The Stacks project

Proof. Since an equivalence commutes with all colimits, we see that Lemmas 56.3.1 applies. Let $K$ be the $A \otimes _\mathbf {Z} B$-module such that $F$ is isomorphic to the functor $M \mapsto M \otimes _ A K$. Let $K'$ be the $B \otimes _\mathbf {Z} A$-module such that a quasi-inverse of $F$ is isomorphic to the functor $N \mapsto N \otimes _ B K'$. By Remark 56.3.4 and Lemma 56.3.3 we have an isomorphism

of $A \otimes _\mathbf {Z} A$-modules. Similarly, we have an isomorphism

of $B \otimes _\mathbf {Z} B$-modules. Choose an element $\xi = \sum _{i = 1, \ldots , n} x_ i \otimes y_ i \in K \otimes _ B K'$ such that $\psi (\xi ) = 1$. Consider the isomorphisms

The composition is an isomorphism and given by

We conclude this automorphism factors as

as a map of $B$-modules. It follows that $K$ is finite projective as a $B$-module.

We claim that $K$ is invertible as a $B$-module. This is equivalent to asking the rank of $K$ as a $B$-module to have the constant value $1$, see More on Algebra, Lemma 15.117.2 and Algebra, Lemma 10.78.2. If not, then there exists a maximal ideal $\mathfrak m \subset B$ such that either (a) $K \otimes _ B B/\mathfrak m = 0$ or (b) there is a surjection $K \to (B/\mathfrak m)^{\oplus 2}$ of $B$-modules. Case (a) is absurd as $K' \otimes _ A K \otimes _ B N = N$ for all $B$-modules $N$. Case (b) would imply we get a surjection

of (right) $A$-modules. This is impossible as the target is an $A$-module which needs at least two generators: $B/\mathfrak m \otimes _ B K'$ is nonzero as the image of the nonzero module $B/\mathfrak m$ under the quasi-inverse of $F$.

Since $K$ is invertible as a $B$-module we see that $\mathop{\mathrm{Hom}}\nolimits _ B(K, K) = B$. Since $K = F(A)$ the action of $A$ on $K$ defines a ring isomorphism $A \to B$. The lemma follows. $\square$