The Stacks project

Lemma 56.3.8. Let $A$ and $B$ be rings. If

\[ F : \text{Mod}_ A \longrightarrow \text{Mod}_ B \]

is an equivalence of categories, then there exists an isomorphism $A \to B$ of rings and an invertible $B$-module $L$ such that $F$ is isomorphic to the functor $M \mapsto (M \otimes _ A B) \otimes _ B L$.

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

\[ \psi : K \otimes _ B K' \longrightarrow A \]

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

\[ \psi ' : K' \otimes _ A K \longrightarrow B \]

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

\[ K \xrightarrow {\psi ^{-1} \otimes \text{id}_ K} K \otimes _ B K' \otimes _ A K \xrightarrow {\text{id}_ K \otimes \psi '} K \]

The composition is an isomorphism and given by

\[ k \longmapsto \sum x_ i \psi '(y_ i \otimes k) \]

We conclude this automorphism factors as

\[ K \to B^{\oplus n} \to K \]

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

\[ A = K \otimes _ B K' \longrightarrow (B/\mathfrak m \otimes _ B K')^{\oplus 2} \]

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. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0GP0. Beware of the difference between the letter 'O' and the digit '0'.