Noether’s Problem – 1. Introduction

This blog post is meant to be the introduction of a series of posts about Noether’s Problem, which, roughly put, asks to study the subfield of a rational function field invariant under some automorphisms (namely to determine its rationality). Since there are quite a number of variants of Noether’s Problem, and also different methods of studying them, it seems necessary to give an outline of what I hope to achieve in these posts.

I studied Noether’s Problem as part of a project (for a course) in which my main goal was to try to understand the ideas behind results as Swan’s [Swa69] (counter)example and Lenstra’s [Len74] solution for (finite) abelian groups (in some special cases). Besides understanding these ideas and results, I also wanted to present them in the most straightforward way possible. This will also be the goal of this series of posts: giving an accessible introduction to studying Noether’s Problem for (finite) cyclic groups over {{\mathbb Q}} where the results of Swan and Lenstra will be the main focus.

The prerequisites needed to be able to follow these posts should be relatively minimal. Basic group theory, Galois theory and some relatively basic facts about modules should essentially be all that is needed (which is almost all covered in the two algebra courses from my mathematics education up until now). Where I feel it might be useful and/or necessary (also for myself), I will expand on some of the more theoretic concepts or try to provide some references.

1. The Problem

Let {k} be an arbitrary field. For an arbitrary finite group {G} we can then define the rational function field {k(x_g\>|\> g\in G)}, where the {x_g} are just symbols or variables functioning as transcendent elements (over {k}). Hence, there is an obvious action of {G} on {k(x_g\>|\> g\in G)}: each {h\in G} permutes the elements {x_g} as {h(x_g)=x_{gh}}. Noether’s Problem then asks the following question:

Noether’s Problem

When is the fixed field {k(G):=k(x_g\>|\> g\in G)^G} rational (over k)?

Where by rational we mean purely transcendental, i.e., isomorphic to a rational function field over {k} with the same transcendency degree {|G|}. Thus, there exists a generating set {\mathcal{B}} of algebraically independent elements such that {k(\mathcal{B})=k(x_g\>|\> g\in G)}.

The origin of the name of Noether’s Problem is more commonly traced back to a paper from Noether [Noe17] from 1917. We shall not go into the history of the problem (more can be found in [JLY02]) but we should probably remark on its connection with the still open inverse Galois problem. Namely, the rationality of {{\mathbb Q}(G)} implies a positive answer to the inverse Galois problem for the group {G} (see [JLY02] again). Not surprisingly then, solving Noether’s Problem was once considered as a possible method for solving the inverse Galois problem. In order not to get lost in the many variants of Noether’s Problem (of which I am not very knowledgeable), I will already set the context for the series:

Assumptions about {k} and {G}

In all that follows we will assume that the field {k} is of characteristic 0 and {G} is a finite cyclic group (as such, we will mostly omit “finite”).

Let us then first consider an important example as an introduction to the problem.

Example 1 ({{\mathbb Q}({\mathbf C}_{2})} is rational)

Let {k={\mathbb Q}} and {G={\mathbf C}_2} (the cyclic group of order two). Then the rational function field {k(x_g\>|\> g\in G)} becomes {{\mathbb Q}(x_1, x_2)}, with the action of {G} corresponding to the transposition {x_1\leftrightarrow x_2} (where {x_1} and {x_2} are transcendental over {{\mathbb Q}}). Noether’s Problem asks whether the fixed field {{\mathbb Q}({\mathbf C}_2)} of this rational function field under the action of {G} is rational.

For example, {x_1x_2} is clearly fixed under this transposition. As well as {\frac{x_1x_2}{x_1+x_2}}. However, since {G={\mathbf C}_2=\mathbf{S}_2}, we can apply the fundamental theorem of symmetric polynomials (which has accessible proofs, mostly by induction) that says that {R[x_1,\ldots,x_n]^{\mathbf{S}_n}=R[e_1,\ldots,e_n]}, where {R} is a commutative ring, {\mathbf{S}_n} is the symmetric group on {n} letters and {e_i} are the elementary symmetric polynomials (in the {n} variables {x_1,\ldots,x_n}). Thus, {{\mathbb Q}({\mathbf C}_2)={\mathbb Q}(x_1x_2,x_1+x_2)} and the fixed field is rational.

After seeing this example, there are two questions that naturally arise (at least for me):

  1. Can we exploit the fundamental theorem of symmetric polynomials further to give an answer to Noether’s Problem for other (classes of) cyclic groups?
  2. If not, how can we, in general, determine the (non-)rationality of certain fields?

As far as I know, the answer to question 1 seems to be “no”. This might make more sense if we just try another (seemingly) easy example. The next obvious group to look at is {G={\mathbf C}_3}, which seems like an easy step up from {G={\mathbf C}_2} in Example 1. The corresponding problem is then to determine the rationality of {{\mathbb Q}({\mathbf C}_3)}, the fixed field of {{\mathbb Q}(x_1,x_2,x_3)} under the action of {G=\langle \sigma\rangle}, with {\sigma:x_1\mapsto x_2\mapsto x_3\mapsto x_1}. Again we can see that {x_1+x_2+x_3}, {x_1x_2x_3}, etc., are elements of {{\mathbb Q}({\mathbf C}_3)}. We could try to remember some results from Galois theory to help us out. For example, a basic property of a finite Galois extension {E/F} is that the roots of the minimal polynomial of an element {a} in {E} over {F}, are the conjugates of {a} under the Galois group {\mathrm{Gal}(E/F)} of the extension. Hence, constructing the minimal polynomial of an element in {E}, will give us coefficients in {F}. Here {x_1}, {x_2} and {x_3} are all roots of the same minimal polynomial. Hence (x-x_1)(x-x_2)(x-x_3) =x^3 - (x_1 + x_2 + x_3)x^2 + x(x_1x_2+x_1x_3+ x_2x_3) - x_1x_2x_3 gives us {\{x_1 + x_2 + x_3, x_1x_2+x_1x_3 + x_2x_3, x_1x_2x_3\}} as elements of {{\mathbb Q}({\mathbf C}_3)}. But these are just the elementary symmetric polynomials in three variables! Thus we are back to square one. Although this was a fairly weak attempt, we will not try to develop these ideas any further.

Question 2 really asks a more difficult question than Noether’s Problem itself and it is one I struggled with myself initially. In particular, what is an “easy” (yet nontrivial) example of a non-rational subfield of a rational function field? (There is not really one, see for example this mathoverflow.net thread.) As far as I know, there is also not really something we could call a “general strategy” and thus we might feel kind of “lost”. Hopefully, the next post will address those feelings and make the problem feel somewhat more accessible.

When we restricted ourselves to fields of characteristic zero and cyclic groups, one might have thought (or hoped) that this would reduce the problem to something relatively simple (where an elementary approach might yield some promising results). This, however, does not seem to be the case. We shall give a rough outline of the history of the problem (relevant to our case of the problem) on which this series will be loosely based.

Relevant history of Noether’s Problem

In 1915 Fischer [Fi15] proved that {k(G)} is rational when {k} contains “enough” roots of unity (we will state his more general result in the following part). But it was not until 1955 that Masuda [Mas55] proved a theorem which allowed him to prove the rationality of {{\mathbb Q}({\mathbf C}_n)} for {n\>{\leqslant}\> 7}. In it he used Fischer’s theorem and introduced the idea of descent, central in our strategy for attacking the problem. Counterexamples seemed hard to find, Swan [Swa69] gave the first important example of a non-rational fixed field in 1969: {{\mathbb Q}({\mathbf C}_{47})} is not rational. Finally, in 1974 Lenstra [Len74] (other strong results had also been proved by Endo, Miyata and Voskresenskii) proved that for abelian {G} and arbitrary {k} the rationality is equivalent with two concrete conditions: 1. some well defined ideals have to be principal 2. some Galois extension, dependent only on the exponent of the group, has to be cyclic. The latter condition will be rather trivial to check; the former, however, certainly will not.

To make all of this slightly more concrete, the table below gives the results known until now (to the best of my knowledge) of Noether’s Problem for the case of {k={\mathbb Q}} and {G} a cyclic group of prime order, {G={\mathbf C}_{p}}. The biggest prime for which we know that {{\mathbb Q}({\mathbf C}_{p})} is rational, is {p=71}; for most of the primes {p} greater than 100, we do not know (in general, the higher the prime, the more difficult). In a way, the objective of these posts is to be able to color this table (at least as shown below).

Known results of Noether's Problem.

Green: {{\mathbb Q}({\mathbf C}_{p})} is rational. Red: {{\mathbb Q}({\mathbf C}_{p})} is not rational. Others: not prime or unknown. The squares with a black border are from own calculations, those without are from the tables of Endo and Miyata [EM71].

In the next post we will prove Fischer’s theorem since its proof will be of significant importance for our general solution.



[EM71] Endo, S.; Miyata, T.; Invariants of Finite Abelian Groups, 1973.
[Fi15] Fischer, E.; Die Isomorphie der Invariantenkorper der endlichen Abelschen Gruppen linearen transformationen, 1915.
[JLY02] Jensen, Ledet, Yui; Generic Polynomials: Constructive Aspects of the Inverse Galois Problem, 2002.
[Len74] Lenstra, H. W., Jr.; Rational Functions Invariant under a Finite Abelian Group, 1974.
[Mas55] Masuda, K.; On a Problem of Chevally, 1955.
[Noe17] Noether, E.; Gleichungen mit vorgeschriebener Gruppe, 1917.
[Sw69] Swan, Richard G.; Invariant Rational Functions and a Problem of Steenrod, 1969.

1 Comment

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One response to “Noether’s Problem – 1. Introduction

  1. Pingback: Noether’s Problem – 2. Fischer | MathBB

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