Simultaneous semisimplicity of commuting endomorphisms We say that an endomorphism $f$ of a $\mathbb{k}$-vector space $V$ is semisimple if the resulting $\mathbb{k}[x]$-module structure on $V$ is semisimple.

Question. Let $V$ be a $\mathbb{k}$-vector space and let $f_1, \dotsc, f_n$ be semisimple endomorphisms of $V$ such that $f_i$ commutes with $f_j$ for all $i, j = 1, \dotsc, n$. Is the resulting $\mathbb{k}[x_1, \dotsc, x_n]$-module structure on $V$ semisimple?

The question is motivated by the following special case:
Suppose that the field $\mathbb{k}$ is algebraically closed and that the vector space $V$ is finite-dimensional. Under this assumptions, semisimplicity is equivalent to (simultaneous) diagonalizability.
The question is then equivalent to the well-known theorem that finitely many diagonalizable, commuting endomorphisms are simultaneously diagonalizable.
 A: Here is a counterexample.  Let $k$ be a field of characteristic $p$ that is not perfect and let $a\in k$ have no $p$th root.  Let $K=k[x]/(x^p-a)$, let $V=K\otimes_k K$ and let $f_1,f_2:V\to V$ be multiplication by $x\otimes 1$ and $1\otimes x$, respectively.  Each of $f_1$ and $f_2$ is semisimple over $k$, since their minimal polynomial $x^p-a$ is irreducible over $k$.  However, as a $k[x_1,x_2]$-module, $V$ is the quotient $k[x_1,x_2]/(x_1^p-a,x_2^p-a)\cong K[x_2]/(x_2^p-a)$ which is not semisimple since $x_2^p-a$ factors as $(x_2-x)^p$ over $K$.
However, this is essentially the only sort of counterexample.  In particular, let us suppose not just that $V$ is semisimple as a $k[x]$-module with respect to each of $f_1,\dots,f_n$, but that it is a sum of simple $k[x]$-modules which are separable extensions of $k$ (in particular, this is automatic if $k$ is perfect).  Equivalently, $V$ is a torsion $k[x]$-module with respect to each $f_i$ and the minimal polynomial of $f_i$ on any finite-dimensional invariant subspace of $V$ is separable.  Then I claim that $V$ is semisimple over $k[x_1,\dots,x_n]$.
By induction on $n$, we may assume $V$ is semisimple over $k[x_1,\dots,x_{n-1}]$.  So, we can decompose $V$ as a direct sum $\bigoplus_m V_m$, where $m$ ranges over the set of maximal ideals of $k[x_1,\dots,x_{n-1}]$ and each $V_m$ is a direct sum of copies of $K_m=k[x_1,\dots,x_{n-1}]/m$.  Since $f_n$ commutes with $f_1,\dots,f_{n-1}$, it maps each $V_m$ to itself, so it suffices to show each $V_m$ is semisimple over $k[x_1,\dots,x_n]$.  Since $m$ acts trivially on $V_m$, we can think of its $k[x_1,\dots,x_n]$-module structure as a $K_m[x_n]$-module structure.
So, here's the situation.  We have a vector space $V_m$ over a field $K_m$ which is an extension of $k$, and an endomorphism $f_n$ of $V_m$ which is semisimple and separable (in the above sense) over $k$, and we wish to show $f_n$ is also semisimple over $K_m$.  If $f_n$ were not semisimple over $K_m$, then there would be a finite-dimensional $f_n$-invariant $K_m$-subspace $W\subseteq V_m$ such that the minimal polynomial $p\in K_m[t]$ of $f_n$ on $W$ over $K_m$ is not squarefree.  By hypothesis, the minimal polynomial $q\in k[t]$ of $f_n$ on $W$ over $k$ is separable.  In particular, this means $q$ is squarefree as an element of $K_m[t]$.
But $p$ is a factor of $q$ and is not squarefree, so this is a contradiction.
