Relative Frobenius

Posted on 28 June 2015 by Remy

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 $f \colon X \ra Y$ of schemes is smooth of relative dimension $r$ if all of the following hold:

$f$ is locally of finite presentation;
$f$ is flat;
all nonempty fibres have dimension $r$ ;
$\Omega_{X/Y}$ is locally free of rank $r$ .

If $f$ is smooth of relative dimension 0, then $f$ 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, $\Omega_{L/K}$ is a vector space of dimension $r > 0$ , with basis given by a $p$ -basis of $L/K$ . Yet the (unique) fibre has dimension 0.

Definition. Let $S$ be a scheme of prime characteristic $p > 0$ . Then the absolute Frobenius on $S$ is given by the morphism $\Frob_S \colon S \rA S$ which is the identity on the underlying topological space, and is given by $x \rm x^p$ on $\O_S$ . This definition makes sense because for a ring $A$ of characteristic $p$ , the Frobenius $\Frob_A \colon A \ra A$ induces the identity on $\Spec A$ .

Definition. Suppose that $X \ra S$ is a morphism of schemes of characteristic $p$ . Then the absolute Frobenius $\Frob_X$ factors through $\Frob_S$ , and therefore induces a morphism $\Frob_{X/S} \colon X \ra X^{(p)}$ in the following diagram

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where the square is a pullback (i.e. $X^{(p)} = X \times_S S$ , where $S$ is viewed as an $S$ -scheme along $\Frob_S$ ). The morphism $\Frob_{X/S}$ is called the relative Frobenius of $X$ over $S$ .

Lemma. Assume $f \colon X \ra S$ is étale, with $S$ a scheme of characteristic $p$ . Then $\Frob_{X/S}$ is an isomorphism. In other words, the square

is a pullback.

Proof. Note that $\Frob_S$ is universally bijective, hence so is $g \colon X^{(p)} \ra X$ . Similarly, $\Frob_X$ is universally bijective. Therefore so is $\Frob_{X/S}$ , since $g \circ \Frob_{X/S} = \Frob_X$ .

On the other hand, $f$ is étale, hence by base change so is $X^{(p)} \ra S$ . But any map between schemes étale over $S$ 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 $\Frob_{X/S}$ is étale.

Now $\Frob_{X/S}$ is étale and universally bijective, so the result follows from my previous post. $\qedsymbol$

Remark. Recall (see Tag 054L) that if $f \colon X \ra S$ is smooth of relative dimension $r$ , then around every $x \in X$ there exist ‘smooth coordinates’ in the following sense: there exist affine opens $U \sbq X$ , $V \sbq Y$ with $f(U) \sbq V$ , such that $f|_U \colon U \ra V$ factors as

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where $\pi$ is étale. In particular, this forces $\Omega_{U/V} = \bigoplus_{i=1}^r dx_i \O_U$ , by the first fundamental exact sequence.

Corollary. Assume $f \colon X \ra S$ is smooth of relative dimension $r$ , with $S$ a scheme of characteristic $p$ . Then $\Frob_{X/S}$ is locally free of rank $p^r$ .

Proof. The question is local on both $X$ and $S$ . By the remark above, we may assume $X$ is étale over $\A^r_S$ , with both $X$ and $S$ affine. We have a diagram

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where the horizontal compositions are the absolute Frobenii on $X$ and $\A^r_S$ respectively. Here, $\pi^{(p)}$ 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 $X = \A^r_S$ , since the result is stable under base change.

But in this case, if $S = \Spec A$ , then $X = \Spec A[x_1, \ldots, x_r]$ , and $X^{(p)} = \Spec A[y_1,\ldots,y_r]$ , with the relative Frobenius given by the $A$ -linear (!) map

$\begin{align*} A[y_1,\ldots,y_r] &\rA A[x_1,\ldots,x_r]\\ y_i &\rM x_i^p. \end{align*}$

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

$\{x_i^j\ |\ i\in\{1,\ldots,r\}, j\in\{0,\ldots,p-1\}\}.$

$\qedsymbol$

Étale and universally injective

Posted on 28 June 2015 by Remy

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 $f \colon X \ra Y$ of schemes is étale if $f$ is flat and locally of finite presentation, and $\Omega_{X/Y} = 0$ .

Lemma. Suppose $f \colon X \ra S$ is étale and universally injective. Then $f$ is an open immersion.

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

The question is local on $S$ , so we may assume $S$ is affine, and hence so is $X$ . Say $f$ is induced by $g \colon A \ra B$ . Now $f$ is proper and affine, hence finite. Moreover, since $B$ is finitely presented and finite as $A$ -algebra, and $B$ is a finitely presented $B$ -module, it is also a finitely presented $A$ -module (Tag 0564).

Now $B$ is flat of finite presentation over $A$ , 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 $B$ is free of rank $r$ .

Now let $\bar{s} \ra S$ be a geometric point; that is, let $A \ra \bar{K}$ be a map to an algebraically closed field. Then the tensor product $B \tens_A \bar{K}$ is étale of dimension $r$ over $\bar{K}$ . Hence, $X_{\bar s} = X \times_{S} \bar{s}$ is a union of $r$ points. Since $f$ is universally bijective, we have $r = 1$ . Then the result follows from my previous post. $\qedsymbol$

Locally free algebras

Posted on 28 June 2015 by Remy

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’

Monthly Archives: June 2015

Relative Frobenius

Étale and universally injective

Locally free algebras