Let $X$ be a projective variety defined over a number field $K$, and $p \in \textrm{Spec }\mathcal{O}_K$ a maximal ideal, so that reduction mod $p$ makes sense, and the resulting scheme (mod $p$) $\bar{X}$ is smooth over the relevant finite field. Assume that $X$ smooth over $K$.

1.) In the case that $X$ is a curve, is there a short argument to show that the geometric genus of $X$ and of $\bar{X}$ are the same? Certainly if $X$ is a plane curve this is clear.

2.) The hodge numbers $h^{p,q}_X = \dim H^p(X, \Omega^q)$ make sense in all characteristics. Are the hodge numbers preserved under reduction mod $p$, that is, $h^{p,q}_X = h^{p,q}_\bar{X}$?

3.) The Weil conjectures tell us that we can recover the Betti numbers of $X$ (considered as a complex manifold) from the zeta function of $\bar{X}$. There are many smooth projective varieties that have reduction $\bar{X}$ mod $p$ and the Weil conjectures tell us that all of them have the same Betti numbers. Can one prove this without using the Weil conjectures, perhaps with Etale cohomology?

4.) More generally, if $\mathcal{L}$ is a locally free sheaf on $X$, and $\bar{\mathcal{L}}$ denotes the reduction mod $p$, I would guess that the numbers $\dim H^p(X, L)$ and $\dim H^p(\bar{X}, \bar{L})$ don't match up - but I don't have a good example.

I am interested in proofs (not using the Weil conjectures if possible).

  • 1
    $\begingroup$ For 2.), the answer is NO! and for 3.) the answer is the smooth-proper base change theorem, which gives an isomorphism on the cohomology groups (see any book on etale cohomology). $\endgroup$
    – anon
    Dec 17, 2012 at 19:56
  • $\begingroup$ @anon, Thanks for your answer to (3). Is there a typical example I can keep in mind for 2? $\endgroup$
    – LMN
    Dec 17, 2012 at 20:07
  • $\begingroup$ Dear LMN, This answer at MSE may be helpful: math.stackexchange.com/a/51153/221 Regards, $\endgroup$
    – Emerton
    Jan 30, 2013 at 5:29

4 Answers 4


As already pointed out, the Hodge numbers may go up under reduction modulo $p$. On the other hand, let me also point out that the situation can be controlled:

1.) For all $p$, where $\overline{X}_p$ is smooth, the $\ell$-adic Betti numbers of $X$ and $\overline{X}_p$ are the same.

2.) Now, by the universal coefficient formula relating crystalline and deRham cohomology, we have for all $i$ short exact sequences

$$ 0 \to H^i_{cris}(\overline{X}_p/W)/p\to $$

$$ H^i_{dR}(\overline{X}_p/k_p)\to {\rm Tor}_1^{W(k_p)}(H_{cris}^{i+1}(\overline{X}_p/W),k_p)\to 0 $$

where $k_p={\cal O}_K/p$. Now, if $\overline{X}_p$ has torsion-free crystalline cohomology, then the term on the right is zero, and the term on the left is a $k_p$-vector space of dimension equal to the $i$.th $\ell$-adic Betti number. Then, the Fr\"olicher spectral sequence relating Hodge- and deRham-cohomology degenerates at $E_2$, we have that $\sum_{p+q=i}h^{p,q}$ is equal to the $i$.th $\ell$-adic Betti number. Thus, simply for dimension reasons, the Hodge numbers of $X$ and $\overline{X}_p$.

The upshot is that torsion in crystalline cohomology of $\overline{X}_p$ detects and controls the differences in Hodge numbers of $X$ and $\overline{X}_p$. For almost all $p\in {\rm Spec} {\cal O}_K$, the reduction $\overline{X}_p$ will be smooth and will have torsion-free crystalline cohomology.

  • $\begingroup$ Neat! Thanks for putting this in perspective. $\endgroup$
    – LMN
    Dec 19, 2012 at 17:51

Since people have addressed (2-4), I'll address (1). There is indeed a fast argument.

Namely, you have a flat family of curves $X\to \operatorname{Spec}(\mathcal{O}_{K, p})$. You are interested in comparing the geometric genus of the generic fiber $X_K$ to that of the special fiber $X_{\mathbb{F}_q}$. Note that by constancy of Euler characteristic in flat families, $$1-p_a({X_K})=\chi(\mathcal{O}_{X_K})=\chi(\mathcal{O}_{X_{\mathbb{F}_q}})=1-p_a(X_{\mathbb{F}_q}).$$ where $p_a$ denotes arithmetic genus. So the arithmetic genera of the fibers are equal. Since you've required that the fibers $X_K$ and $X_{\mathbb{F}_q}$ are smooth, their geometric genera equal their arithmetic genera, which completes the argument.

ADDED (12/19/2012): I'd also like to add a comment about the Hodge decomposition for curves, to complement Christian's excellent answer. Namely, one can see that the Hodge-theoretic predictions one would make from the situation over $\mathbb{C}$ are true for smooth projective curves over any base. A rigorous version of this is given (with a sketch proof) as Theorem 2 in these notes of mine.


For some explicit counterexamples to (2) and therefore also (4) again, see J. Suh, Plurigenera of general type surfaces in mixed characteristic, Compositio (2008). The only thing that you can say for sure is that $h^{pq}(\bar X)\ge h^{pq}(X)$ by upper semicontinuity of cohomology.

  • 3
    $\begingroup$ one should also mention Serre's article {\em Sur la topologie des vari\'et\'es alg\'ebriques en caract\'eristique $p$} from 1958, where the first examples of this kind were constructed. $\endgroup$ Dec 19, 2012 at 10:08

This is only a comment but I'm a new user and can't make comments.

For (4), the dimensions of the $H^i$ might not even be the same if $i=0$. let $X$ be an elliptic curve and choose two distinct points $P$ and $Q$ on the generic fibre whose reductions mod $p$ are the same. Let $L$ be the line bundle $P-Q$. This has no global sections on the generic fibre, because any function with a simple pole at $P$ and a simple zero at $Q$ would be an isomorphism between the curve and projective 1-space. However mod $p$ the line bundle becomes trivial, so has global sections.

  • $\begingroup$ wccanard, are you suggesting that the pullback of the linear equivalence class of $P - Q$ is $\bar{P} - \bar{Q}$? Is it clear that this respects linear equivalence? $\endgroup$
    – LMN
    Dec 17, 2012 at 21:45
  • 2
    $\begingroup$ If it didn't respect linear equivalence, what would the pullback mean? $\endgroup$
    – Will Sawin
    Dec 17, 2012 at 21:55
  • $\begingroup$ Will, I don't understand your comment. You can pullback line bundles by base changing $X \rightarrow Spec O_k$ by $Spec F_p \rightarrow Spec O_k$. Is the answer to the question completely clear to you? $\endgroup$
    – LMN
    Dec 17, 2012 at 21:58
  • 2
    $\begingroup$ @Will: It's perfectly possible to pull back cycles along, say, a flat map, without imposing an equivalence relation such as rational equivalence. The fact that pullback respects e.g. rational equivalence of cycles does require proof, though it's not difficult. @LMN: See Fulton, Intersection Theory, first chapter, for example. But the case of divisors on a normal scheme is pretty easy--I suggest you try it yourself. $\endgroup$ Dec 17, 2012 at 23:24
  • $\begingroup$ Right, I just figured it out. $\endgroup$
    – LMN
    Dec 18, 2012 at 2:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.