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How do I embed $M/(N \cap N')$ as a submodule of $(M/N) \oplus(M/N')$? My thought it the following...

Simply send $m + N \cap N'$ to $(m + N, m + N').$ There is no ambiguity as $N \cap N'$ is a submodule of both $N$ and $N'.$ It is also injective by obvious reasons. Is this the correct way of looking at this?

green frog
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  • $m + N \cap N' = m' + N \cap N' \implies m - m' \in N \cap N' \subset N, N'$ so $m + N = m' + N$ sim. $N'$. So yes, the subset causes it. – Daniel Donnelly Nov 30 '17 at 17:52

1 Answers1

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Yes.

Alternatively and more explicitly, you could just note that the natural map

$f:m\mapsto (m+N, m+N')$ is a homomorphism, and by the first homomorphism theorem $M/\ker(f)\cong Im(f)$, but $\ker(f)=N\cap N'$. So the image of $f$ in $(M/N)\times (M/N')$ is isomorphic to $M/(N\cap N')$.

rschwieb
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