Semi simplicity over commutative algebras over non-algebraically closed fields I have a question:
If k is an arbitrary field then is it true that if $M$ a finite dimensional  $k[x, y]$ is semisimple as a $k[x]$ module  and  also as a $k[y]$  module then it is semisimple (as a $k[x, y]$ module?).........(*)
I came to this while exploring simple representations of the Jordan Quiver and then going for the double loop case and first studying the commutative subcase.
I have shown and it is known that the simple representations of the Jordan Quiver are all finite dimensional (for arbitrary fields) and correspond to Transformations whose minimal and characteristic polynomial are same and irreducible. To be clear, also there no infinite dimensional simples (this i have shown).
Then I came to investigate the simples and over $k[x, y]$:
I cold show using  simultaneous diagnolization and such results in linear algebra that if $k$ is algebraically closed then (*) holds for algebraically closed case.
The idea is to view $x$ and $y$ as transformations $T_1$ and $T_2$ and then pick a basis which diagonalizes both simultaneously. (semisimple is equivalent to diagonlizability since the simples are 1 dimensional). So, this breaks up $V$ into 1 dimensional $k[x, y]$ modules and so this is automatically a decomposition into simples.
Furthermore, I investigated what are ALL simples over $k[x, y]$. if they are all 1 dimensional then the above result is 'more useful'. And indeed the this is true.
Infact over any finitely generated commutative algebra over algebraically closed fields, the only simples are 1 dimensional even if we allow for infinite dimensional ones.
I had checked this explicitly(using matrices )  and found that there are no simples of dimension 2. The proof of the above general result I found online. Basically the idea is to view simples as quotients with maximal ideals and use Zariski lemma.
So, this is for algebraically closed.
Obviously for non-algebraically closed we do have simples of dimension > 1.
To by original question (*) It is equivalent to prove the case when $x = T_1$ has only one irreducible factor because the space annhilated by one irreducible factor is invariant under both the transformations and I have not made much progress from here.
 A: If $X, Y \colon V \to V$ are semisimple linear transformations of a finite-dimensional $k$-vector space, where $k$ is algebraically closed, then each diagonalises, i.e. we have
$$ V = \bigoplus_{\lambda \in k} \ker (X - \lambda) = \bigoplus_{\lambda \in k} \ker (Y - \lambda).$$
If $XY = YX$ then we can show that $Y$ restricts to each eigenspace $\ker (X - \lambda)$ of $X$, because if $v \in \ker(X - \lambda)$ then
$$ (X - \lambda)Yv = Y(X - \lambda)v = Y0 = 0, $$
hence $Yv \in \ker(X - \lambda)$ as well. This gives simultaneous diagonalisation and all that good stuff.
If $k$ is not algebraically closed, all that changes is that the eigenspaces cannot necessarily all be written as $(X - \lambda)$ for some $\lambda \in k$, but might be some higher degree polynomial $p \in k[x]$, so instead we have (for semisimple X) that
$$ V = \bigoplus_{p(x) \in k[x]} \ker p(X). $$
For example, if $k = \mathbb{R}$ and $X$ is a 90 degree rotation, then the polynomial in the above sum would be $x^2 + 1$. Provided that $Y$ is semisimple and commutes with $X$, the same proof of simultaneous "diagonalisability" (there may be higher degree polynomials, not just eigenvalues) still goes through, and shows that $V$ is semisimple as a $k[x, y]$ module.
