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I put the paragraph to clarify because it is a vector space. I have a question with the proposition, I don´t know why he concludes the red line assertion, only knowing that there exist a surjective morphism onto $M/mM $ Sorry for my stupid questions D:

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up vote 3 down vote accepted

This is not involved with the structure of module or vector space over $k$, it derives simply from easy properties of abelian groups.

From the surjectivity of the composite $N \to M \to M / \mathfrak{m} M$, you have that every class in $M / \mathfrak{m} M$ comes from an element of $N$, i.e. for every $x \in M$, there exists $y \in N$ such that $x + \mathfrak{m} M = y + \mathfrak{m} M$ as elements of $M/ \mathfrak{m} M$, hence $x -y \in \mathfrak{m} M$, therefore $x \in y + \mathfrak{m} M \subseteq N + \mathfrak{m} M$. Since this holds for every $x \in M$, you have $M \subseteq N + \mathfrak{m} M$. But the opposite containment $\supseteq$ is obvious.

If you want, convince yourself that, if $G$ is an abelian group and $H, K$ are subgroups of $G$, then $G = H + K$ holds if and only if the composite $H \hookrightarrow G \twoheadrightarrow G/K$ is surjective.

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Here is a slightly more general point of view, in which the notation is compatible with your particular case .

Suppose you have a ring $A$, an $A$- module $M$ and a submodule $P\subset M$ (in your case $P={\mathfrak m}M$ ). Call $\pi$ the quotient map $\pi:M\to M/P$.
There is a bijective correspondence between submodules $\Sigma \subset M/P$ and submodules $S\subset M$ containing $P$ , in which $\Sigma/P\subset M/P \;$ corresponds to $P\subset \Sigma\subset M$.

Now, even if $N \subset M$ does not contain $P$, we still have a submodule $\pi(N)\subset M/P$.
Question: to what submodule of $M$ containing $P$ does $\pi(N)$ correspond ?
Answer: To $\;N +P \;$ (which does contain $P \;$).
Consequence: If $\pi(N)=\pi(M)=M/P \; $ , then $N+P=M$, just as the book says

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