Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

enter image description here

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:

share|improve this question
add comment

2 Answers

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.

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.