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How does a working mathematician usually think about algebraic geometry in characteristic $p$? For the sake of concreteness, and to make things more "geometric" (whatever that means), let's say we work over an algebraically closed field of characteristic $p$. It seems to me that (motivated by the étale topology, say), authors tend to use the geometry of $\mathbb{C}$ as a template, envisioning $\mathbb{A}^1$ as something like the complex plane. Then, using this as a model, one accounts for extra phenomena that occur in characteristic $p$; namely, extra automorphisms due to frobenii, as well as some extra singularities that don't occur in characteristic $0$.

Is this picture of characteristic $p$ geometry accurate? To make this question a little less soft, I'll ask the following naive, and certainly false question: Is the category of varieties over $\overline{\mathbb{F}}_p$ equivalent to the category of varieties over $\mathbb{C}$ along with extra morphisms coming from Frobenii (as well as objects obtained from glueing via these extra morphisms)?

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    $\begingroup$ I tend to think of characteristic $p$ algebraic geometry as both similar and different to the complex algebraic geometry, but this probably doesn't answer your question...Turning to the less soft question, supersingular elliptic curves have endomorphism rings much larger than elliptic curves over $\mathbb{C}$, so I don't see how to make an equivalence of categories along the lines you're suggesting. $\endgroup$ Commented Aug 31, 2023 at 2:06
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    $\begingroup$ Are you asking for the best visual image of the geometric space of $\mathbb{A}^1$ in characteristic $p$? Or are you asking how we use special features of algebraic geometry in positive characteristic to help us prove theorems? In many ways a closer "geometric analogy" of $\mathbb{A}^n$ over a field like $\overline{\mathbb{F}}_p$ is a fibration over a punctured complex disk where the fibers are isomorphic to affine space. Frobenius is analogous to analytic continuation along a (fraction of a) loop in the disk. $\endgroup$ Commented Aug 31, 2023 at 11:17
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    $\begingroup$ @JasonStarr I'm primarily interested in just visualizing $\mathbb{A}^1$, along with any other varieties over a field of characteristic $p$. Though admittedly I hope that such a visualization would help to understand theorems in positive characteristic. Could you elaborate on the analogy you laid out? Or point me toward a reference which discusses this analogy in detail; I've never heard of this interpretation. $\endgroup$ Commented Aug 31, 2023 at 20:34
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    $\begingroup$ It is not an interpretation; it is just an analogy. The absolute Galois group of $\mathbb{F}_p$ equals the absolute Galois group of $\mathbb{C}((t))$. Also, both fields are "quasi algebraically closed" fields. $\endgroup$ Commented Sep 1, 2023 at 10:09
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    $\begingroup$ @LSpice Every pseudo-finite field is quasi-algebraically closed, but not vice versa, since e.g. algebraically-closed fields are quasi-algebraically-closed but not pseudo-finite. $\endgroup$
    – Will Sawin
    Commented Jun 24 at 16:58

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The basic strategy you give is correct: To first approximation, algebraic geometry over $\overline{\mathbb F_p}$ is like algebraic geometry over $\mathbb C$ except for extra phenomena. Usually, I think of these extra phenomena as "inseparable maps" and "wild ramification".

Inseparable maps include Frobenius, but not every inseparable map is a Frobenius. They all have properties which are unexpected from a characteristic zero point of view (e.g. rank of the derivative of the map a a generic point does not equal the dimension of the image).

Wildly ramified covers also behave differently from characteristic zero maps, but in a more subtle way. Perhaps the biggest is that to specify a finite branched cover in characteristic zero it suffices up to only a finite ambiguity to specify the branch locus, while in characteristic $p$ the ambiguity can be infinite.

Of course as one gets more experience in algebraic geometry in characteristic $p$ one increasingly thinks about characteristic $p$ problems by reference to previous characteristic $p$ situations one has encountered, but I think "characteristic zero plus inseparable maps and wild ramification" is a good starting point.

In terms of visualizing, the only reasons to divert from visualizing algebraic curves as complex surfaces and so on are the same reasons that exist in characteristic $0$: sometimes, to fit more in your visual imagination, you want to visualize algebraic curves as curves, algebraic surfaces as surfaces, and so on, and sometimes you want to visualize arbitrarily-high-dimensional varieties, which means you have to imagine them as two- or maybe three-dimensional.

Let me also briefly answer your formal question in the negative: The set of maps $\mathbb P^1_{\mathbb C}\to\mathbb P^1_{\mathbb C}$ is uncountably large, and adding extra morphisms couldn't make this countable, but the set of maps $\mathbb P^1_{\overline{\mathbb F_p}}\to\mathbb P^1_{\overline{\mathbb F_p}}$ is countable.

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Since you've mentioned $\mathbb{A}^1$, we can look at group schemes $G \subset \mathbb{G}_a$ sitting inside the additive group scheme, and contrast what is going on over characteristic 0 versus over some $\mathbb{F}_p \subset k = k^{\text{alg}}$ of positive characteristic. This should give a rather elementary exposition in how those worlds may differ.

It is easy to see that in characteristic 0, $\mathbb{G}_a$ has no nontrivial sub group schemes, for if $G(k)$ had some $k$-rational point $a \neq 0$, it must have all of its multiples $na$ for $n \in \mathbb{Z}$ corresponding to a polynomial vanishing at infinitely many points, thus $f \equiv 0$ and $G = V(f) = \mathbb{G}_a$ is the whole thing.

In contrast, the procedure for positive characteristic would give you a constant group scheme $(\mathbb{Z}/p\mathbb{Z})_k$ and furthermore at every closed point we can concentrate $\alpha_p \cong \operatorname{Spec}k[T]/(T^p)$. In fact, we can characterise these group schemes entirely by additive polynomials, which have their own arithmetic under addition and composition sometimes denoted by $k\{F,\sigma\}$.

The reason I brought this up is because even at this fundamental level there are some genuinely new objects that appear in mixed characteristic. I hope this is somewhat close to what you're looking for!

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    $\begingroup$ Welcome to MO! I think your argument for why $\mathbb G_a$ has no non-trivial subgroup schemes in characteristic $0$ skips over the important point that such a subgroup scheme must be smooth, hence have a non-trivial point over an algebraically closed field. To me this is even more important than the distinction between the additive subgroup generated by a non-trivial point being infinite vs. finite, but perhaps if one dug deeply enough into Oort's proof (I've never read it!) one would see that the latter remained the essential idea. $\endgroup$
    – LSpice
    Commented Jun 24 at 15:29
  • $\begingroup$ Another direction of pathologies that can arise even for smooth group schemes can only be seen over an imperfect field. My favourite is, taking $p = 2$ for convenience, the image of $\left\{\begin{pmatrix} a & b \\ t b & a \end{pmatrix}\right\}$ in $\operatorname{PGL}_2$, where $t$ is a non-square. This is a 1-dimensional, smooth, connected, commutative, unipotent group scheme, but it's not $\mathbb G_a$. $\endgroup$
    – LSpice
    Commented Jun 24 at 15:45
  • $\begingroup$ @LSpice Why does one need smoothness for the existence of a point? Isn't that Hilbert's Nullstellensatz? $\endgroup$ Commented Jul 30 at 11:24
  • $\begingroup$ @red_trumpet, re, the existence of a point follows from the Nullstellensatz, but is trivial for an algebraic group, because the identity is always a point. The question is whether there is another point, which is what is needed to make the argument in the answer go, and what I meant by "non-trivial". $\endgroup$
    – LSpice
    Commented Jul 30 at 12:58

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