I have already posted this on stackexchange

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 distinct 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.

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    $\begingroup$ (Finite) semi-simple modules over a commutative $k$-algebra are direct sums of finite extensions of $k$. If this holds over $k[x]$, it holds over $k[x,y]$. $\endgroup$
    – abx
    Apr 28, 2022 at 7:08
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    $\begingroup$ @abx Maybe I'm misunderstanding what you're saying, but there are certainly non-semisimple $k[x,y]$-modules that are semisimple over $k[x]$. $\endgroup$ Apr 28, 2022 at 8:06
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    $\begingroup$ If an operator on a vector space is semisimple (i.e., the vector space equals the direct sum of its eigenspaces), then its restriction to every invariant subspace is also semisimple (and the subspace equals the direct sum of its eigenspaces). Since the operator of $y$ commutes with the operator of $x$, the operator of $y$ preserves every eigenspace of $x$ (and vice versa). If the operator of $y$ is semisimple, then its restriction to each $x$-eigenspace is semisimple. Then the $y$-eigenspaces inside the $x$-eigenspaces are $k[x,y]$-submodules. $\endgroup$ Apr 28, 2022 at 10:40
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    $\begingroup$ By the way, you can find this argument in textbooks about representations of semisimple Lie algebras (I am currently teaching out of one such textbook). This is precisely the argument that finite dimensional representations of a semisimple complex Lie algebra have weight decompositions for every "toral" subalgebra. $\endgroup$ Apr 28, 2022 at 10:42
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    $\begingroup$ I missed that you are not using "semisimple operator" to mean "diagonalizable operator", but to mean something else. My comment really shows that two commuting diagonalizable operators are simultaneously diagonalizable (a standard result that gets used constantly in the theory of finite dimensional Lie algebras). $\endgroup$ Apr 28, 2022 at 13:10

1 Answer 1


There is an inseparable field extension $L/k$ with $L = k(a)$, such that the ring $L\otimes_k L$ has nilpotents, so is not semisimple. See for example


Now there is a surjective ring homomorphism $k[x,y]\to L\otimes_k L$ sending $x$ to $a\otimes 1$ and $y$ to $1\otimes a$. This turns $L\otimes_k L$ into a module for $k[x,y]$ which is not semisimple. But the images of $k[x]$ and $k[y]$ are copies of $L$, so for these it is semisimple.

  • $\begingroup$ thanks. Do you know if there is any counterexample in characteristic $0$? $\endgroup$ Apr 28, 2022 at 18:27
  • $\begingroup$ The assertion is true if $k$ has characteristic 0, or more generally is a perfect field, because the quotients of the algebras $k[x]$ and $k[y]$ by their intersections with the annihilator of $M$ are semisimple, hence separable over $k$, so their tensor product is again separable over $k$, so semisimple. But the action of $k[x,y]$ on $M$ factors through this tensor product. $\endgroup$
    – wcb
    Apr 29, 2022 at 7:04

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