Automorphisms over finite field that do not lift to an automorphism in characteristic zero

My main question is the following: is there an automorphism of the affine space $$\mathbb{A}^n$$ (automorphism of an algebraic variety) defined over a finite field which does not lift to an automorphism defined over a field in characteristic zero?

Let us first consider finite fields of the form $$\mathbb{F}_p$$ for some prime $$p$$. If you take an automorphism $$f$$ of the affine space that is defined over $$\mathbb{Q}$$ and such that $$p$$ does not divide any of the denominators of $$f$$ and of $$f^{-1}$$, you may consider the restriction of $$f$$ and $$f^{-1}$$ to automorphisms defined over $$\mathbb{F}_p$$, by considering the coefficients modulo $$p$$, and get automorphisms defined over $$\mathbb{F}_p$$. Is every automorphism of $$\mathbb{A}^n$$ over $$\mathbb{F}_p$$ obtained by this way? (one can do similar constructions for other finite fields and other fields of caracteristic zero).

For each prime $$p$$, every linear automorphism of $$\mathbb{A}^n$$, given by $$(x_1,\ldots,x_n)\mapsto (a_{11}x_1+\cdots +a_{1n}x_n,\ldots,a_{n1}x_1+\cdots +a_{nn}x_n)$$ for some matrix $$(a_{ij})\in \mathrm{GL}_n(\mathbb{F}_p)$$ comes from an element of $$\mathrm{GL}_n(\mathbb{Q})$$ whose entries are integers and whose determinant is not divisible by $$p$$. Its inverse in $$\mathrm{GL}_n(\mathbb{Q})$$ has thus all denominators that are not multiple of $$p$$. Hence, every linear automorphism of $$\mathbb{A}^n$$ over $$\mathbb{F}_p$$ comes from a linear automorphism of $$\mathbb{A}^n$$.

Similarly, if you take an elementary automorphism of $$\mathbb{A}^n$$ defined over $$\mathbb{F}_p$$, i.e. given by $$(x_1,\dots,x_n)\mapsto (x_1+a(x_2,x_3,\ldots,x_n),x_2,\ldots,x_n)$$ for some polynomial $$a\in \mathbb{F}_p[x_2,\ldots,x_n]$$, it comes from an automorphism of $$\mathbb{A}^n$$ defined over $$\mathbb{Q}$$ (here even over $$\mathbb{Z}$$). In particular, every "tame" automorphism (generated by linear and elementary automorphisms) of $$\mathbb{A}^n$$ over $$\mathbb{F}_p$$ comes from a tame automorphism of $$\mathbb{A}^n$$ over $$\mathbb{Q}$$. In dimension $$n=1$$ and $$n=2$$, every automorphism is tame, so every automorphism of $$\mathbb{A}^n$$ over $$\mathbb{F}_p$$ comes from an automorphism defined over $$\mathbb{Q}$$. The question is then more interesting for $$n\ge 3$$. It would give examples of non-tame automorphisms, a fact not know until now in positive characteristic (see the famous article Shestakov and Umirbaev - The tame and the wild automorphisms of polynomial rings in three variables for examples in characteristic zero).

This seems hard to answer in general, but can we then replace the variety $$\mathbb{A}^n$$ by an affine (or projective) algebraic variety, defined over $$\mathbb{F}_p$$ and find some automorphisms over $$\mathbb{F}_p$$ that do not come from automorphisms in characteristic zero?

EDIT: Will Sawin gave a nice answer for elliptic curves, where some given automorphism do not lift to $$\mathbb{Q}$$. Are there examples where some given automorphisms do not lift to any field of caracteristic zero?

• Could you clarify what you mean with "restriction of an automorphism defined over a field in charactersitic zero"? How would you restrict an automorphism over $\mathbb Q$ (say) to one over $\mathbb F_p$? Jan 9 '21 at 14:46
• The Frobenius is not an automorphism. Jan 9 '21 at 22:14
• @Jérémy: if you really mean "over $\mathbb{Z}$" then already the automorphism $x \mapsto 2x$ of $\mathbb{A}^1$ over $\mathbb{F}_p$ for $p$ odd is not the restriction of an automorphism of $\mathbb{A}^1$ over $\mathbb{Z}$. (But it is a restriction (suitably interpreted) of an automorphism over a field of characteristic zero, namely $\mathbb{Q}$, which is what you originally wrote. So it would be good to clarify what you mean here.) Jan 9 '21 at 22:28
• By an automorphism of a variety we usually mean an automorphism over the field. For affine varieties this is equivalent to a ring automorphism that preserves every element of the base field. Jan 10 '21 at 14:49
• If $E$ is an elliptic curve then $E$ minus a point is an affine variety, whose automorphisms are the automorphisms of $E$. The elliptic curve $y^2 =x^3-x$ has an automorphisms of order $4$ over $\mathbb F_p$ when $p \equiv 1 \mod 4$, but not over $\mathbb Q$ - in fact, no elliptic curve over $\mathbb Q$ has an automorphism of order $4$. Jan 10 '21 at 14:53

Some simple examples are provided by affine elliptic curves.

Recall that a smooth genus zero curves $$E$$ over a field $$k$$ with a rational point $$P$$ has a natural group structure, where $$P$$ is the identity. Thus automorphisms of $$E$$ fixing $$P$$ are automorphisms of the group.

Automorphisms of the affine curve $$E -\{P\}$$ extend to automorphisms of $$E$$ fixing $$P$$, since affine curves have a canonical projective closure.

A lot is known about automorphisms of elliptic curves (as groups), and we can use this to construct examples, depending on how the problem is formulated.

The main subtlety in the formulation is that an elliptic curve over $$\mathbb F_p$$ can have many lifts to $$\mathbb Q$$. Thus whether an automorphism lifts might depend on the choice of a lift of the curve.

However, it is possible to construct automorphisms that do not lift regardless of the lift of a curve. This is because, over $$\mathbb Q$$, every elliptic curve has automorphism group of order $$2$$ – the automorphisms have finite order because they fix a point on a genus $$1$$ curve, and they act faithfully on the tangent space at the identity, which is $$\mathbb Q$$, but the only finite order elements of $$\mathbb Q^\times$$ are $$\pm 1$$.

However, the curve defined by $$y^2=x^3-x$$ has an automorphism of order $$4$$ over $$\mathbb F_p$$ for any $$p \equiv 1 \mod 4$$ and the curve defined by $$y^2 = x^3-1$$ has an automorphism of order $$6$$ over $$\mathbb F_p$$ for any $$p \equiv 1 \mod 6$$.

You could instead ask for lifts to any field of characteristic $$0$$, rather than just $$\mathbb Q$$. In this case, we would still have examples as long as you let me choose the lifts: choosing a lift of $$y^2=x^3-1$$ with nonzero $$j$$-invariant would guarantee the order $$4$$ automorphism doesn't lift, and similarly for $$y^2=x^3-x$$ and $$j\neq 1728$$.

If you merely want there to exist a lift of a curve to characteristic $$0$$ where all the automorphisms lift, there are still counterexamples. Supersingular curves in characteristic $$2$$ and $$3$$ have $$24$$ and $$12$$ automorphisms (over an algebraically closed field), respectively, which is more than elliptic curves can have in characteristic zero over an algebraically closed field.

However, if you wanted to find examples of a variety with some fixed automorphism that don't lift to any lift of that variety over any characteristic zero field, I believe you would have to look beyond elliptic curves.

• Nice detailed answer, thanks. I would also be interested in other varieties than elliptic curves, but these are some very natural examples. Thanks. Jan 10 '21 at 16:21
• As you pointed out, the examples you gave do not lift to $\mathbb{Q}$ but lift to some subfields of $\mathbb{C}$ (you simply need some roots of unity, right?) When you say " if you wanted to find examples of a variety with some fixed automorphism that don't lift to any lift of that variety over any characteristic zero field, I believe you would have to look beyond elliptic curves." would you have some other examples in mind? Jan 10 '21 at 17:42
• @JérémyBlanc I don't have other examples in mind, but I'm sure they exist. (Well, technically speaking, a variety that doesn't lift would be an example, but other than those...) Jan 10 '21 at 17:44
• @JérémyBlanc Maybe the paper John L. linked to arxiv.org/abs/1406.2761 contains an example... Jan 10 '21 at 17:45
• It is a conjecture of Oort that every automorphism of a smooth proper curve of characteristic $p$ lifts to some field of characteristic $0$ - see the file "Lifting an automorphism of a curve to characteristic zero" on his webpage. webspace.science.uu.nl/~oort0109 Progress might have been made since 2003, though. Jan 10 '21 at 18:05