Locally free algebras

This is the first in a three-part post about a proof that I contributed to the Stacks project. The result was already there, but I found a slightly easier proof. This post contains a preliminary lemma; the second post contains the result; and the third one contains the application that I was interested in.

Remark. If $M$ is a (locally) free $A$ -module of rank 1, then $\End_A(M) = A$ . Multiplication by $a \in A$ is injective on $M$ if and only if $a$ is not a zero-divisor, and it is surjective if and only if $a \in A\x$ . In particular, if it is surjective, it is also injective.

Lemma. Suppose $f \colon A \ra B$ is a ring homomorphism, such that $B$ is locally free of rank 1 over $A$ . Then $f$ is an isomorphism.

Proof. The question is local on $A$ , so (after replacing $A$ with a suitable localisation) we may assume that $B$ is free of rank 1. Let $b$ be a basis element.

Then we can write $1 = f(a)b$ for some $a \in A$ , hence $b \in B\x$ . Also, we can write $b^2 = f(c)b$ for some $c \in A$ , hence $b = f(c)$ . Therefore, $f$ is surjective, so by the remark above, it is an isomorphism. $\qedsymbol$

Using a different argument, we can also prove:

Lemma. Suppose $f \colon A \ra B$ is a ring homomorphism, such that $B$ is locally free of rank $r > 0$ over $A$ . Then $f$ is injective.

Proof. Since $B$ is locally free of rank $r$ , it is faithfully flat over $A$ . Thus it suffices to prove that $f \tens 1 \colon B \ra B \tens_A B$ is injective. But this map admits a contraction $B \tens_A B \ra B$ given by $b_1 \tens b_2 \rm b_1b_2$ . $\qedsymbol$

Lovely little lemmas

Or maybe ‘amusing observations’

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