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During my revision, I came across this exercise:

Let $R$ be a ring and $I$ an ideal of $R$. If $M$ is a simple left $R/I$-module, then $M$ is a simple left $R$-modules with operation define as follows: $rm=(r+I)m$, for all $r \in R$ and $m \in M$.

I was thinking to use the fact that $M \cong (R/I)/J$, for a maximal left ideal $J$ of $R/I$. Then I can use the "Correspondence Theorem" to get that maximal ideals of $(R/I)/J$ correspond to maximal ideals of $R/I$ containing $J$. And these correspond to maximal ideals of $R$ containing I, so $M=R/L$ for a maximal ideal $L$ of R. Am I on the right track? I don't think I am because I am not using the information on the operation of $R$ (unless, this bit of information is redundant).

2 Answers2

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For any ring $R$ and any (two-sided) ideal $I$ of $R$, left $(R/I)$-modules are literally the same as left $R$-modules which are annihilated by $I$. This is because, given an $(R/I)$-module $M$, $M$ becomes an $R$-module via the map $R\rightarrow R/I$, and clearly, for this $R$-action, $I$ annihilates $M$ (because the elements of $I$ constitute the kernel of the ring map which we are using to define the action on $M$). To go the other way, if $M$ is a left $R$-module annihilated by $I$, then you can verify that if $r,r^\prime\in R$ are congruent modulo $I$, that is, $r-r^\prime\in I$, then $rm=r^\prime m$ for any $m\in M$, so it makes sense to define $(r+I)m=rm$, and this turns $M$ into an $(R/I)$-module.

An important fact about this correspondence (or equivalence or whatever you want to call it) is that $R$-submodules of $M$ are the same as $(R/I)$-submodules, i.e., an additive subgroup $N$ of $M$ is an $R$-submodule if and only if it is an $(R/I)$-submodule. This is because the map $R\rightarrow R/I$ is surjective, and the action of $r\in R$ is the same as the action of $r+I\in R/I$. So $M$ is simple as an $R$-module if and only if it is simple as an $(R/I)$-module.

Keenan Kidwell
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I think you really showed good sense with that approach, but maybe in this case you can get by with a simpler argument.

An $S$ module is simple iff for every $m,n$ in the module, there exists $s$ such that $sm=n$.

In our case, we're given that for every $m,n\in M$, there exists $r$ such that $(r+I)m=n$. But in the proposed module action, this says exactly that $r\cdot m=n$. So, we've satisfied the condition to show that $M$ is a simple $R$ module under this particular action.

In fact this hints at why the $R/I$ submodules of an $R/I$ module $M$ are exactly the $R$ submodules with the proposed action in general:

Given an additive subgroup $N<M$, these conditions are the same:

$$\forall r\in R,n\in N\ \ (r+I)n\in N $$ $$\forall r\in R,n\in N,\ \ r\cdot n\in N$$

rschwieb
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  • Very nice! One question: $S$ module $M$ is simple iff for every $m,n \in M$, there is $s \in S$ such that $sm=n$. The first implication is trivial, but how do I show "if for every $m,n \in M$, there is $s \in S$ such that $sm=n$, then $M$ is simple"? –  Apr 01 '13 at 13:21
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    @dado You'll be able to handle it after this, if you haven't already: suppose $0<N<M$ and $m\in M\setminus N$, $n\in N\setminus{0}$. What would it mean to find $r$ such that $rn=m$? – rschwieb Apr 01 '13 at 13:27
  • Got it, thank you very much!!! –  Apr 01 '13 at 13:28