I'm currently studying about the generalisation of Galois Theory for commutative rings. I'm trying to find a good example that illustrates how this generalisations allows a Galois extension of a commutative ring $S$ over $R$ to have different associated Galois groups (non-isomorphic).
I've been trying to find a scenary where this happends. As I know, the natural case in which this occurs is when having the commutative ring $S_n=Fe_1\oplus Fe_2\oplus\cdots\oplus Fe_n$, where $e_i=(0,\dots,0,1,0,\dots,0),$ where $1$ is in the $i$-th place (the $e_i$ are orthogonal idempotents). I've found I can always define a Galois extension from $S_n$ over $F(e_1+\cdots + e_n)$ with the group $G=\langle\sigma\rangle$, where $\sigma$ is any $n$-cycle (for simplicity I take the natural one $(123...n)$).
But I've tried up to $n=5$ and for these examples it doesn't seem I can find another group (non isomorphic to $\langle \sigma \rangle$) valid as Galois group of the extension.
So my question is, which is the easiest possible example (lowest possible $n$) of this kind where I can find two different and non-isomorphic Galois groups for a Galois extension of commutative ring $S_n$? I have assumed the easiest possible example where this happends is for a ring of the kind of those $S_n$, but I may be wrong, so in that case, which is an easier example?
If I'm not wrong,
On $\Bbb{R}^4$
there is the $\Bbb{Z/4Z}$ action generated by $\sigma:(a,b,c,d)\to (d,a,b,c)$
but also a $\Bbb{Z/2Z}\times \Bbb{Z/2Z}$ action generated by $\rho:(a,b,c,d)\to (b,a,d,c),\phi:(a,b,c,d)\to (c,d,a,b)$
($\phi\rho=\rho \phi$ so it is clear that every element has order $2$).
The subring fixed by $\langle \sigma \rangle$ is $\Bbb{R}(1,1,1,1)$, same for the subring fixed by$\langle \rho,\phi \rangle$.
For the "invertibility of the matrix of conjugates" condition: in both case take $x_1=y_1=(1,0,0,0)$, $x_2=y_2=(0,1,0,0)$,$x_3=y_3=(0,0,1,0)$, $x_4=y_4=(0,0,0,1)$ so that $\sum_{i=1}^4 x_i y_i=(1,1,1,1)$ and $x_i g(y_i)=0$ whenever $g\ne 1$.