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Suppose $A$ is a commutative ring, $\mathfrak{a} \subset A$ some ideal and $M$ an $A$-module.

Is $(A\otimes _A M) / (\mathfrak{a} \otimes _A M) \cong M/ \mathfrak{a} M$ true?

I know that there is an isomorphism between $A\otimes _A M$ and $M$ induced by $A \times M \rightarrow M, (a,m) \mapsto am$, but I suspect that the similar map $\mathfrak{a} \otimes _A M \rightarrow \mathfrak{a} M$ is only surjective.

Now I think I proved the statement in question via considering the kernel of the surjection $A \otimes M \rightarrow M \rightarrow M/\mathfrak{a} M$ which should be equal to $\mathfrak{a} \otimes _A M$, but I am not sure since I'm pretty bad with the tensor product so far. Thanks in advance for giving me reassurance/proving me wrong :-)

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Perhaps some parenthesis are in order... –  Asaf Karagila Jan 2 '12 at 22:15
    
Right, I hope this is better. –  Stefan Jan 2 '12 at 22:19
2  
Unless $M$ is flat, it is not true that the map $\pi : {\frak{a}} \otimes_A M \rightarrow A \otimes_A M$ is injective (right ?). So by definition $(A \otimes_A M)/({\frak{a}} \otimes M)$ must be $(A \otimes_A M)/\pi({\frak{a}} \otimes M)$. But you proved that via the canonical isomorphism $A \otimes M \cong M$, you have $Im(\pi) \cong {\frak{a}}M$. –  user10676 Jan 2 '12 at 23:46

1 Answer 1

up vote 7 down vote accepted

The isomorphism $(A\otimes _A M) / (\mathfrak{a} \otimes _A M) \cong M/ \mathfrak{a} M$ is neither true nor false!
Indeed, the left side does not make sense, since the $A$-module $\mathfrak{a} \otimes _A M$ is not a submodule of $A\otimes _A M$. Let's analyze this.

Start with the exact sequence $0\to \mathfrak{a}\to A\to A/ \mathfrak{a} \to 0$.
Tensoring it with $M$ you get the exact sequence $\mathfrak{a}\otimes_A M \to A \otimes _A M \to A/ \mathfrak{a}\otimes_A M \to 0$ or
$\mathfrak{a}\otimes_A M \stackrel {f}{\to} M \to M/ \mathfrak{a} M \to 0$ so that you get the correct isomorphism
$$M/f(\mathfrak{a}\otimes_A M)\simeq M/ \mathfrak{a} M$$

The point to keep in mind is that $f:\mathfrak{a}\otimes_A M \stackrel {f}{\to} M:\alpha\otimes m\mapsto \alpha.m$ is not injective in general (unless you know that $M$ is flat over $A$) and it is thus false to consider that $\mathfrak{a}\otimes_A M$ is a submodule of $M$ (or of $A\otimes_AM)$.

An explicit counterexample to the injectivity of $f$
Let $p\in \mathbb Z$ be a prime, $A=\mathbb Z/(p^2)$, $M=\mathbb Z/(p)$ and $\mathfrak{a}=pA$.
The morphism $f:\mathfrak{a}\otimes _A M\to M:pa\otimes m\mapsto pam=0$ is zero hence $f$ is not injective because $\mathfrak{a}\otimes _A M=\mathfrak{a}\otimes _{A/p} M\neq 0$ .
[The last equality is due to the fact that $p$ kills both $\mathfrak{a}$ and $M$, so that the tensor products over $A$ or over $A/p$ are the same. The inequality is due to the fact that $A/p=\mathbb F_p$, a field: over a field the tensor product of two nonzero vector spaces is nonzero.]

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I think your first exact sequence is off. –  Sean Tilson Jan 3 '12 at 2:58
    
Thanks @Sean, corrected. –  Georges Elencwajg Jun 25 '12 at 5:23

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