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Given two $k$-algebras $A$ and $B$ and a surjection $A\to B$, then every $B$-module is also an $A$-module.

Can we use this to show that quotients of semisimple algebras are semisimple? An algebra $A$ is said to be semisimple if every $A$-module is a direct sum of simple modules.

EDIT: Consider the surjective projection $\pi :A\to A/B$ for $A$ semi-simple and $B\subset A$ an ideal.

If we choose an $A/B$-moduoe $M$, it is also an $A$-module via $a.m:=\pi (a).m$. $(1)$

Since $A$ is semi-simple, get $M=\bigoplus_{i\in I}M_i$ with simple $A$-Modules $M_i$. These $M_i$ are also $A/B$-modules via $\pi(a).m:=a.m$. This clearly gives a module structure but it is not clear that this is well defined since an element in $A/B$ might have more than one pre-image, so: Let $\pi(a)=\pi(a')$ then we need to show $a.m=a'.m$. We get $a-a'\in \ker(\pi)=B$. That is $0=\pi(a-a').m=(a-a’).m=a.m-a'.m$, so we are done.

Now we are left to show that those $M_i$ are simple as $A/B$-modules: Those modules aren't zero since they are simple as $A$-modules. Besides there can't be any non-trivial $A/B$-submodules because: Assume there is such an $A/B$-submodule $U$. Then $A.U=\pi(A).U=A/B.U\subset U$. (The first equation is $(1)$ and the second is the surjectivity of $\pi$.) Thus $U$ would be a non-trivial $A$-submodule

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  • $\begingroup$ What's $A/B$? You can form a quotient by an ideal. $\endgroup$ – egreg Nov 9 '17 at 23:29
  • $\begingroup$ I have changed that! Now does the proof work like this? $\endgroup$ – user500228 Nov 10 '17 at 7:04
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Sure. Suppose $\phi:A\twoheadrightarrow B$ is a surjection.

Given a $B$-module, look at it as an $A$ module $M$, decompose it into a sum of simple $A$ modules, then look at them again as $B$ modules.

This is possible because the definition of the lifted action makes $\ker\phi$ annihilate $M$ when considered as an $A$ module, so it also annihilates the simple components, making it possible to define the $B$ action on them as $m\cdot\phi(r):=mr$.

The general idea is this: if the ideal $I\subseteq ann(M)$ for some $R$ module $M$, then $M$ has a natural $R/I$ module structure, and $M$ has the same submodules considered as an $R$ module as it does as an $R/I$ module.

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  • $\begingroup$ So it is not only true that every B-module is an A-module but also that every A-module is an B-module? Or just in the quotient case? $\endgroup$ – user500228 Nov 8 '17 at 14:17
  • $\begingroup$ @VML No, it's not true that every $A$ module is a $B$ module. As I said, the $A$ modules annihilated by $\ker \phi$ have a natural $B$ module structure. $\endgroup$ – rschwieb Nov 8 '17 at 16:18
  • $\begingroup$ I added two questions in my question maybe you can answer them please? :) $\endgroup$ – user500228 Nov 9 '17 at 22:17
  • $\begingroup$ @VML $B$ is not the kernel of the homomorphism I'm talking about. You need to rethink those questions. $\endgroup$ – rschwieb Nov 9 '17 at 22:35
  • $\begingroup$ But $B$ is the kernel of the projection. Is my proof correct, that the action is well-defined? $\endgroup$ – user500228 Nov 9 '17 at 22:41

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