Lemma 15.6.9. Let $A, A', B, B', I$ be as in Situation 15.6.1. The category of finite projective $B'$-modules is equivalent to the full subcategory of $\text{Mod}_ B \times _{\text{Mod}_ A} \text{Mod}_{A'}$ consisting of triples $(N, M', \varphi )$ with $N$ finite projective over $B$ and $M'$ finite projective over $A'$.

Proof. Recall that a module is finite projective if and only if it is finitely presented and flat, see Algebra, Lemma 10.78.2. Using Lemmas 15.6.8 and 15.6.7 we reduce to showing that $N' = N \times _{\varphi , M} M'$ is a $B'$-module of finite presentation if $N$ finite projective over $B$ and $M'$ finite projective over $A'$.

By Lemma 15.6.7 the module $N'$ is finite over $B'$. Choose a surjection $(B')^{\oplus n} \to N'$ with kernel $K'$. By base change we obtain maps $B^{\oplus n} \to N$, $(A')^{\oplus n} \to M'$, and $A^{\oplus n} \to M$ with kernels $K_ B$, $K_{A'}$, and $K_ A$. There is a canonical map

$K' \longrightarrow K_ B \times _{K_ A} K_{A'}$

On the other hand, since $N' = N \times _{\varphi , M} M'$ and $B' = B \times _ A A'$ there is also a canonical map $K_ B \times _{K_ A} K_{A'} \to K'$ inverse to the displayed arrow. Hence the displayed map is an isomorphism. By Algebra, Lemma 10.5.3 the modules $K_ B$ and $K_{A'}$ are finite. We conclude from Lemma 15.6.7 that $K'$ is a finite $B'$-module provided that $K_ B \to K_ A$ and $K_{A'} \to K_ A$ induce isomorphisms $K_ B \otimes _ B A = K_ A = K_{A'} \otimes _{A'} A$. This is true because the flatness assumptions implies the sequences

$0 \to K_ B \to B^{\oplus n} \to N \to 0 \quad \text{and}\quad 0 \to K_{A'} \to (A')^{\oplus n} \to M' \to 0$

stay exact upon tensoring, see Algebra, Lemma 10.39.12. $\square$

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