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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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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.

rschwieb
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  • 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? –  Nov 08 '17 at 14:17
  • @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. – rschwieb Nov 08 '17 at 16:18
  • I added two questions in my question maybe you can answer them please? :) –  Nov 09 '17 at 22:17
  • @VML $B$ is not the kernel of the homomorphism I'm talking about. You need to rethink those questions. – rschwieb Nov 09 '17 at 22:35
  • But $B$ is the kernel of the projection. Is my proof correct, that the action is well-defined? –  Nov 09 '17 at 22:41
  • @VML You need to make up your mind if you want B to be the image or the kernel. From what you originally said, $R/I\cong B$ for some ideal $I$. – rschwieb Nov 09 '17 at 23:13
  • I again edited my question and gave my proof of “quotients of semi-simple modules are semi-simple”. My last question is: is this proof correct? –  Nov 09 '17 at 23:25
  • @VML For the problem I mentioned before, which you don't appear to have changed at all, your explanation using "A/B" doesn't make any sense. – rschwieb Nov 10 '17 at 00:29
  • I see the problem, $B$ need to be an ideal not an subalgebra. I changed that, is there still a problem in the proof? –  Nov 10 '17 at 07:06
  • @VML Actually, you need to show that $a\cdot m=a'\cdot m'$ for any $(a,m)\equiv (a', m')\in A/B\times M$, but what you have written works for both halves, and you can easily fix that. I'd say you are fudging too much with the justification in the last paragraph for why there aren't other submodules. It doesn't have anything to do with surjectivity, afaik. You should just prove the lemma (if you aren't allowed to use it, already) that the $A/B$ modules correspond to $A$ modules annihilated by $B$. – rschwieb Nov 10 '17 at 11:41
  • I am sorry for my maybe stupid questions. I have now written a few lines fro the last pragraph and explained where I have used the surjectivity of $\pi$.Where is the problem? For your first remark: Well, what may be the problem if I define an $A/B$ action like $x.m=a.m$ for $\pi(a)=m$: $x$ might have two pre-images $\pi(a)=\pi(a')=x$. We want that the action is well defined, i.e. that those different pre-images have same action, i.e. $a.m=a'.m$. So I do not see why I should show $a.m=a'.m'$. :( –  Nov 10 '17 at 13:54
  • @VML A left $R$ module $M$ has, by definition, a function $R\times M\to M$ defining its action. To demonstrate this function is well-defined, you'd have to take two equal elements in its domain and show they yield the same image. It's not any harder than what you already did, it's more of a detail. – rschwieb Nov 10 '17 at 13:58
  • It was not the problem showing this I just didn't get the reason why I should do this, but it's okay now :) What do you say to my last paragraph? Does it work like this? –  Nov 10 '17 at 15:48
  • @VML yes, I think that looks OK now. I see what you mean by surjectivity helping here now. – rschwieb Nov 10 '17 at 16:12
  • Alright, then nothing is left but thanking you a lot! –  Nov 10 '17 at 17:27