# $\alpha (M/N) = (\alpha M + N) / N$

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

There will be many ways, for example, define a map $f:\alpha M + N \to \alpha (M/N)$ 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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## 1 Answer

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. I am going to denote the ideal by $\mathfrak{a}$. 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 $\mathfrak{a}M+N$.

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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
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