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$M$ is an $R$-module and $N$, $P$, $Q$ are submodules of $M$.

If $Q\subset N $ then $N\cap(P+Q)=(N\cap P)+(N\cap Q)$. Why?

If we don't have the condition $Q\subset N $, is the equality $N\cap(P+Q)=(N\cap P)+(N\cap Q)$ true?

Please tell with example?

hbghlyj
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iliya
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    What have you tried? Try to look at some simple examples of modules on your own. – Martin Brandenburg Nov 25 '13 at 18:34
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    I dont understand why this question is downvoted this is not trivial question maybe it lacks context but its not the worst question i have seen on this site. Secondly even though this is standard fact its proof is really hard to find if you dont know where to look. – user52045 Nov 25 '13 at 19:09

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First the example: take two dimensional vector space over field $k$, $k^2$ with base $e_1$, $e_2$. Now if we set $N:=(e_1+e_2)$, $P=(e_1)$ and $Q=(e_2)$ equality does not hold.

The proof assuming that $Q\subset N$:

"$\subset$" Take any $x\in N\cap(P+Q)$ then $N\ni x=p+q$ for some $p\in P$ and $q\in Q$. We know that $Q\subset N$ so $q\in N$ and $p=x-q\in N$. This is what we wanted to show $p\in N\cap P$ and $q\in N\cap Q$ so $x=p+q\in N\cap P+N\cap Q$.

"$\supset$" Take $x\in N\cap P+N\cap Q$ then $x=p+q$ for some $p\in N\cap P$ and $q\in N\cap Q$. Obviously $x= p+q\in N$ and $x= p+q\in P+Q$ and we are done.

user52045
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  • i dont undrestand example, please tell more? – iliya Nov 26 '13 at 06:15
  • $P+Q=(e_1)+(e_2)=k^2$ so left hand site is $(P+Q)\cap N= k^2\cap (e_1+e_2)=(e_1+e_2)$. Now $N\cap P=(e_1)\cap (e_1+e_2)=0$ and $N\cap Q=(e_2)\cap (e_1+e_2)=0$ so right hand site is $0$. – user52045 Nov 26 '13 at 08:31