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 .