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

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