My brain has refused to understand tensor products for a very long time, so I wrote up a solution for this classical exercise accompanied by my elementary questions.
Exercise: Let $A$ be a ring, $\mathfrak{a}$ an ideal, $M$ an $A$-module. Show that $A/\mathfrak{a} \otimes M \cong M/\mathfrak{a} M$.
Proof: Consider long exact sequence
$$ 0 \rightarrow \mathfrak{a} \rightarrow A \rightarrow A/\mathfrak{a} \rightarrow 0 $$
It is exact, because it is exact at
- $\mathfrak{a}$, since the image of $0$ is mapped to $0$ in $A$ under inclusion.
- $A$, since $\mathfrak{a}$ is by definition the kernel of the quotient map $A\rightarrow A/\mathfrak{a}$,
- $A/\mathfrak{a}$, since the quotient map is surjective and everything is mapped to $0$.
If we tensor this sequence with $M$, we get
$$ 0\otimes M\rightarrow \mathfrak{a}\otimes M \rightarrow A\otimes M \rightarrow A/\mathfrak{a}\otimes M \rightarrow 0\otimes M$$
- We can say $0\otimes M = 0$, because $\forall m \in M : B(0,m) = B(0+0,m) = 2B(0,m) \implies B(0,m) = 0$. And since all elements of $0 \otimes M$ are linear combinations of pure tensors $0\otimes m$, they are all equal to $0$? There is probably a simpler way to see this...
- Next $\mathfrak{a}\otimes M = \mathfrak{a}M$ because $\forall a \in\mathfrak{a}: B(a,m)=aB(1,m)=B(1,am)$, so $\mathfrak{a}\otimes M \cong 1\otimes \mathfrak{a}M \cong \mathfrak{a}M$?
- Now $A\otimes M \cong M$ by the same principle as above. In my mind, if $M$ is an $A$-module, then it contains a copy of $A$ by definition if we consider scalar multiplication of $A \times \{1\} \rightarrow M$, so tensoring $M$ by something that is already part of $M$ does nothing?
Since tensor product is right-exact, they are supposed to preserve the exactness of the right part of the diagram, implying that this diagram is exact...?
$$\mathfrak{a} M \rightarrow M \rightarrow A/\mathfrak{a}\otimes M \rightarrow 0$$
I'm confused, if I have an exact long sequence and tensor it, how do I understand what part of the sequence stays exact? Is it enough to exclude the left-most object from the diagram? What if this object isn't zero?
Assuming this diagram is exact, the map $M \rightarrow A/\mathfrak{a}\otimes M$ is surjective and $ker(M \rightarrow A/\mathfrak{a}\otimes M) = \mathfrak{a} M$, so it follows from isomorphism theorems that $A/\mathfrak{a}\otimes M = M/ker(M \rightarrow A/\mathfrak{a}\otimes M) = M/\mathfrak{a} M$ and we're done. Or are we? In this step I am confused by the silent treatment of the map $M \rightarrow A/\mathfrak{a}\otimes M$, so I tried to calculate it by hand.
What exactly is the map $M\rightarrow A/\mathfrak{a}\otimes M$? It comes from the quotient map $A \rightarrow A/\mathfrak{a}, a \mapsto a +\mathfrak{a}$ which was then tensored with $M$ and it became $(a\otimes m) \mapsto (a+\mathfrak{a}\otimes m)$. We've shown that it is the same as $(a\otimes m) \mapsto (1\otimes am)\mapsto (1\otimes am + \mathfrak{a}m)$, so the kernel is $\mathfrak{a}m$ and the final answer is $A/\mathfrak{a} \otimes M \cong M/\mathfrak{a}M$...?