Question involving direct summand

Question

Lets $L,N$ submodules of $M$ such that $L \cap N$ is direct summand of $L$ and $N$. Show that $L \cap N$ is direct summand of $L+N$.

$\textbf{My attempt:}$

Well, if $L \cap N$ is direct summand of $L$ and $N$ then there exists $L'$ and $N'$ such that $L \simeq L \cap N \oplus L'$ and $N \simeq L \cap N \oplus N'$. I was thinking about $L+N \simeq L \cap N \oplus (L' \oplus N')$, but I don't know if it is correct and how to prove it.

Another way is using splitting sequences. Can you help me???

Answer 1

The idea with splitting sequences is, at least to me, much easier.

We know that $L \cap N \hookrightarrow L$ and $L \cap N \hookrightarrow N$ are direct summands, and so we have splittings

• $\pi_L : L \to L \cap N$
• $\pi_R : N \to L \cap N$

with the bonus property that including and splitting is the identity (this is more or less the definition of a splitting):

• $L \cap N \hookrightarrow L \overset{\pi_L}{\longrightarrow} L \cap N$ is the identity
• $L \cap N \hookrightarrow N \overset{\pi_N}{\longrightarrow} L \cap N$ is the identity

But to show that $L \cap N$ is a summand of $L+N$, we want to show that there is such a splitting $\pi_{L+N} : L+N \to L \cap N$... Do you see a good way to build such a splitting? I'll give another hint under the fold.

By the universal property of direct sums, we have a map $\pi_L + \pi_N : L+N \to L \cap N$ which sends $l + n \mapsto \pi_L(l) + \pi_N(n)$. Notice it's important that $\pi_L$ and $\pi_N$ agree on the intersection, but of course they do since they are both the identity on the intersection. Can you show that this is the desired splitting? That is, can you show that $L \cap N \hookrightarrow L + N \overset{\pi_L + \pi_N}{\longrightarrow} L \cap N$ is the identity?

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I hope this helps ^_^