Why $M/\mathfrak{a}M \oplus M/\mathfrak{b}M \simeq M/(\mathfrak{a \cap b})M$? Let $M$ be an $A$-module and let $\mathfrak{a}$ and $\mathfrak{b}$ be coprime ideals of A.
I must show that $M/ \mathfrak{a}M \oplus M/ \mathfrak{b}M \simeq M/ (\mathfrak{a \cap b})M$.
My attempt is the following:
Let $x \in M/ \mathfrak{a}M \oplus M/ \mathfrak{b}M$,then $x = [y]+[z]$, where $[y] = y+\mathfrak{a}M $ and $[z]=z + \mathfrak{b}M $, $y,z \in M$.
So, $x = y+z+ \mathfrak{a}M +\mathfrak{b}M $.
$\mathfrak{a}M +\mathfrak{b}M =\{z | z=am_1+bm_2, a \in \mathfrak{a}, b \in \mathfrak{b}   \} $. But then I don't know how to continue.
Is this approach correct? Or is there another way to prove it?
Thanks
 A: There is an obvious homomorphism $\varphi\colon M\to M/\mathfrak{a}M\oplus M/\mathfrak{b}M$, namely $\varphi(x)=(x+\mathfrak{a}M,x+\mathfrak{b}M)$.
The kernel is obviously $\mathfrak{a}M\cap\mathfrak{b}M$. You want to prove that $\mathfrak{a}M\cap\mathfrak{b}M=(\mathfrak{a}\cap\mathfrak{b})M$, using the known fact that $\mathfrak{a}\cap\mathfrak{b}=\mathfrak{a}\mathfrak{b}$ when $\mathfrak{a}+\mathfrak{b}=A$.

 We can write $1=a+b$, with $a\in\mathfrak{a}$ and $b\in\mathfrak{b}$. If $x\in\mathfrak{a}M\cap\mathfrak{b}M$, we have $$x=ax+bx\in (\mathfrak{a}\mathfrak{b})M$$ because $x\in\mathfrak{b}M$ implies $ax\in\mathfrak{a}\mathfrak{b}M$ and $x\in\mathfrak{a}M$ implies $bx\in\mathfrak{b}\mathfrak{a}M$. The other inclusion $\mathfrak{a}\mathfrak{b}M\subseteq\mathfrak{a}M\cap\mathfrak{b}M$ is obvious. Therefore $\ker\varphi=(\mathfrak{a}\cap\mathfrak{b})M$.

Now you want to show that $\varphi$ is also surjective, which is the (abstract version of the) Chinese remainder theorem. If $x,y\in M$, then
$$
(x+\mathfrak{a}M,y+\mathfrak{b}M)=(z+\mathfrak{a}M,z+\mathfrak{b}M)
$$

 where $z=bx+ay$.

A: Hint:
Consider the short exact sequence:
$$0\longrightarrow A/\mathfrak a\cap\mathfrak b\longrightarrow A/\mathfrak a\times A/\mathfrak b\longrightarrow A/\mathfrak a+\mathfrak b\longrightarrow 0 $$
and tensor by $M$.
A: In fact it's a bit easier to go backwards. Let $[x]\in M/(\mathfrak{a}\cap\mathfrak{b})M$. Since $\mathfrak{a}$ contains the ideal $\mathfrak{a}\cap\mathfrak{b}$, we can restrict $[x]$ to a class $[x_1]\in M/\mathfrak{a}M$, and similarly to a class $[x_2]\in M/\mathfrak{b}M$. We have to show this map is an isomorphism.
Why is it injective? Well, suppose $[x_1] = 0$ and $[x_2] = 0$. That means $x$ was originally in $\mathfrak{a}M$ as well as in $\mathfrak{b}M$, so you are done.
Why is it surjective? Take $([x_1],[x_2])$ on the LHS, and lift them to $x_1,x_2\in M$. Because $\mathfrak{a}$ and $\mathfrak{b}$ are coprime, there are $a\in \mathfrak{a}$ and $b\in\mathfrak{b}$ for which $a+b = 1$. What happens to the element $x = x_2a+x_1b$?
(EDIT: I should really say that you should define the map from $M$ and then use one of the isomorphism theorems, that makes it much clearer)
