# A basic isomorphism of modules (useful for Corollary 2.7 of Atiyah and MacDonald).

Let $M$ be an $A$-module, $N$ a submodule of $M$, $\mathfrak{a} \subseteq A$ an ideal such that $M = \mathfrak{a}M + N$. Then

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

I am having troubles in understanding why this isomorphism holds...any hint? Of course I tried many times using the classical theorems of isomorphism of modules, but always failing...I think there should be a point that I am missing...

Ps: my question is useful for a better understanding of Corollary 2.7 from Atiyah and MacDonald - I'll add this information maybe useful for someone in the future :)

Let us try this :

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

Now, we always have $\mathfrak{a}(M/N)\subseteq M/N$ so the only thing to show to get the equality is to show the reverse inclusion, take $m_0\in M$ then there exists $a\in\mathfrak{a}$, $m\in M$ and $n\in N$ such that :

$$m_0=am+n$$

Now modulo $N$ we have $[m_0]=[am]$ hence $[m_0]\in \mathfrak{a}M/N=\mathfrak{a}(M/N)$ and you are done. Here you need something that, I think is assumed but I state it explicitely : $N$ is a $\mathfrak{a}$-module (otherwise you could not write something like $\mathfrak{a}(M/N)$). Finally any $[m_0]\in M/N$ is in $\mathfrak{a}(M/N)$ and you are done.

• Wow is was so easy...!!! Thank you a lot!!! :-) – user233650 May 26 '15 at 11:29

It's not an isomorphism, it's equality!

The module $\mathfrak{a}(M/N)$ is generated by the elements of the form $a(x+N)$, for $a\in\mathfrak{a}$ and $x\in M$.

Clearly $a(x+N)=ax+N\in (\mathfrak{a}M+N)/N$, so one inclusion is settled up.

Conversely, an element in $(\mathfrak{a}M+N)/N$ is of the form $$y+z+N$$ for some $y\in\mathfrak{a}M$ and $z\in N$. Then $y+z+N=y+N$ and $$y=\sum_{i=1}^n a_ix_i$$ for $a_i\in\mathfrak{a}$ and $x_i\in M$ ($i=1,2,\dots,n$). Therefore $$y+N=\sum_{i=1}^n a_i(x_i+N)\in\mathfrak{a}(M/N)$$

• Thank you @egreg, absolutely very useful!!! I accepted the answer from Clément Guérin because he answered before, but of course I appreciate you explanation! Cheers – user233650 May 26 '15 at 11:33