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 $f \colon X \ra Y$ 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

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

$\qedsymbol$

2 thoughts on “Relative Frobenius”

Urs Hartl on 10 February 2017 at 15:05 said:

Dear Remy,

thank you for posting this lemma. An important consequence of it is that for a Deligne-Mumford stack locally of finite type over S the relative Frobenius is finite. Namely by your lemma this holds over an étale presentation of the stack. I suggest to add this result to the stacks project. Would you like to take care of this?

Reply ↓
- Remy on 23 June 2017 at 04:22 said:
  
  Dear Urs,
  
  Thanks for your comment. I am not personally very involved with the Stacks project. But you can always ask Johan de Jong to include it (or even send him a draft that he can incorporate in the Stacks project).
  
  Reply ↓

Lovely little lemmas

Or maybe ‘amusing observations’

2 thoughts on “Relative Frobenius”

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