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 is a (locally) free
-module of rank 1, then
. Multiplication by
is injective on
if and only if
is not a zero-divisor, and it is surjective if and only if
. In particular, if it is surjective, it is also injective.
Lemma. Suppose is a ring homomorphism, such that
is locally free of rank 1 over
. Then
is an isomorphism.
Proof. The question is local on , so (after replacing
with a suitable localisation) we may assume that
is free of rank 1. Let
be a basis element.
Then we can write for some
, hence
. Also, we can write
for some
, hence
. Therefore,
is surjective, so by the remark above, it is an isomorphism.
Using a different argument, we can also prove:
Lemma. Suppose is a ring homomorphism, such that
is locally free of rank
over
. Then
is injective.
Proof. Since is locally free of rank
, it is faithfully flat over
. Thus it suffices to prove that
is injective. But this map admits a contraction
given by
.