Recall first the famous Noether's problem: let $G$ be a finite group acting linearly on a finite dimensional $F$-vector space $V$.
This induces an action on $F[V]$ and $F(V)$.
Nother's problem: is $F(V)^G/F$ purely transcendental ?
It is known for a long time that Noether's problem has a negative answer, but despite my search in book or on online documents, I was not able to find counterexamples not using heavy machineries such as ramification theory, algebraic geometry or homology theory.
So my question is:
Question (strong version). Does there exist a counterexample to Noether's problem which is not using heavy machineries ?
I am also interested in the following weak version:
Question (weak version). Can we find a finite group $G$ acting by $F$-algebra automorphism on $F(X_1,\ldots,X_n)$ such that $F(X_1,\ldots,X_n)^G/F$ is not purely transcendental, and for which the proof of the latter fact is not using heavy machineries ??
I know that the terms "not using heavy machineries" seem awfully vague, so let me give two examples to make understand what I have in mind in slightly different contexts.
Example 1. An example of $F$-linear action of $G$ on $V$ such that $F[V]^G$ is not isomorphic to a polynomial $F$-algebra.
Let $G=\mathbb{Z}/2\mathbb{Z}$ acts linearly on $F^2$ by $\bar{m}\cdot v=(-1)^m v$. Thus $F$-linear action induces an action of $F[X,Y]$ such that $\bar{1}\cdot X=-X, \bar{1}\cdot Y=-Y$.
Assuming $2\neq 0$ in $F$, easy computations show that $F[X,Y]^G=F[X^2,XY,Y^2]$, that $X^2,XY$ and $Y^2$ are pairwise non associate irreducible elements. Since $(XY)^2=X^2 Y^2=(XY)(XY)$, it shows that $F[X,Y]^G$ is not a UFD, so cannot be isomorphic to a polynomial algebra. Note however, that $F(X,Y)^G=F(X^2,XY)$ is purely transcendental over $F$, so it is not an answer to the strong version of my question.
Example 2. An example of a non-linear action of $G$ on $F(X,Y)$ such that $G$ restricts to an action on $F$, $G$ acts by $F^G$-automorphisms and $F(X,Y)^G/F^G$ is not rational.
Let $G=\mathbb{Z}/2\mathbb{Z}$ acts on $\mathbb{C}(X,Y)$ by $\bar{1}\cdot{z}=z^*$ (complex conjugation), $\bar{1}\cdot X=X,\bar{1}\cdot Y=Y^{-1}$. Then $u=\frac{Y-Y^{-1}}{2}$ and $v=i\frac{Y+Y^{-1}}{2}$ are $G$-invariants and satisfy $u^2+v^2+1=0$. Once again, easy computations show that this cannot happen in a purely transcendental extension of $\mathbb{R}$. Note that this is not an answer to the weaker version of my question, since the action of $G$ on $\mathbb{C}$ is not trivial.
I hope that these two examples will clear up a bit what I have in mind.