# 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?

• You should be more specific to avoid trivial answers, like the "theorem" stating that "not all morphisms are separable". Sep 9, 2012 at 18:38
• I agree this should be made more specific to exclude trivial examples, e.g. "there exist non-smooth Fermat hypersurfacees". Sep 9, 2012 at 18:55
• How about the rationality of the zeta function? Sep 10, 2012 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 $0$.

• 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). Sep 9, 2012 at 20:28
• (deleted a mistaken comment)
– user5117
Sep 9, 2012 at 21:18
• what is the policy regarding deleting things? sometimes it makes subsequent comments confusing I reckon. Sep 9, 2012 at 21:40
• @Olivier Is it true as it stands that it is not true? Sep 10, 2012 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. Sep 10, 2012 at 12:29

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.)

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 $0$.

• 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. Sep 11, 2012 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 Sep 11, 2012 at 12:24

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.

Here is a paper by Rachel Pries and Katherine Stevenson that addresses the question:

http://www.math.colostate.edu/~pries/Preprints/11dgroupreportv036.pdf

Full reference: Pries, R. and Stevenson, K. "A survey of Galois theory of curves in characteristic p" , Fields Institute Communications 60 , American Mathematical Society, Providence RI, (2011), 169-191.

I am not well-versed in this area, but I found this paper quite readable. Some of the examples in the paper (particularly the first half of the paper) have been brought up in answers already in this thread.

The stack $\overline{\mathcal{M}}_{g,n}$ of Deligne-Mumford stable curves and its coarse moduli space $\overline{M}_{g,n}$ are defined over $\mathbb{Z}$. Therefore they are defined over any commutative ring and in any characteristic.

Let $\pi:\mathcal{U}\rightarrow\overline{\mathcal{M}}_{g,n}$ be the universal curve, $\omega_{\pi}$ the relative dualizing sheaf and $\Sigma$ the union of the sections of $\pi$. Then $\mathcal{L}:=\pi_{*}\omega_{\pi}(\Sigma)$ is a line bundle on $\overline{\mathcal{M}}_{g,n}$.

If $p:\overline{\mathcal{M}}_{g,n}\rightarrow\overline{M}_{g,n}$ is the coarse moduli space then $p_{*}\mathcal{L}$ is semi-ample in positive characteristic but this fails in characteristic zero.