Lemma 38.23.2 (05PE)—The Stacks project

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Table of contents
Part 2: Schemes
Chapter 38: More on Flatness
Section 38.23: Flattening a map
Lemma 38.23.2 (cite)

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Lemma 38.23.2. Let $A$ be a ring. Let $u : M \to N$ be a surjective map of $A$ -modules. If $M$ is projective as an $A$ -module, then there exists an ideal $I \subset A$ such that for any ring map $\varphi : A \to B$ the following are equivalent

$u \otimes 1 : M \otimes _ A B \to N \otimes _ A B$ is an isomorphism, and
$\varphi (I) = 0$ .

Proof. As $M$ is projective we can find a projective $A$ -module $C$ such that $F = M \oplus C$ is a free $A$ -module. By replacing $u$ by $u \oplus 1 : F = M \oplus C \to N \oplus C$ we see that we may assume $M$ is free. In this case let $I$ be the ideal of $A$ generated by coefficients of all the elements of $\mathop{\mathrm{Ker}}(u)$ with respect to some (fixed) basis of $M$ . The reason this works is that, since $u$ is surjective and $\otimes _ A B$ is right exact, $\mathop{\mathrm{Ker}}(u \otimes 1)$ is the image of $\mathop{\mathrm{Ker}}(u) \otimes _ A B$ in $M \otimes _ A B$ . $\square$

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