## Characteristic zero and characteristic $p$ in algebraic geometry

Are there non-trivial (i.e. excluding concepts that can be defined only for $p>0$) statements in algebraic geometry that hold for all fields of characteristic $p$ for all prime $p$ but are known to be false in characteristic zero?

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You should be more specific to avoid trivial answers, like the "theorem" stating that "not all morphisms are separable". – Dan Petersen Sep 9 at 18:38
I agree this should be made more specific to exclude trivial examples, e.g. "there exist non-smooth Fermat hypersurfacees". – algori Sep 9 at 18:55
How about the rationality of the zeta function? – Olivier Sep 10 at 7:48

Here are two examples.

The moduli space of dimension $g$ principally polarized abelian varieties $A_g$ contains complete codimension $g$ subvarieties in any positive characteristic $p$ (for instance, the locus of abelian varieties with no nontrivial $p$-torsion points), but not in characteristic $0$ (by Keel and Sadun arXiv:math/0204229).

The other example is also a resul of Keel (arXiv:math/9901149). It states that a nef and big line bundle $L$ on a projective variety over a field of positive characteristic is semi-ample if and only if its restriction to the exceptional locus (i.e. the union of subvarieties $Z$ such that $L|_Z$ is not big) is semi-ample. This criterion is not true in characteristic $0$.

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No : it is true as it stands. However, over $\overline{F}_p$, it is possible to deduce a statement that is very easy to apply in practice. You need only to check that the restriction of $L$ to the exceptional locus is numerically trivial (because, over $\overline{F}_p$ it implies that it is torsion, hence semi-ample). – Olivier Benoist Sep 9 at 20:28
(deleted a mistaken comment) – Artie Prendergast-Smith Sep 9 at 21:18
what is the policy regarding deleting things? sometimes it makes subsequent comments confusing I reckon. – Jacob Bell Sep 9 at 21:40
@Olivier Is it true as it stands that it is not true? – Wilberd van der Kallen Sep 10 at 12:11
@Wilberd Yes it is. In a deleted comment, Artie was wondering whether it was true over any field of positive characteristic or only over $\overline{F}_p$. My comment above was answering this question. – Olivier Benoist Sep 10 at 12:29
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Tamagawa has shown that a smooth curve (of genus $\neq 1$) over $\bar{\mathbb{F}}_p$ is determined by its profinite $\pi_1$ up to finite indeterminacy, that is, that the map $$\pi_1: M_{g,n}(\bar{\mathbb{F}}_p )\longrightarrow \mbox{Profinite groups up to isomorphism}$$ is finite-to-one. This is clearly false over, say, $\bar{\mathbb{Q}}$. It's not correct to conclude therefore that there are just fewer curves in characteristic $p$, in some sense. The truth is that $\pi_1$ simply retains more geometric information.

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Perhaps this is an example of the contrapositive of a statement in char 0 that fails in all positive characteristics. The affine line has nontrivial \'etale covers over every field of positive characteristic, yet it is algebraically simply connected in characteristic $0$.

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In the same spirit: Kedlaya proved in 2004 that over any field $k$ of positive characteristic, any geometrically reduced purely $n$-dimensional projective variety is a finite cover of $\mathbb P^n_k$, étale away from a hyperplane. This is false in characteristic $0$ because the affine space is simply connected. – Qing Liu Sep 11 at 8:05
In fact, in positive characteristic, any irreducible affine variety of positive dimension has non-trivial étale cover. See mathoverflow.net/questions/16047 – Qing Liu Sep 11 at 12:24

One other example: in characteristic $p>0$ there exist non trivial embeddings of the affine line $\mathbb{A}^1$ into the affine plane $\mathbb{A}^2$ (i.e. embeddings that are not equal to the composition of $x\mapsto (x,0)$ with an automorphism of $\mathbb{A}^2$). For example,

$x\mapsto (x^{p^2},x^{p^2+p}+x)$

is an embbeding because $k[x^{p^2},x^{p^2+p}+x]=k[x]$ (take $f=x^{p^2}$ and $g=x^{p^2+p}+x$, then $g^p-f^{p+1}=x^p$). It is not trivial because the degree of one component does not divide the other one.

In characteristic zero, there is no non-trivial embedding of $\mathbb{A}^1$ into $\mathbb{A}^2$. This is the famous Abhyankar-Moh theorem (Abhyankar, S.; Moh, T. T., Embeddings of the line in the plane. J. Reine Angew. Math. 276 (1975), 148–166.)

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