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Is there an easy way to see the following:

Given a $k$-algebra $A$, with $k$ a field, and a finite dimensional semisimple $A$-module $M$. Look at the natural map $A \to \mathrm{End}_k(M)$ that sends an $a \in A$ to $$ M \to M: m \mapsto a \cdot m. $$ Then the image of $A$ is a finite-dimensional semisimple algebra.

Zev Chonoles
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Boris Datsik
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2 Answers2

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Here's one way to look at it: Notice that the kernel of the map is exactly $ann(M)$, which necessarily contains the Jacobson radical $J(A)$ of $A$. Since $A/J(A)$ and all of its quotients are semiprimitive, it follows that $A/ann(M)$ is semiprimitive.

Now $M$ as a faithful $A/ann(M)$ module. Since the simple submodules of $M$ remain the same during this passage, $M$ is also still semisimple over this new ring. You can see in this question why a ring with a faithful module of finite length must be Artinian. Now we have that $A/ann(M)$ is Artinian and semiprimitive: so it is semisimple.

I see I overlooked a simple way for concluding that the image is finite dimensional. Of course our image ring is a subalgebra of $End(M_k)$ which is finite dimensional... so the subring is finite dimensional as well. The argument I gave before essentially proves a more general case: "If $M$ is a semisimple $R$ module of finite length, the image of the natural map is a semisimple ring."

Community
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rschwieb
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  • Instead of «the Jacobson radical is the smallest ideal I such that A/I is semisimple» it would be better to say «since $A/J$ is semisimple and every quotient of a semisimple algebra is semisimple...» for the first variant leaves open the possibility that ideals containing the radical have quotients which are not semisimple (of course, this does not happen) – Mariano Suárez-Álvarez Jun 13 '13 at 21:27
  • Is it a problem that $A/J$ is only semisimple if $A$ is artinian? Would it be better to say that $A/J$ is $J$-semisimple, and so $A/ann(M)$ is $J$-semisimple, but also finite-dimensional, so semisimple? Or am I missing something. – Boris Datsik Jun 14 '13 at 06:38
  • @BorisDatsik I have no idea why I thought I had two approaches. It's clear I needed the second half with the first. The latest edit should clarify things. In a nutshell $M$ being semisimple allows us to conclude $A/ann(M)$ is semiprimitive, and the finite dimensionality allows us to conclude $A/ann(M)$ is Artinian. – rschwieb Jun 14 '13 at 10:42
  • @BorisDatsik What you described is a valid approach too, sticking to the context of finite dimensional algebras. (BTW, I am using "semiprimitive" for "J-semisimple", to keep terms farther apart from each other.) – rschwieb Jun 14 '13 at 11:00
  • A very nice answer – Mod.esty Nov 19 '20 at 15:47
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Here's another slightly different approach. Let $B$ be the image algebra, which is a finite dimensional algebra over $k.$ Note that $M$ is a $B$-module and its $B$-submodules are precisely the $A$-submodules of $M.$ Hence $M$ is also a semisimple $B$-module, and the Jacobson radical $J(B)$ annihilates $M.$ Thus $J(B) = \{0 \} $ and $B$ is semisimple. ($B$ is a genuine subalgebra of ${\rm End}_{k}(M),$ so is faithfully embedded there, and the elements of $J(B)$ are annihilating $M$, so only the zero endomorphism can be in $J(B) ).$

Geoff Robinson
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