# $\mathfrak{a}(M/N) = (\alpha M + N) / N$

I want to prove $\mathfrak{a}(M/N) = (\mathfrak{a}M + N) / N$, where $M$ is an $A$-module and $\mathfrak{a}$ is an ideal of $A$.

There will be many ways, for example, define a map $f:\mathfrak{a}M + N \to \mathfrak{a}(M/N)$ and show that $f$ is an $A$-homomorphism and $\ker(f)=N$.

But what is the best simple way to prove it? I don't want to define a map and prove it a homomorphism. It looks similar to 2nd ismomorphism but it is a little different.

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I guess it depends on what you mean by "best simple", but for me it amounts to unwinding the definition and apply one of the basic theorem about quotient modules. The module $\mathfrak{a}(M/N)$ is generated by elements of the form $a \bar{m}$ for $a\in \mathfrak{a}$. By definition, $a \bar{m}=\overline{am}$. In other words, $\mathfrak{a}(M/N)$ is the image of $\mathfrak{a}M$ under the canonical map $M\to M/N$. Now via the one-to-one correspondence between submodules of $M/N$ and submodules of $M$ containing $N$, you see that $\mathfrak{a}(M/N)$ corresponds to $\mathfrak{a}M+N$.
Thank you, it's what I wanted. I understood like this: $\mathfrak{a}(M/N)$=$f(\mathfrak{a}M)$=$\mathfrak{a}M/\ker(f)$=$\mathfrak{a}M/ (\mathfrak{a}M \cap N)$=$(\mathfrak{a}M+N)/N$ by 1st,2nd isomorphism theorem and where $f:\mathfrak{a}M \hookrightarrow M \to M/N$ – Gobi May 26 '11 at 12:32