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?
 A: I think the point is to look at the two long exact sequences associated to $0 \to A/T \to B \to T^\prime \to 0$ and $0 \to T \to A \to A/T \to 0$, then use the fact that $N \otimes_A T,\, N \otimes_A T^\prime$ are A-torsion together with the fact that $\pi_A \colon \operatorname{Gr}A \to \operatorname{QGr}A$ is exact and kills A-torsion.  Proposition 2.4 (5) comes into play when looking at the long exact sequence associated to the first short exact sequence.
Applying $\pi_A$ to the long exact sequence
$$\cdots \to N \otimes_A T \to N \to N \otimes_A A/T \to 0$$
yields the long exact sequence of $\operatorname{QGr} A$
$$\cdots \to \pi_A(N \otimes_A T) \cong 0 \to \pi_A(N) \overset{\sim}\to \pi_A(N \otimes_A A/T) \to 0$$
and the isomorphism $\pi_A(N) \cong \pi(N \otimes_A A/T)$.
Now, the long exact sequence associated to $0 \to A/T \to B \to T^\prime \to 0$ is 
$$\cdots \to \operatorname{\underline{Tor}}^A_1(N,T^\prime) \to N \otimes_A A/T \to N \otimes_A B \to N \otimes_A T^\prime \to 0.$$
By 2.4(5), $\operatorname{\underline{Tor}}^A_1(N,T^\prime)$ is $B$-torsion, hence $A$-torsion.  Applying $\pi_A$ yields the long exact sequence of $\operatorname{QGr}A$
$$\cdots \to \pi_A\left(\operatorname{\underline{Tor}}^A_1(N,T^\prime)\right) \cong 0 \to \pi_A(N \otimes_A A/T) \overset{\sim}\to \pi_A(N \otimes_A B) \to \pi_A(N \otimes_A T^\prime) \cong 0$$
and the isomorphism $\pi_A(N \otimes_A B) \cong \pi(N \otimes_A A/T)$.
Therefore 
$$\pi_A(N) \cong \pi_A(N \otimes_A A/T) \cong \pi_A(N \otimes_A B).$$
