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I am having difficulty in understanding the proof of Corollary 3 on Pg. 44 of Miles Reid's Undergraduate Commutative Algebra

Corollary 3 Let $(A,m)$ be a local ring, $M$ an $A$-module, and $N \subset M$ a submodule; suppose that $M/N$ is finite over $A$, and that $M = N+mM$; then $N = M$.

Proof Since $m(M/N)$ = $M/N$, ...

Taking quotient by $N$ on both sides in $M = N+mM$, we get $M/N = (N+mM)/N = mM/(N \cap mM)$, but I can't see how this is $m(M/N)$. Maybe I am missing something easy here, but could someone elaborate on this?

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Whether $M=N+mM$ or not, it is always true that for any ideal $I$, we have $$I(M/N)\simeq(IM+N)/N$$ the homomorphism being given by $$\sum_i a_i(m_i+N)\longmapsto\Bigl(\sum_i a_im_i\Bigl)+N$$ One can easily check it is well-defined and bijective.

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  • $\begingroup$ Well I actually wanted a clarification on this itself. How does $\sum a_i(m_i+N) = 0$ in $I(M/N)$ imply $\sum a_i m_i \in N$? $\endgroup$ – Seven Oct 11 '15 at 17:23
  • $\begingroup$ $\sum_i a_i(m_i+N)=0$ means $\sum_i a_i(m_i+N)=N$ really. $\endgroup$ – Bernard Oct 11 '15 at 17:34
  • $\begingroup$ Okay. I was thinking that the $0$ in $I(M/N)$ is $IN$. Instead of looking at $I(M/N)$ as a module on its own, I should just consider it as a submodule of $M/N$, and then their $0$'s coincide, is that right? $\endgroup$ – Seven Oct 11 '15 at 17:40
  • $\begingroup$ Yes. $IN$ is the $0$ of $IM/IN$, which is a different module. $\endgroup$ – Bernard Oct 11 '15 at 17:56

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