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I am having hard time understanding some details in Proposition 2.8 which is on page 22 of Atiyah and Macdonald's book: Introduction to Commutative Algebra.

How the writers are claiming that being the composite map $$ N\rightarrow{M}\rightarrow{M/mM} $$ onto implies that $N+mM=M$?

  • I was trying to use isomorphism theorems but I couldn't get it. –  Apr 14 '14 at 09:03
  • N is just a submodule of M which is generated by $x_i's$. –  Apr 14 '14 at 09:06
  • Why we do have $n+mM=a$? I understand for each $a\in{M/mM}$ there is $n\in{N}$ such that$\pi(i(n))=a$, where $\pi$ is natural projection and $i$ is injection. –  Apr 14 '14 at 09:12

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Let $\iota$ be the map $N \to M$, let $\pi:M \to M/\mathfrak{m}M$ be the quotient map, and suppose $\pi \circ \iota$ is onto. In this problem $N$ is a submodule of $M$, so $\iota(n)$ is just $n$ thought of as an element of $M$. Let $x \in M$, then $x + \mathfrak{m}M \in \operatorname{im}(\pi\circ\iota)$, so $x + \mathfrak{m}M = \pi(\iota(n)) = n + \mathfrak{m}M$ for some $n \in N$. Therefore $x-n \in \mathfrak{m}M$, and $x = n + y$ for some $y \in \mathfrak{m}M$. We have written an arbitrary element of $M$ as a sum of something in $N$ and something in $\mathfrak{m}M$, therefore $M = N+\mathfrak{m}M$.

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Matthew Towers
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Given $x/mM\in M/mM$ you have $n\in N$ such that :

$\pi(i(n))=x+mM=n+mM$ for $x\in M $ and $n\in N$

i.e., $x=n+mM+mM=n+mM$ this mean $M\subseteq N+mM$

It is upto you to look for the other way.... (Which is trivial)