# If module $M = N + mM$, then why is $m(M/N) = M/N$?

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?

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.
• Well I actually wanted a clarification on this itself. How does $\sum a_i(m_i+N) = 0$ in $I(M/N)$ imply $\sum a_i m_i \in N$? – Seven Oct 11 '15 at 17:23
• $\sum_i a_i(m_i+N)=0$ means $\sum_i a_i(m_i+N)=N$ really. – Bernard Oct 11 '15 at 17:34
• 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? – Seven Oct 11 '15 at 17:40
• Yes. $IN$ is the $0$ of $IM/IN$, which is a different module. – Bernard Oct 11 '15 at 17:56