# Relative Frobenius

This is the third 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. The proof is given in my previous post. In this post, I will present the application that caused me to look at the result in the first place.

Remark. Recall that a morphism of schemes is smooth of relative dimension if all of the following hold:

• is locally of finite presentation;
• is flat;
• all nonempty fibres have dimension ;
• is locally free of rank .

If is smooth of relative dimension 0, then is étale. In this case, the third condition follows from the other ones.

Example. To show that the third condition is really necessary, consider any finite inseparable field extension. This is clearly flat of finite presentation. Moreover, is a vector space of dimension , with basis given by a -basis of . Yet the (unique) fibre has dimension 0.

Definition. Let be a scheme of prime characteristic . Then the absolute Frobenius on is given by the morphism which is the identity on the underlying topological space, and is given by on . This definition makes sense because for a ring of characteristic , the Frobenius induces the identity on .

Definition. Suppose that is a morphism of schemes of characteristic . Then the absolute Frobenius factors through , and therefore induces a morphism in the following diagram

where the square is a pullback (i.e. , where is viewed as an -scheme along ). The morphism is called the relative Frobenius of over .

Lemma. Assume is étale, with a scheme of characteristic . Then is an isomorphism. In other words, the square

is a pullback.

Proof. Note that is universally bijective, hence so is . Similarly, is universally bijective. Therefore so is , since .

On the other hand, is étale, hence by base change so is . But any map between schemes étale over is étale (see Tag 02GW, or for a nice geometric proof taken from Milne’s book on étale cohomology, see Corollary 1.1.9 of my Master’s Thesis), so in particular is étale.

Now is étale and universally bijective, so the result follows from my previous post.

Remark. Recall (see Tag 054L) that if is smooth of relative dimension , then around every there exist ‘smooth coordinates’ in the following sense: there exist affine opens , with , such that factors as

where is étale. In particular, this forces , by the first fundamental exact sequence.

Corollary. Assume is smooth of relative dimension , with a scheme of characteristic . Then is locally free of rank .

Proof. The question is local on both and . By the remark above, we may assume is étale over , with both and affine. We have a diagram

where the horizontal compositions are the absolute Frobenii on and respectively. Here, denotes the unique map making the top right square commutative. (Exercise: use the various universal properties to show that the top left square commutes).

The bottom right square and the right large rectangle are pullback squares, hence so is the top right square. The top large rectangle is a pullback by the lemma above. Hence, since the top right square is a pullback, so is the top left square. Hence, it suffices to prove the case , since the result is stable under base change.

But in this case, if , then , and , with the relative Frobenius given by the -linear (!) map

But in this case the result is clear: an explicit basis is

# Étale and universally injective

This is the second 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 the actual result, and the next post will contain a fun application.

Remark. Recall that a morphism of schemes is étale if is flat and locally of finite presentation, and .

Lemma. Suppose is étale and universally injective. Then is an open immersion.

Proof. Since étale morphisms are open and surjectivity is stable under base change, we may assume that is universally bijective; then we have to prove that is an isomorphism. Since étale morphisms are open, is in fact a universal homeomorphism. By Tag 04DE, is affine.

The question is local on , so we may assume is affine, and hence so is . Say is induced by . Now is proper and affine, hence finite. Moreover, since is finitely presented and finite as -algebra, and is a finitely presented -module, it is also a finitely presented -module (Tag 0564).

Now is flat of finite presentation over , hence locally free (actually, we need the slightly stronger result that I mention in the first remark; see Tag 00NX for statement and proof). Since the question is local, we may assume is free of rank .

Now let be a geometric point; that is, let be a map to an algebraically closed field. Then the tensor product is étale of dimension over . Hence, is a union of points. Since is universally bijective, we have . Then the result follows from my previous post.

# 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 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 .