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We know if $R$ is semi simple ring and $M$ is $R$-module then for any submoldule $N$ of $M$ there exists submodule $N'$ such that $N \oplus N'=M$.

Now I have trouble with sum and its supposed answer.

let $R$ be ring (Not semi simple) and $M$ is $R$-module and $N$ is submodule of $M$. Does exist submodule $N'$ such that $N+N'=M$?

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Take $N'=M$, or do you want a proper submodule? –  Asaf Karagila Aug 21 '12 at 17:55
    
@AsafKaragila YES, it is trivial if N is not proper –  Babak Miraftab Aug 21 '12 at 18:04
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2 Answers 2

up vote 2 down vote accepted

No. Take for example $R={\bf Z}=N$, $M=\bf Q$. Assume by contradiction that we have $N'$ a proper subgroup of $\bf Q$ such that $N'+{\bf Z}=\bf Q$.

  1. Any subgroup of $\bf Q$ is of the form $\langle \frac{p}{q_n}\vert n\in{\bf N}\rangle$ with $p,q_n\in \bf N$, $q_{n}\vert q_{n+1}$, and $(p,q_n)=1$ for all $n$, so suppose $N'$ is of that form.
  2. If $p=1$, then ${\bf Z}\subseteq N'$, so $N'=N'+\bf Z$, so $p>1$.
  3. $N'+\mathbf Z=\mathbf Q$, so in particular, there are $m,n,k\in \bf N$ such that $m\frac{p}{q_n}-k=\frac{1}{p}$. But then $p\vert q_n$, which is a contradiction, so there is no such $N'$.
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If $M$ is an $R$-module such that there is a unique maximal proper submodule $N \subsetneq M$, then for any other proper submodule $N' \subsetneq M$ we must have $N' \subseteq N$ and therefore $N + N' = N \subsetneq M$.

For instance, if $R$ itself is a (say, commutative) local ring, then one may take $M = R$ and $N$ to be the unique maximal ideal of $R$.

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:YOU suppose that M is local I want to know if M is not local what can we say? –  Babak Miraftab Aug 21 '12 at 18:38
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I believe that Manny is trying to point out a counterexample, to show that in general there need not exist such an $N'$. –  Cameron Buie Aug 21 '12 at 18:41
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