Union of hyperplanes over a finite field

Posted on 22 February 2022 by Remy

The following lemma is a (presumably well-known) result that Raymond Cheng and I happened upon while writing our paper Unbounded negativity on rational surfaces in positive characteristic (arXiv, DOI). Well, Raymond probably knew what he was doing, but to me it was a pleasant surprise.

Lemma. Let $q$ be a power of a prime $p$ , and let $x_0,\ldots,x_n \in \bar{\mathbf F}_q$ . Then $x_0,\ldots,x_n$ satisfy a linear relation over $\mathbf F_q$ if and only if

$\det \begin{pmatrix} x_0 & x_1 & \cdots & x_n \\ x_0^q & x_1^q & \cdots & x_n^q \\ \vdots & \vdots & \ddots & \vdots \\ x_0^{q^n} & x_1^{q^n} & \cdots & x_n^{q^n} \end{pmatrix} = 0.$

Proof. If $\sum_{i=0}^n c_ix_i = 0$ for $(c_0,\ldots,c_n) \in \mathbf F_q^n - \{0\}$ , then $c_i^{q^j} = c_i$ for all $i,j \in \{0,\ldots,n\}$ since $c_i \in \mathbf F_q$ . As $(-)^q \colon \bar{\mathbf F}_q \to \bar{\mathbf F}_q$ is a ring homomorphism, we find

$\begin{pmatrix} x_0 & x_1 & \cdots & x_n \\ x_0^q & x_1^q & \cdots & x_n^q \\ \vdots & \vdots & \ddots & \vdots \\ x_0^{q^n} & x_1^{q^n} & \cdots & x_n^{q^n} \end{pmatrix}\begin{pmatrix} c_0 \\ c_1 \\ \vdots \\ c_n \end{pmatrix} = 0,$

so the determinant is zero. Conversely, the union of $\mathbf F_q$ -rational hyperplanes $H \subseteq \mathbf P^n_{\mathbf F_q}$ is a hypersurface $Y$ of degree $|\check{\mathbf P}^n(\mathbf F_q)| = q^n + \ldots + q + 1$ (where $\check{\mathbf P}^n$ denotes the dual projective space parametrising hyperplanes in $\mathbf P^n$ ). Since the determinant above is a polynomial of the same degree $q^n + \ldots + q + 1$ that vanishes on all $\mathbf F_q$ -rational hyperplanes, we conclude that it is the polynomial cutting out $Y$ , so any $[x_0:\ldots:x_n] \in \mathbf P^n(\bar{\mathbf F_q})$ for which the determinant vanishes lies on one of the hyperplanes. $\qedsymbol$

Of course when the determinant is zero, one immediately gets a vector $(c_0,\ldots,c_n) \in \bar{\mathbf F}_q^{n+1} - \{0\}$ in the kernel. There may well be an immediate argument why this vector is proportional to an element of $\mathbf F_q^{n+1}$ , but the above cleverly circumvents this problem.

For concreteness, we can work out what this determinant is in small cases:

$n=0$ : a point $x_0 \in \bar{\mathbf F}_q$ only satisfies a linear relation over $\mathbf F_q$ if it is zero.
$n=1$ : the polynomial $x_0x_1^q-x_0^qx_1$ cuts out the $\mathbf F_q$ -rational points of $\mathbf P^1$ .
$n=2$ : the polynomial
$x_0x_1^qx_2^{q^2}+x_1x_2^qx_0^{q^2}+x_2x_0^qx_1^{q^2}-x_0^{q^2}x_1^qx_2-x_1^{q^2}x_2^qx_0-x_2^{q^2}x_0^qx_1$
cuts out the union of $\mathbf F_q$ -rational lines in $\mathbf P^2$ . This is the case considered in the paper.

Hodge diamonds that cannot be realised

Posted on 30 September 2020 by Remy

In Paulsen–Schreieder [PS19] and vDdB–Paulsen [DBP20], the authors/we show that any block of numbers

$\left(\begin{array}{ccccc} & & h^{n,n} & & \\ & \iddots & & \ddots & \\ h^{n,0} & & & & h^{0,n} \\ & \ddots & & \iddots & \\ & & h^{0,0} & & \end{array}\right) \in \big(\mathbf Z/m\big)^{(n+1)^2}$

satisfying $h^{0,0} = 1$ , $h^{p,q} = h^{n-p,n-q}$ , and $h^{p,q} = h^{q,p}$ (characteristic $0$ only) can be realised as the modulo $m$ reduction of a Hodge diamond of a smooth projective variety.

While preparing for a talk on [DBP20], I came up with the following easy example of a Hodge diamond that cannot be realised integrally, while not obviously violating any of the conditions (symmetry, nonnegativity, hard Lefschetz, …).

Lemma. There is no smooth projective variety (in any characteristic) whose Hodge diamond is

$\begin{array}{ccccc} & & 1 & & \\ & 1 & & 1 & \\ 0 & & 1 & & 0 \\ & 1 & & 1 & \\ & & 1.\!\! & & \end{array}$

Proof. If $\operatorname{char} k > 0$ , we have $\sum_{p+q = i}h^{p,q} \geq h_{\operatorname{dR}}^i \geq h_{\operatorname{cris}}^i$ , with equality for all $i$ if and only if the Hodge–de Rham spectral sequence degenerates and $H_{\operatorname{cris}}^i$ is torsion-free for all $i$ . Because $H^2_{\operatorname{cris}}$ contains an ample class, we must have equality on $h^2$ , hence everywhere because of how spectral sequences and universal coefficients work.

Thus, in any characteristic, we conclude that $h^1_{\operatorname{Weil}}(X) = 2$ , so $\dim \mathbf{Pic}_X^0 = 1$ and the same for $\dim \mathbf{Alb}_X$ . Thus, $X \to \mathbf{Alb}_X$ is a fibration, so a fibre and a relatively ample divisor are linearly independent in the Néron–Severi group, contradicting the assumption $h^{1,1}(X) = 1$ . $\qedsymbol$

Remark. In characteristic zero, the Hodge diamonds

$\begin{array}{ccccc} & & 1 & & \\ & a & & a & \\ 0 & & 1 & & 0 \\ & a & & a & \\ & & 1 & & \end{array}$

cannot occur for any $a \geq 1$ , by essentially the same argument. Indeed, the only thing left to prove is that the image $X \to \mathbf{Alb}_X$ cannot be a surface. If it were, then $X$ would have a global 2-form; see e.g. [Beau96, Lemma V.18].

This argument does not work in positive characteristic due to the possibility of an inseparable Albanese map. It seems to follow from Bombieri–Mumford’s classification of surfaces in positive characteristic that the above Hodge diamond does not occur in positive characteristic either, but the analysis is a little intricate.

Remark. On the other hand, the nearly identical Hodge diamond

$\begin{array}{ccccc} & & 1 & & \\ & a & & a & \\ 0 & & 2 & & 0 \\ & a & & a & \\ & & 1 & & \end{array}$

is realised by $C \times \mathbf P^1$ , where $C$ is a curve of genus $a$ . This is some evidence that the full inverse Hodge problem is very difficult, and I do not expect a full classification of which Hodge diamonds are possible (even for surfaces this might be out of reach).

References.

[Beau96] A. Beauville, Complex algebraic surfaces. London Mathematical Society Student Texts 34 (1996).

[DBP20] R. van Dobben de Bruyn and M. Paulsen, The construction problem for Hodge numbers modulo an integer in positive characteristic. Forum Math. Sigma (to appear).

[PS19] M. Paulsen and S. Schreieder, The construction problem for Hodge numbers modulo an integer. Algebra Number Theory 13.10, p. 2427–2434 (2019).

An interesting Noether–Lefschetz phenomenon

Posted on 20 January 2020 by Remy

The classical Noether–Lefschetz theorem is the following:

Theorem. Let $X \subseteq \mathbf P^3_{\mathbf C}$ be a very general smooth surface of degree $d \geq 4$ . Then the natural map $\Pic(\mathbf P^3) \to \Pic(X)$ is an isomorphism.

If $\mathscr X \to S$ is a smooth proper family over some base $S$ (usually of finite type over a field), then a property $\mathcal P$ holds for a very general $X = \mathscr X_s$ if there exists a countable intersection $U = \bigcap_i U_i \subseteq S$ of nonempty Zariski opens $U_i$ such that $\mathcal P$ holds for $X_s$ for all $s \in U$ .

In general, Hilbert scheme arguments show that the locus where the Picard rank is ‘bigger than expected’ is a countable union of closed subvarieties $Z_i$ of $S$ (the Noether–Lefschetz loci), but it could be the case that this actually happens everywhere (i.e. $U = \varnothing$ ). The hard part of the Noether–Lefschetz theorem is that the jumping loci $Z_i$ are strict subvarieties of the full space of degree $d$ hypersurfaces.

If $\mathscr X \to S$ is a family of varieties over an uncountable field $k$ , then there always exists a very general member $\mathscr X_s$ with $s \in S(k)$ . But over countable fields, very general elements might not exist, because it is possible that $\bigcup Z_i(k) = S(k)$ even when $\bigcup Z_i \neq S$ .

The following interesting phenomenon was brought to my attention by Daniel Bragg (if I recall correctly):

Example. Let $k = \bar{\mathbf F}_p$ (the algebraic closure of the field of $p$ elements, but the bar is not so visible in MathJax), let $S = \mathcal A_1 = \mathcal M_{1,1}$ (or some scheme covering it if that makes you happier) with universal family $\mathscr E \to S$ of elliptic curves, and let $\mathscr X = \mathscr E \times_S \mathscr E$ be the family of product abelian surfaces $E \times E$ . Then the locus

$NL(S) = \left\{s \in S\ \big| \ \operatorname{rk} \Pic(\mathscr X_s) > 3\right\}$

is exactly the set of $k$ -points (so it misses only the generic point).

Indeed, $\Pic(E \times E) \cong \Pic(E) \times \Pic(E) \times \End(E)$ , and every elliptic curve $E$ over $k$ has $\operatorname{rk} \End(E) \geq 2$ . But the generic elliptic curve only has $\End(E) = \mathbf Z$ . $\qedsymbol$

We see that the Noether–Lefschetz loci might cover all $k$ -points without covering $S$ , even in very natural situations.

Finite domains are fields

Posted on 23 June 2017 by Remy

This is one of the classics.

Lemma. Let $R$ be a finite commutative ring. Then every element is either a unit or a zero-divisor.

Proof. If $x \in R$ is not a zero-divisor, then the map $x \colon R \to R$ is injective. Since $R$ is finite, it is also surjective, so there exists $y \in R$ with $xy = 1$ . $\qedsymbol$

Corollary 1. Let $R$ be a finite commutative ring. Then $R$ is its own total ring of fractions.

Proof. The total ring of fractions is the ring $R[S^{-1}]$ , where $S$ is the set of non-zerodivisors. But that set consists of units by the lemma above, so inverting them doesn’t change $R$ . $\qedsymbol$

Corollary 2. Let $R$ be a finite domain. Then $R$ is a field.

Proof. In this case, the total ring of fractions is the fraction field. Therefore, $R$ is its own fraction field by Corollary 1. $\qedsymbol$

Sites without a terminal object

Posted on 3 April 2017 by Remy

Let $\mathcal C$ be a site with a terminal object $X$ . Then the cohomology on the site is defined as the derived functors of the global sections functor $\Gamma(X,-)$ . But what do we do if the site does not have a terminal object?

The solution is to define $H^i(\mathcal C,-)$ as $\Ext{\mathcal O}{i}(\mathcal O,-)$ , where $\mathcal O$ denotes the structure sheaf if $\mathcal C$ is a ringed site. If $\mathcal C$ is not equipped with a ring structure, we take $\mathcal O$ to be the constant sheaf $\underline{\mathbb Z}$ ; this makes $\mathcal C$ into a ringed site.

Lemma. Let $\mathcal C$ be a site with a terminal object $X$ . Then the above definitions agree, i.e.

$H^i(X,-) = \Ext{\mathcal O}{i}(\mathcal O,-).$

Proof. Note that $\Hom_{\mathcal O}(\mathcal O, \mathscr F) = \Gamma(X, \mathscr F)$ , since any map $\mathcal O(X) \to \mathscr F(X)$ can be uniquely extended to a morphism of (pre)sheaves $\mathcal O \to \mathscr F$ , and conversely every such morphism is determined by its map on global sections. The result now follows since $\Ext{\mathcal O}{i}(\mathcal O, -)$ and $H^i(X,-)$ are defined as the derived functors of $\Hom_{\mathcal O}(\mathcal O,-)$ and $\Gamma(X,-)$ respectively. $\qedsymbol$

Remark. From this perspective, it seems quite magical that for a sheaf $\mathscr F$ of $\mathcal O_X$ -modules on a ringed space $(X,\mathcal O_X)$ , the cohomology groups $\Ext{\mathcal O_X}{i}(\mathcal O_X,\mathscr F)$ and $\Ext{\underline{\Z}}{i}(\underline{\Z},\mathscr F)$ agree. It turns out that this is true in the setting of ringed sites as well; see Tag 03FD.

So why is this useful? Let’s give some examples of sites that do not have a terminal object.

Example. Let $G$ be a group scheme over $k$ . Then we have a stack $BG$ of $G$ -torsors. The objects of $BG$ are pairs $(U,P)$ , where $U$ is a $k$ -scheme and $P$ is a $G$ -torsor over $U$ . Morphisms $(U,P) \to (U',P')$ are pairs $(f,g) \colon (U,P) \to (U',P')$ making the diagram

Rendered by QuickLaTeX.com

commutative. This forces the diagram to be a pullback, since all maps between $G$ -torsors are isomorphisms.

The (large) Zariski site on $BG$ is defined by declaring coverings $\{(U_i, P_i) \to (U,P)\}$ to be families such that $\{U_i \to U\}$ is a Zariski covering (and similarly for the étale and fppf sites).

Now does the category $BG$ have a terminal object? This would be a $G$ -torsor $P_0 \to U_0$ such that every other $G$ -torsor $P \to U$ admits a unique map to it, realising $P$ as the pullback of $P_0$ along $U \to U_0$ . But this object would exactly be the classifying stack $U_0 = BG$ , which does not exist as a scheme (or algebraic space). The fact that a terminal object does not exist is the whole reason we need to define it as a stack in the first place!

Example. Let $X/k$ be a variety in characteristic $p > 0$ ; for simplicity, let’s say $k = \mathbb F_p$ . Then consider the crystalline site of $X/\Spec(\Z/p^n\Z)$ . Roughly speaking, its objects are triples $(U,T,\delta)$ , where $U \to X$ is an open immersion, $U \to T$ is a thickening with a map to $\Spec{\F_p} \to \Spec{\Z/p^n\Z}$ , and $\delta$ is a divided power structure on the ideal sheaf $\mathcal I_U \subseteq \mathcal O_T$ (with a compatibility condition w.r.t. $\Spec{\F_p} \to \Spec{\Z/p^n\Z}$ ). There is a suitable notion of morphisms.

This site does not have a terminal object, basically because there are many thickenings on $U = X$ with the respective compatibilities. (I am admittedly no expert, and it could very well be true that this is not 100% correct. However, I am certain that the crystalline site in general does not have a terminal object.)

Odd degree Betti numbers are even

Posted on 19 May 2016 by Remy

In characteristic 0, it follows from the Hodge decomposition and Hodge symmetry that the Betti numbers $h^i(X) = \dim H^i(X^{\operatorname{an}},\mathbb C)$ of a smooth proper complex variety $X/\mathbb C$ are even when $i$ is odd. In characteristic $p$ however, both Hodge-de Rham degeneration and Hodge symmetry fail (and de Rham cohomology is not a Weil cohomology theory), so we cannot use this method to obtain the result.

On the other hand, in the projective case, we can use hard Lefschetz plus the explicit description of the Poincaré pairing to conclude (we get a perfect alternating pairing, so the dimension has to be even). This leaves open the proper (non-projective) case in positive characteristic. This was settled by Junecue Suh [1]. I will explain the case for finite fields; one can easily reduce the general case to this case.

Notation. Throughout, $K$ will be a $p$ -adic field with ring of integers $W = \O_K$ , residue field $k$ of size $q$ , and (normalised) valuation $v$ such that $v(q) = 1$ (this is the $q$ -valuation on $K$ ).

Throughout, $X$ will be a smooth proper variety over $k$ . We will write $h^i(X)$ for the Betti numbers of $X$ . It can be computed either as the dimension of $H^i\et(\bar X, \Q_\ell)$ , or that of $H^i_{\operatorname{crys}}(X/W)[\frac{1}{p}]$ .

Remark. Recall that if $f$ is the characteristic polynomial of Frobenius acting on $H^i\et(\bar X, \mathbb Q_\ell)$ for $\ell \neq p$ , and $\alpha \in \bar{\mathbb Q}$ is the reciprocal of a root of $f$ , then for every complex embedding $\sigma \colon \bar \Q \to \C$ we have

(1) $\begin{equation*} |\sigma(\alpha)| = q^{\frac{i}{2}}. \end{equation*}$

The same holds for the eigenvalues of Frobenius on crystalline cohomology (in fact, the characteristic polynomials agree). All reciprocal roots are algebraic integers, and $f \in \mathbb Z[t]$ .

Defintion. An algebraic integer $\alpha \in \bar \Q$ is a $q^i$ -Weil integer if it satisfies (1) (for every embedding $\sigma \colon \bar \Q \to \C$ ).

Lemma. Let $f \in \mathbb Q[t]$ be a polynomial, and let $S$ be the multiset of reciprocal roots of $f$ . Assume all $\alpha \in S$ are $q^i$ -Weil integers. Then $v(S) = i - v(S)$ (counted with multiplicity).

Proof. If $\alpha \in S$ , then $\frac{q^i}{\alpha}$ is the complex conjugate with respect to every embedding $\sigma \colon \bar \Q \to \C$ . Thus, it is conjugate to $\alpha$ , hence a root of $f$ as well (with the same multiplicity). Taking valuations gives the result. $\qedsymbol$

Theorem. Let $X$ be smooth proper over $k$ , and let $i$ be odd. Then $h^i(X)$ is even.

Proof. The Frobenius-eigenvalues whose valuation is not $\frac{i}{2}$ come naturally in pairs $(\alpha, \frac{q^i}{\alpha})$ . Now consider valuation $\frac{i}{2}$ . Note that the $p$ -valuation of the semilinear Frobenius $F$ equals the $q$ -valuation of the $K$ -linear Frobenius $F^r$ (which is the one used in computing the characteristic polynomial $f$ ). The sum of the $p$ -valuations of the roots should be an integer, because $f$ has rational coefficients. Thus, there needs to be an even number of valuation $\frac{i}{2}$ eigenvalues, for otherwise their product would not be a rational number. $\qedsymbol$

References.

[1] Suh, Junecue, Symmetry and parity in Frobenius action on cohomology. Compos. Math. 148 (2012), no. 1, 295–303. MR2881317.

Number of points modulo q is a stable birational invariant

Posted on 19 March 2016 by Remy

This post is about a (very weak) shadow in characteristic $p$ of the Larsen–Lunts theorem. See my previous post for the statement and sketch of the proof of Larsen–Lunts.

Remark. In characteristic $p$ , we do not even know the weakest form of resolution of singularities (e.g. find a smooth proper model for any function field). Thus, we certainly do not know the Larsen–Lunts theorem. However, we can still try to prove corollaries (and if they fail, we know that resolution must fail).

Today, I want to talk about the following statement:

Theorem. (Ekedahl) Let $k = \mathbb F_q$ . Let $X$ and $Y$ be smooth proper varieties, and assume $X$ and $Y$ are stably birational. Then $|X(k)| \equiv |Y(k)| \pmod{q}$ .

Remark. This would follow immediately from Larsen–Lunts if we knew a sufficiently strong form of resolution of singularities. Indeed, the map

$K_0(\operatorname{Var}_k) \to \Z/q\Z$

given by counting $\F_q$ -points modulo $q$ factors through $K_0(\operatorname{Var}_k)/(\mathbb L)$ since $|\mathbb A^1(\F_q)| = q$ . Hence, by Larsen–Lunts, it factors through $\mathbb Z[\operatorname{SB}]$ .

It turns out that the theorem is true without assuming resolution of singularities, and the proof is due to Ekedahl (although in his paper he never explicitly states it in this form). The reader should definitely check out Ekedahl’s article (see references below), because his proof is more beautiful than the one I present here, and actually proves a bit more.

We will need one fairly deep theorem:

Theorem. Let $X$ be a variety of dimension $n$ over $k = \F_q$ . Let $\alpha$ be an eigenvalue of Frobenius on $H^i_c(\bar X\et, \Q_\ell)$ . Then $\alpha$ and $q^n\alpha^{-1}$ are both algebraic integers.

The first part (integrality of $\alpha$ ) is fairly well-known. For the second part (integrality of $q^n\alpha^{-1}$ ), see SGA 7 $_{\text{II}}$ , Exp. XXI, Corollary 5.5.3(iii).

The statement that appears in Ekedahl’s article is the following:

Theorem. (Ekedahl’s version) Let $k = \mathbb F_q$ . Let $X$ and $Y$ be smooth connected varieties (not necessarily proper!), and assume $X$ and $Y$ are birational. If $\alpha$ is an eigenvalue of Frobenius on $H^i(\bar X, \Q_\ell)$ which is not an eigenvalue on $H^i(\bar Y, \Q_\ell)$ , then $\alpha$ is divisible by $q$ .

This statement should be taken to include multiplicities; e.g. a double eigenvalue for $X$ which is a simple eigenvalue for $Y$ is also divisible by $q$ . By symmetry, we also get the opposite statement (with $X$ and $Y$ swapped). Thus, the eigenvalues (with multiplicities) that are not divisible by $q$ are the same for $X$ and $Y$ .

Proof. We immediately reduce to the case where $X \sbq Y$ is an open immersion, with complement $Z$ . We have a long exact sequence for étale cohomology with compact support:

$\cdots \to H^{i-1}_c(Z) \to H^i_c(X) \to H^i_c(Y) \to H^i_c(Z) \to H^{i+1}(X) \to \cdots.$

If $\alpha$ is an eigenvalue on some $H^i_c(Z)$ , then $q^{n-1}\alpha^{-1}$ is an algebraic integer (see above). Hence, for any valuation $v$ on $\bar \Q$ with $v(q) = 1$ , we have $v(\alpha) \leq n-1$ . We conclude that the eigenvalues for which some valuation is $> n-1$ on $H^i_c(X)$ and $H^i_c(Y)$ agree. Hence, by Poincaré duality, the eigenvalues of $H^{2n-i}(X)$ and $H^{2n-i}(Y)$ for which some valuation is $< 1$ agree. These are exactly the ones that are not divisible by $q$ . $\qedsymbol$

The theorem I stated above immediately follows from this one:

Proof. Since $|X\times\P^n(\F_q)| = (q^n + \ldots + 1) |X(\F_q)|$ , we may replace $X$ by $X \times \P^n$ . Thus, we can assume $X$ and $Y$ are birational; both of dimension $n$ .

By the Weil conjectures, we know that

$|X(\F_q)| = \sum_{i=0}^{2n} \sum_\alpha \alpha^i,$

where the inner sum runs over all eigenvalues of Frobenius. If we reduce mod $q$ , then we only need to consider eigenvalues that are not divisible by $q$ . By Ekedahl’s version of the theorem, the set (with multiplicities) of such $\alpha$ are the same for $X$ and $Y$ . $\qedsymbol$

Historical remark. Although the theorem above was essentially proven in 1983 (but not explicitly stated), a separate proof for threefolds appeared in a paper by Gilles Lachaud and Marc Perret in 2000. It uses Abhyankar’s results on resolution of singularities, and is much closer to the proof of Larsen–Lunts than Ekedahl’s proof was. In 2002, Bruno Kahn provided a different proof for the general case using some (fairly advanced?) motive machinery (‘almost without cohomology’).

References.

Torsten Ekedahl, Sur le groupe fondamental d’une variété unirationelle. Comptes rendus de l’académie des sciences de Paris, Serie I: mathématiques, 297(12), p. 627-629 (1983).

Bruno Kahn, Number of points of function fields over finite fields. arXiv:math/0210202

Gilles Lachaud and Marc Perret, Un invariant birationnel des variétés de dimension 3 sur un corps fini. Journal of Algebraic Geometry 9 (2000), p. 451-458.

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

Rendered by QuickLaTeX.com

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

Rendered by QuickLaTeX.com

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

Rendered by QuickLaTeX.com

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