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

Question

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?

Answer 1

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.