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Suppose I have an $R$-module $M$ and an onto ring homomorphism $\varphi:S\rightarrow R$. This homomorphism induces an $S$-module structure on $M$ by the action: $$s\cdot m := \varphi(s)\cdot m $$ As far as I understand, there can be no module isomorphism between $M$ as an $R$-module and $M$ as an $S$-module since they are over different rings, but is there some other sense in which it is possible to show that both structures are 'equivalent'? e.g. that one of them is a (semi)simple module if and only if the other is?

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    $\begingroup$ The correct way to think of this situation is simply that $M$ is an $S$-module that happens to factor through $S \to R$. Namely, the surjection $S \to R$ induces a fully faithful embedding of $R$-mod into $S$-mod, so any property (such as semi-simplicity) that is preserved under this embedding of categories is the same for $M$ whether thought of as an $R$-module or an $S$-module. $\endgroup$ – Stephen Apr 24 '20 at 18:02
  • $\begingroup$ @Stephen Unfortunately I am yet unfamiliar with category theory. Should I first learn some of it if I wish to understand what's going on? Or can this also be understood in more 'elementary' terms? $\endgroup$ – Bar Alon Apr 24 '20 at 18:32
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It would be nice if you learned some category theory since these are the sort of things for which it is particularly useful.

However it is not too hard to prove this by hand.

First note that $R\cong S/I$ for some ideal $I$. So the question of whether $M$ being simple as an $R$ module iff simple as an $S$ module translates to asking whether every $S/I$ submodule of $M$ is an $S$ submodule and vice versa. This is sort of obvious since the $S/I$ action on $M$ is defined via equivalence classes in $S/I$.

Now the question of semi-simplicity is the same as asking whether if $M\cong M_1\oplus M_2$ as an $S/I$ module, then the same isomorphism holds as $S$ modules and vice versa.

Well first note that the splitting is as abelian groups. No in particular $M_1\cap M_2=\{0\}$ and every element of $m\in M$ can be written $m=m_1+m_2$ for $m_1\in M_1$ and $m_2\in M_2$.

So if the splitting occurs as $S/I$ (resp. $S$) modules, then you only need to check that the $S$ (resp. $S/I$) action preserves the submodules $M_1$ and $M_2$, but this was checked previously.

Edit: If you were to think about this categorically, then you would be showing something much stronger.

For example if $M$ to $N$ are $S$ modules which are annihilated by $I$ i.e. $IM=0=IN$ then we see that any morphism $S$ modules between them gives a morphism of $S/I$ modules. Then we also note that any morphism of $S/I$-modules lifts to a morphism of $S$ modules. In particular $M\cong N$ as $S$-modules if and only if congruent as $S/I$ modules.

This is what Stephens comment above about fully faithful embedding means.

At the level of modules, you can prove most properties in a piecemeal fashion, but a categorical approach gives a much more powerful perspective.

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  • $\begingroup$ Would I be correct in assuming that for any particular property of modules (such as semisimplicity) an elementary proof similar to the one in your answer is easy to construct, but that in order to understand the correspondence in general we need to understand categories? $\endgroup$ – Bar Alon Apr 26 '20 at 12:02
  • $\begingroup$ @BarAlon I have edited my answer to respond to your comment. $\endgroup$ – Shubhankar Sahai Apr 28 '20 at 20:47

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