$\operatorname{Tor}^A_1(A/\mathfrak a,N) = 0$?

Question

Let $A$ be a commutative ring with unity, and $\mathfrak a$ be an ideal.

Then we have the exact sequence: $$0 \to \mathfrak a \to A \to A/\mathfrak a \to 0$$

Tensoring with $N$, an $A$-module, gives us an exact sequence: $$\operatorname{Tor}^A_1(A,N) \to \operatorname{Tor}^A_1(A/\mathfrak a,N) \to \mathfrak a \otimes_A N \to A \otimes_A N \to (A/\mathfrak a) \otimes_A N \to 0$$

Since $A$ is projective, $\operatorname{Tor}^A_1(A,N) = 0$; doing canonical reductions gives: $$0 \to \operatorname{Tor}^A_1(A/\mathfrak a,N) \to \mathfrak a N \to N \to (A/\mathfrak a) \otimes_A N \to 0$$

Since $\mathfrak a N$ is a submodule of $N$ and the map is just inclusion, the map is injective, so $\operatorname{Tor}^A_1(A/\mathfrak a,N) = 0$.

But this result would be ridiculous for a lot of reasons. I can't spot the error somehow. Am I missing something obvious?

Answer 1

The error is that $\mathfrak{a}\otimes_A N$ is not isomorphic to $\mathfrak{a}N$ in general.

As a simple example, let $A=\Bbb Z$, $\mathfrak{a}=2\Bbb Z$ and $N=\Bbb Z/2\Bbb Z$.

Answer 2

$\mathfrak{a}\otimes_A N$ is not the same thing as $\mathfrak{a}N$. Indeed, they will fail to be isomorphic via the natural map between them iff $\operatorname{Tor}^A_1(A/\mathfrak a,N)$ is nontrivial, as your argument shows. The image of the map $\mathfrak{a}\otimes_A N \to N$ is $\mathfrak{a}N$ but it is not always injective.