# Isomorphism modulo torsion is stable under tensor products

Let $\phi : A \to B$ be a homomorphism of grades $k$-algebras and $N$ be a graded $A$-right module. Assume that $\mathrm{ker}(\phi)$ and $\mathrm{coker}(\phi)$ are torsion $A$-right modules, i.e. $\phi$ is an isomorphism modulo $A$-torsion. Why is then $N \to N \otimes_A B$ an isomorphism modulo $A$-torsion?

This is used in a proof in the paper "Noncommutative projective schemes" by Artin-Zhang (Prop. 2.5). The authors claim that this follows from the fact that $\mathrm{Tor}^A_i(N,L)$ is $B$-torsion if $L$ is an $(A,B)$-bimodule which is $B$-torsion. But I don't see how one can use this. Of course it is tempting to write down a short exact sequence of $A$-modules and then consider the associated long exact sequence of Tor-groups, but neither $0 \to \mathrm{ker}(\phi) \to A \to \mathrm{im}(\phi) \to 0$ nor $0 \to \mathrm{im}(\phi) \to B \to \mathrm{coker}(\phi) \to 0$ seem to be helpful. The authors just write down the exact sequence $0 \to \mathrm{ker}(\phi) \to A \to B \to \mathrm{coker}(\phi) \to 0$, but this doesn't produce a sequence of Tor-groups. Therefore, I don't understand the argument.

I understand why $N \to N \otimes_A B$ is an epimorphism modulo $A$-torsion: The cokernel is $N \otimes_A \mathrm{coker}(\phi) = \mathrm{Tor}_0^A(N,\mathrm{coker}(\phi))$, and now we use the fact above (with $A=B$). Why is the kernel also torsion?

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