I've been trying to determine the rationality of certain fields of invariants coming from G-lattices. More precisely, letting $G$ be a finite group, $L=\mathbb{Z}^n$ a free abelian group with a $G$ action, we set $F[L]$ be the group algebra (Laurent polynomials in n indeterminates) and $F(L)$ to be its field of fraction and then ask whether $F(L)^G$ is rational over $F^G$.
If $L$ is a permutation lattice then this is the usual Noether probelm. In my work I came across a family of lattices which are usually not permutation but are very close to being permutation. My question is if there is some way to move from this type of lattices to permutation lattices in a "nice" way.
To illustrate I added my smallest example below.
Let $L$ be the free abelian group on generators {$a,\hat{a},b,\hat{b},u$} (written multiplicativly) . Let $K=$ {$e,\sigma,\tau,\sigma\tau $} be the Klein four group and define a $K$ action on $L$ as follows:
$\sigma$: switches between $a$ and $\hat{a}$, fixes $b$ and $\hat{b}$ and $\sigma(u)=ab\hat{a}\hat{b}/u$.
$\tau$: switches between $b$ and $\hat{b}$, fixes $a$ and $\hat{a}$ and $\tau(u)=ab\hat{a}\hat{b}/u$.
The K-lattice $L_0$ generated by {$a,\hat{a},b,\hat{b}$} is easily seen to be a permutation lattice, so my hope is to find some sort of rationality condition (pure, stable, retract) of $F(L)^K$ over $F(L_0)^K$, so I can then use the solution to the usual Noether's problem (if it exists).
The only theorem that (I found and) is close to helping me in this case is a generalization of Luroth theorem which says that if the G action on $u$ in $F(L)$ is something like $g(u)=a_g u+b_g$ where $a_g,b_g \in F(L_0)$ then $F(L)^G$ is rational over $F(L_0)^G$. This is similar to the case above except that the action on $u$ is of the form $\sigma(u) = a_\sigma u^{-1}$.
In this particular example, we also have that $L$ is a permutation lattice over the subgroup {$\sigma\tau,e$}, because $\sigma\tau(u)=u$. I tried to first take the $\sigma\tau$ invariants and then the rest of the group but still didn't get anything helpful, though this way still seems promising.
