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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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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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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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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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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