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 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 be an arbitrary field. For an arbitrary finite group we can then define the rational function field , where the are just symbols or variables functioning as transcendent elements (over ). Hence, there is an obvious *action* of on : each permutes the elements as . Noether’s Problem then asks the following question:

**Noether’s Problem**

* When is the fixed field rational (over k)? *

Where by *rational* we mean *purely transcendental*, i.e., isomorphic to a rational function field over with the same transcendency degree . Thus, there exists a generating set of algebraically independent elements such that .

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