(u+1)\equiv....\equiv y(u+n-1)\equiv 0\text{ (mod $p$)}$, $y(u+n)\equiv 1\text{ (mod $p$)}$, it follows that $C_{i}\equiv\frac{1}{\alpha_{i}^{u}f^{\prime}(\alpha_{i})}\text{ (mod $p$)}$, i.e. $a_{i}^{u}\equiv 1\text{ (mod $p$)}$. However, since $(\alpha_{i}\to\alpha_{i}^{p})$ is a generating permutation for the [Galois] group of the congruence $f(x)\equiv 0\text{ (mod $p$)}$ and its order is hence equal to the smallest number $\mathfrak{f}$, for which $a_{i}^{p^{\mathfrak{f}}}\equiv\alpha_{i}\text{ (mod $p$)}$ holds, it follows that $\mathfrak{f}$ is the smallest exponent for which $p^{\mathfrak{f}}\equiv 1\text{ (mod $u$)}$ holds.