Around abelian schemes over the integers

This is the transcription to blog format of a talk I gave at the UC Berkeley Student Arithmetic Geometry Seminar about several topics related to Fontaine’s famous result that there are no abelian schemes over $\mathbb{Z}$.

Statement of the result and some consequences

Two statements of the main result

The goal of this post is to discuss the consequences of, and nearby results, to the following famous theorem of Fontaine:

Theorem 1(Fontaine, ’81): There does not exist a non-trivial abelian variety $A/\mathbb{Q}$ with everywhere good reduction.

Recall that we say that an abelian variety $A/\mathbb{Q}$ has good reduction at a prime $p$ if there exists an abelian scheme $\mathscr{A}/\mathbb{Z}_{(p)}$ (equiv. an abelian scheme $\mathscr{A}/\mathbb{Z}_p$) such that $\mathscr{A}_\mathbb{Q}=A$ (resp. $\mathscr{A}_{\mathbb{Q}_p}=A_{\mathbb{Q}_p})$. Recall that an abelian scheme $\mathscr{A}/S$ is a smooth proper group scheme with connected fibers (one usually requires constant fiber dimension, but this is automatic in our case—when $S$ is integral).

Intuitively, this means that an abelian variety has good reduction at $p$ when there is a ‘good notion’ of what the ‘reduction modulo $p$‘ of $A$ means. In particular, not to state the obvious, the reduction being $\mathscr{A}_{\mathbb{F}_p}$.

This result is usually not stated in this format in fact, it’s stated in the following equivalent form:

Theorem 2: There does not exist a non-trivial abelian scheme $\mathscr{A}/\text{Spec}(\mathbb{Z})$.

One might think that these are exactly the same, but there is something to be said. Namely, just because we can find a model $\mathscr{A}_p/\mathbb{Z}_{(p)}$ why does this mean that we can find a global model $\mathscr{A}/\mathbb{Z}$?

Proof(Theorem 1=Theorem 2): Clearly if $\mathscr{A}/\mathbb{Z}$ is an abelian scheme then $A:=\mathscr{A}_\mathbb{Q}$ has everywhere good reduction in particular with model $\mathscr{A}_{\mathbb{Z}_{(p)}}/\mathbb{Z}_{(p)}$ for all $p$. Thus, we see that Theorem 1 implies Theorem 2.

The converse is the non-obvious step. The key observation is that there one needn’t try to glue the models $\mathscr{A}_p/\mathbb{Z}_{(p)}$ together to get some group scheme over $\text{Spec}(\mathbb{Z})$. Instead, there is always a canonically associated smooth group scheme $\mathscr{G}/\text{Spec}(\mathbb{Z})$ associated to an $A/\mathbb{Q}$ which is a candidate for our desired abelian scheme $\mathscr{A}/\text{Spec}(\mathbb{Z})$. This allows us to bypass any nasty gluings and, instead, just compare $\mathscr{G}$ with these various local factors $\mathscr{A}_p$.

This cryptically describe group scheme $\mathscr{G}$ is none other than the Néron model of $A$ over $\text{Spec}(\mathbb{Z})$. Thus, we aim to show that $\mathscr{G}$ is proper which will then imply that its automatically an abelian scheme—the connectedness of the fibers will then follow from standard theory around the theorem on formal functions (e.g. see corollary 2.23 in Illusie’s Topics in Algebraic Geometry).

To see that $\mathscr{G}/\text{Spec}(\mathbb{Z})$ is proper we proceed as follows. Note that it suffices to show that for all primes $p\in\text{Spec}(\mathbb{Z})$ there is a neighborhood $V_p\subseteq\text{Spec}(\mathbb{Z})$ of $p$ such that $\mathcal{G}_{V_p}$ is proper. That said, note that the abelian scheme $\mathscr{A}_p/\mathbb{Z}_{(p)}$ must spread out to an abelian scheme $\mathscr{A}_{U_p}/U_p$ for some open $U_p\subseteq\text{Spec}(\mathbb{Z})$. Note then that $\mathscr{A}_{U_p}$ and $\mathscr{G}_{U_p}$ are both Néron models for $A$ over $U_p$ and thus, by unicity of Néron models, $\mathscr{G}_{U_p}=\mathscr{A}_{U_p}$. Since $\mathscr{A}_{U_p}$ is proper the result follows. $\blacksquare$

Both of these equivalent variations of the theorem of Fontaine are useful in their own right—each shedding a different light on a different aspect of why Fontaine’s theorem is stupendous. Let us explain these two perspectives (which, again, by the above equivalence are purely cosmetic).

To me the fact that there is no abelian variety $A/\mathbb{Q}$ with everywhere good reduction is a statement about Galois representations and their rigidity. Namely, it follows from the Néron-Ogg-Shafarevich that $A$ has good reduction at $p$ if and only if the $\ell$-adic representation

$\rho:G_{\mathbb{Q}}\to \text{GL}(V_\ell(A))$

is unramified at $p$ for all $\ell\ne p$ (equivalently some $\ell\ne p$)–an extension of this to the case when $\ell=p$ is due to Coleman-Iovita in which case unramified is replaced by crystalline (a condition from $p$-adic Hodge theory). Thus, we can interpret the non-existence of such an abelian variety as a sort of non-existence of an extremely nice geometric representation. This is the perspective that the actual proof of Fontaine takes.

The other perspective, the fact that there are no abelian schemes over $\mathbb{Z}$ is more a rigidity statement about varieties over $\mathbb{Z}$. Namely, any abelian scheme over $\mathbb{Z}$ would provide, in some sense, a ‘universally definable’ abelian scheme. Namely, such an abelian scheme $\mathscr{A}$ would give rise to an abelian scheme over any scheme $S$: namely $\mathscr{A}_S$. Considering how different abelian varieties act over different characteristics, and how they act in sometimes incommensurate ways, it’d be surprising that we could find one that placates the $p$‘s for all $p$. The statement of a Fontaine is a confirmation of this suspicion.

A result on curves

Now, while the theorem of Fontaine is astounding in its own right, it has some stupendous consequences that follow with very little work

First and foremost amongst these consequences is the following:

Theorem 3: The only smooth proper curve $C/\mathbb{Z}$ is $\mathbb{P}^1_\mathbb{Z}$.

Here a curve means that every fiber $C_{\mathbb{F}_p}$ is a geometrically connected variety of dimension $1$.

Proof: The clever idea is to realize that if $C/\mathbb{Z}$ is a curve, then $\mathrm{Jac}(C_\mathbb{Q})$, the Jacobian of $C_\mathbb{Q}$, should be an abelian variety with everywhere good reduction. That said, this idea has one tiny hiccup. Namely, what if $\mathrm{Jac}(C)$ is trivial? This happens, for example, if $C=\mathbb{P}^1_\mathbb{Z}$ and, more generally, if $C_\mathbb{Q}$ has genus $0$. Thus, we must do something different in that case.

So, as we saw above, the proof breaks into two cases $g(C)=0$ and $g(C)>0$. Note, here, that $g(C)$ means the genus of any fiber of $C$. This is well-defined (in the sense of being independent of fiber) by smooth proper base change or, much more simply, the fact that the Euler characteristic of the structure sheaf is locally constant function on the base (for flat proper families).

Let us first deal with the case when $g(C)>0$. Then, we claim that the Jacobian $\mathrm{Jac}(C)$ has everywhere good reduction, and since $\dim(\mathrm{Jac}(C_\mathbb{Q}))=g(C)>0$ this contradicts Fontaine’s theorem. There are two ways to see this. Namely, one can use the fact that the Jacobian scheme (defined as the connected component of the relative Picard scheme) is defined for much more general objects than projective varieties over fields. In fact, the curve $C/\mathbb{Z}$ has its own Jacobian $\mathrm{Jac}(C)$ which serves as a model of $\mathrm{Jac}(C_\mathbb{Q})$ showing that it has good reduction.

A simpler way of seeing that $\mathrm{Jac}(C_\mathbb{Q})$ has everywhere good reduction is to recall that the Néron-Ogg-Shafarevich criterion implies that $\mathrm{Jac}(C_\mathbb{Q})$ has good reduction at $p$ if and only if $V_\ell \text{Jac}(C_\mathbb{Q})$ is unramified (for some/any $\ell\ne p$). But,

$V_\ell \text{Jac}(C_\mathbb{Q})=(H^1_{\acute{e}\text{t}}(\text{Jac}(C_\mathbb{Q})_{\overline{\mathbb{Q}}},\mathbb{Q}_\ell))^\vee$

as Galois representations and so, consequently, it suffices to see that the $\ell$-adic cohomology of $\text{Jac}(C_\mathbb{Q})$ is unramified. But, as is well-known, and implicit in the computation of the cohomology of a curve

$H^1_{\acute{e}\text{t}}(C_{\overline{\mathbb{Q}}},\mathbb{Q}_\ell)=H^1_{\acute{e}\text{t}}(\text{Jac}(C_\mathbb{Q})_{\overline{\mathbb{Q}}},\mathbb{Q}_\ell)$

as Galois representations. So, since $C_\mathbb{Q}$ admits a model over some $\mathbb{Z}_{(p)}$ the $G_\mathbb{Q}$ representation $H^1_{\acute{e}\text{t}}(C_{\overline{\mathbb{Q}}},\mathbb{Q}_\ell)$, and thus $H^1_{\acute{e}\text{t}}(\text{Jac}(C_\mathbb{Q})_{\overline{\mathbb{Q}}},\mathbb{Q}_\ell)$ is unramified from where the good reduction of $\mathrm{Jac}(C_\mathbb{Q})$ at $p$ follows.

Remark: The Albanese variety of an $X$, which for curves is the dual of the Jacobian, is the ‘$1$-motive’ associated to $X$. In other words, $X$ and $\mathrm{Alb}(X)$ share the same first cohomologies.

Regardless, we see that if $g(C)>0$ then $\mathrm{Jac}(C_\mathbb{Q})$ has everywhere good reduction and so contradicts Fontaine’s theorem.

So, suppose now that $C/\mathbb{Z}$ is a smooth proper curve of genus $0$. Note then that every geometric fiber of $C$ must be the projective line (since, trivially, all genus $0$ curves over an algebraically closed field are such). Thus, $C/\mathbb{Z}$ is a so-called Brauer-Severi scheme—a scheme which is étale locally on the base isomorphic to projective space (see Grothendieck’s article Le Groupe de Brauer in Dix Exposes for more details). Thus, $C$ defines a class in the Brauer group $\mathrm{Br}(\mathbb{Z})$ which is trivial if and only if $C\cong\mathbb{P}^1_\mathbb{Z}$. But, $\mathrm{Br}(\mathbb{Z})=0$ (see this post) from where the claim follows. $\blacksquare$

There are two natural questions that one might ask following Theorem 3.

First, can we give an analogue of Theorem 2? Namely, is it true that if $C/\mathbb{Q}$ has everywhere good reduction (again that for every $p\in\text{Spec}(\mathbb{Z})$ there exists a smooth proper curve $C_p/\mathbb{Z}_{(p)}$ such that $(C_p)_\mathbb{Q}=C$) then $C=\mathbb{P}^1_\mathbb{Q}$? Again, this may seem obvious, but we have a major snag: Néron models may not exist for arbitrary curves.

That said, I believe a proof can be easily adapted with a little extra work for the genus $0$ case:

Theorem 4: Let $C/\mathbb{Q}$ be a smooth proper geometrically connected curve with everywhere good reduction. Then, $C\cong\mathbb{P}^1_{\mathbb{Q}}$.

Proof: It’s known (see, for example, the ninth chapter of Qing Liu’s text) that if $g(C)>0$ then $C$ has a unique minimal regular model over $\mathcal{C}/\mathbb{Z}$. I claim that $\mathcal{C}$ is smooth. To see this, again, it suffices to work Zariski locally on the base in which case one can imitate the proof as in Theorem 2.

Remark: It’s known, but with considerably more work, that in the above setup $C$ has a Néron model. For example, see this paper.

We are thus reduced to the case when $g(C)=0$. Note that since $C$ has genus $0$ that it defines a class $[C]\in\mathrm{Br}(\mathbb{Q})$ which is trivial if and only if $C\cong\mathbb{P}^1_\mathbb{Q}$. Thus, it suffices to show that $[C]=0$. Now, from global class field theory  the group $\mathrm{Br}(\mathbb{Q})$ fits into the following short exact sequence

$\displaystyle 0\to \mathrm{Br}(\mathbb{Q})\to \bigoplus_{p\in\text{Spec}(\mathbb{Z})}\mathrm{Br}(\mathbb{Q}_p)\oplus\mathrm{Br}(\mathbb{R})\to \mathbb{Q}/\mathbb{Z}\to 0$

where the first map is the obvious one (pullback of the cohomology class/base change of the central simple algebra) and the second one, the so-called invariant map $\mathrm{inv}$, is the ‘sum coordinates map’ once the canonical identifications

$\mathrm{Br}(\mathbb{Q}_p)\cong\mathbb{Q}/\mathbb{Z},\qquad \mathrm{Br}(\mathbb{R})=\frac{1}{2}\mathbb{Z}$

In particular, we see that if $\alpha\in\mathrm{Br}(\mathbb{Q})$ is such that $\alpha_p\in\mathrm{Br}(\mathbb{Q}_p)=0$ for all $p$ prime, then $\alpha=0$. This doesn’t quite follow from the injectivity of the above map, but the injectivity followed by the fact that $\mathrm{inv}((\alpha_v))=0$ (where $v$ ranges over all places of $\mathbb{Q}$). In particular, if $\alpha_p=0$ for all $p$ prime, but $\alpha_\infty\in\mathrm{Br}(\mathbb{R})\ne 0$, then $\mathrm{inv}((\alpha_v))=\frac{1}{2}$ which is impossible. Thus, $(\alpha_v)=0$ and so by the injectivity of the map, $\alpha=0$.

Now, note that for all $p$ prime the image of $[C]$ in $\mathrm{Br}(\mathbb{Q}_p)$ is precisely $[C_{\mathbb{Q}_p}]$. Thus, by the previous paragraph, it suffices to show that $[C_{\mathbb{Q}_p}]=0$ for all $p$. But, by assumption there exists a genus $0$ curve $C_p/\mathbb{Z}_p$ such that $(C_p)_{\mathbb{Q}_p}=C_{\mathbb{Q}_p}$. So, in particular, $[C_{\mathbb{Q}_p}]$ is the image of $[C_p]$ under the map

$\mathrm{Br}(\mathbb{Z}_p)\to\mathrm{Br}(\mathbb{Q}_p)$

That said, since $\mathbb{Z}_p$ is a Henselian local ring with residue field $\mathbb{F}_p$, we know from a theorem of Grothendieck (see Corollary 2.13 of Chapter 4 of Milne’s text on étale cohomology) that

$\mathrm{Br}(\mathbb{Z}_p)=\mathrm{Br}(\mathbb{F}_p)$

and it’s a classic fact (implied, for example, by Wedderburn’s theorem) that $\mathrm{Br}(\mathbb{F}_p)=0$. Thus, we see that $[C_{\mathbb{Q}_p}]=0$ as desired. Since $p$ was arbitrary, we conclude that $[C]=0$ as desired. $\blacksquare$

The second question is whether one can deduce Theorem 1 from Theorem 3 (or, equivalently, Theorem 2 from Theorem 4). Namely, can one deduce that if all curves with good reduction everywhere are $\mathbb{P}^1_\mathbb{Q}$ that all abelian varieties with everywhere good reduction are trivial?

There are two impediments to such a claim

1. Not all abelian varieties are Jacobians.
2. Whether the proof that good reduction at $p$ of $C/\mathbb{Q}$ implies good reduction of $\mathrm{Jac}(C)$ at $p$ is reversible.

Both of these are, as far as I can now tell, difficult to fix. Namely, it’s a classic fact that every abelian variety over a field is a quotient of a Jacobian (cf. Milne’s text on on abelian varieties, specifically III.10). One may be able to finagle this in to something meaningful, but I haven’t thought about it too deeply.

Remark: I must have been sleep deprived when I wrote this initially for I had written the obviously false statement “Every abelian variety is isogenous to a Jacobian”. Thanks to Arithmetica for pointing this out in the comments!

But, 2. is definitively not fixable. Namely, there are curves for which their Jacobian has good reduction at a prime but the curve itself does not (see the nice answer of Noam Elkies here).

Remark: From the last paragraph we know that the $\ell$-adic cohomology of a curve is not enough to detect its reduction type since, after all, a curve and its Jacobian have the same first cohomology groups (and the other cohomology groups of a curve are non-interesting). Thus, one may wonder if there is anything at all analogous to Néron-Ogg-Shafarevich for curves.

Somewhat surprisingly the answer is yes. In fact, instead of caring about the action of the inertia group on the cohomology of the curve (Néron-Ogg-Shafarevich saying that for an abelian variety good reduction is equivalent to triviality of this action) one looks at the ‘action’ of the inertia group on the étale fundamental group of the curve. More specifically, if $C/K$ (here $K=\text{Frac}(R)$ where $R$ is a DVR) is a curve, then a $K$-point $x$ gives a splitting of

$1\to \pi_1(C_{\overline{K}})\to \pi_1(C)\to G_K\to 1$

and thus a conjugation action on $\pi_1(C_{\overline{K}})$. The induced map $G_K\to \text{Out}(\pi_1(C_{\overline{K}})$ is independent of the choice of $K$-point and a Theorem of Oda says that this homomorphism can determine reduction type—in fact, one only needs to check the action on the third term of the derived sequence. See section 10.2 of this for details.

Where this fails

Before we continue on, it’s worth mentioning in what ways Fontaine’s theorem can fail. Specifically, Fontaine’s theorem is stated specifically for $\mathbb{Z}$ and this manifests itself clearly in the proof which relies heavily on the Odlyzko bounds . So, what happens if you’re not dealing with $\mathbb{Q}$? We’ll mention later that Fontaine’s results do extend a little beyond $\mathbb{Q}$, but let’s mention some cases where it does fail.

Perhaps the best way of producing abelian varieties $A/K$, for $K$ a number field, with everywhere good reduction is to deal with abelian varieties with CM (complex multiplication). Recall that $A$ is said to have CM if $\mathrm{End}^0(A):=\mathrm{End}(A)\otimes_\mathbb{Z}\mathbb{Q}$ contains a number field of dimension $2\dim(A)$. Note that some authors might call this rational CM and reserve the phrase ‘CM’ for what we’d call geometric CM (i.e. that $\mathrm{End}^0(A_{\overline{K}})$ contains such a field).

The reason why abelian varieties with CM are good places to look for abelian varieties with everywhere good reduction is the following:

Theorem 5: Let $A/K$ be CM. Then, there exists a finite extension $L/K$ such that $A_L$ has everywhere good reduction.

Proof(Sketch): The abelian variety $A/K$ will have good reduction almost everywhere trivially.  So, we only need to take care of finitely many primes, say the set of which is $S$. Let $\ell$ be a rational prime such that $\mathfrak{p}\nmid\ell$ for all $\mathfrak{p}\in S$. Then, it suffices to show that for each $\mathfrak{p}\in S$ the image $\rho(I_\mathfrak{p})$ is finite where

$\rho:G_K\to\text{GL}(V_\ell A)$

since we can then kill these non-trivial inertial actions by a finite base extension. But, since $A$ has a CM we know that $\rho$ has image lying in $(\mathcal{O}_F\otimes_\mathbb{Z}\mathbb{Z}_\ell)^\times$. In particular, $\rho$ has abelian image and thus factors through $G_K^\mathrm{ab}$.

In particular, $\rho\mid_{I_\mathfrak{p}}$ factors through $I_\mathfrak{p}^\mathrm{ab}$ for all $\mathfrak{p}\in S$. But, by local class field theory we know that $I_\mathfrak{p}$ is isomorphic to $\mathcal{O}_{K_\mathfrak{p}}^\times$ which is a product of a finite group and a free $\mathcal{O}_{K_\mathfrak{p}}$-module. Also, $(\mathcal{O}_F\otimes\mathbb{Z}_\ell)^\times$ is a product of a finite group and a free module over

$\displaystyle A:=\prod_{\mathfrak{q}\mid\ell}\mathcal{O}_{F_\mathfrak{q}}$

for similar reasons.

Since there are no continuous morphisms $\mathcal{O}_{K_\mathfrak{p}}$ to $A$ (since $\ell$ is topologically nilpotent in the latter and not in the former) we see that $\rho\mid_{I_\mathfrak{p}}$ has image in a finite group. $\blacksquare$

An explicit example of a non-CM elliptic curve with everywhere good reduction was given by Tate as follows. Let

$\displaystyle \varepsilon:=\frac{5+\sqrt{29}}{2}$

and $K=\mathbb{Q}(\sqrt{29})$. Then, Tate showed that

$E:y^2x^2+xy+\varepsilon y=x^3$

has everywhere good reduction over $K$. For more explicit examples see this paper of Dembélé and Kumar.

Ideas of proof

Not surprisingly the proof of Theorem 1 is exceedingly difficult. But, before we give the vaguest of ideas of how this is attacked, let us mention the following. The elliptic curve case of Theorem 1 was known well before the general proof of Fontaine. Namely, Tate proved that there are no elliptic curves $E/\mathbb{Q}$ with everywhere good reduction by very elementary means. We outline this now.

It suffices to prove Theorem 2. So, let $\mathcal{E}/\mathbb{Z}$ be an elliptic scheme. It’s known that a sufficient (and perhaps necessary, but I’m not positive) condition for an elliptic scheme $\mathcal{E}'/S$ with $S$ affine to be isomorphic to a Weierstrass form (i.e. a cubic in $\mathbb{P}^2_S$) is that $f_\ast\Omega^1_{\mathcal{E}'/S}$ is free. That said, it’s a vector bundle and since all vector bundles on $\mathbb{Z}$ are free we conclude that $\mathcal{E}$ must be a Weierstrass form.

Remark: The above can be stated in a slightly less obnoxious way for our particular needs. Namely, since $\mathrm{Cl}(\mathbb{Z})=0$ we know that $E_\mathbb{Q}$ has a global minimal Weierstrass form—we can do it at all primes individually and the fact that we’re class number $1$ implies that we can globally work one prime at a time. Then, one can easily see that this global minimal Weierstrass form is our $\mathcal{E}$.

Thus, we know that $\mathcal{E}$ is isomorphic to a cubic in $\mathbb{P}^2_\mathbb{Z}$ with $5$ coefficients $a_1,a_2,a_3,a_4$, and $a_6$. Since $\mathcal{E}$ has everywhere good reduction this implies that $\Delta(a_1,a_2,a_3,a_4,a_6)=\pm 1$. One can then explicitly show that the Diophantine equation

$\Delta(a_1,a_2,a_3,a_4,a_6)=\pm 1$

has no integral solutions. Of course, this does not, at all, extend to higher dimensions.

So, about Fontaine’s proofs. Fontaine has given two proofs of this result. The first is found in l n’y a pas de variété abélienne sur $\mathbb{Z}$ which, essentially, comes down to super, super fine analysis of ramification of finite flat group schemes over $\mathbb{Z}$. It’s this paper that most readily implies that there are no abelian varieties with everywhere good reduction over other fields (e.g. $\mathbb{Q}(i)$). For a nice overview, with ample background, to this proof see these notes of Schoof.

He later gave another more conceptual proof of the result using $p$-adic Hodge theory. Namely, in Schémas lisse et propres sur $\mathbb{Z}$ he proves the following astounding result which greatly generalizes Theorem 1:

Theorem 6(Fontaine, ’93): Let $\mathscr{X}/\mathbb{Z}$ be a smooth proper scheme. Then, for all $i\ne j$ and $i+j\leqslant 3$ the Hodge number $h^{i,j}(X)=0$ where $X:=\mathscr{X}_\mathbb{Q}$.

Here, the Hodge number is as usual:

$h^{i,j}(X):= H^j(X,\Omega^i_{X/\mathbb{Q}})$

This, to me, is absolutely astounding. Namely,  knowing that a scheme is proper and smooth over $\mathbb{Z}$ gives one an astounding control on the cohomology of the generic fiber. This is entirely unexpected, and somewhat mysterious—why $i+j\leqslant 3$?

That said, we can immediately see that Theorem 6 implies Theorem 1 and Theorem 2. Namely, if $\mathscr{X}/\mathbb{Z}$ is an abelian scheme of positive dimension then by Theorem 6

$0=h^{0,1}(X)=H^0(X,\Omega^1_{X/\mathbb{Q}})=H^0(X,\mathcal{O}_X)=1$

which is impossible (here we have used the fact that abelian varieties (more generally group schemes) are parallelizable—they have trivial cotangent sheaf). Similarly, if $\mathscr{X}/\mathbb{Z}$ is a curve then

$0=h^{0,1}(X)=H^0(X,\Omega^1_{X/\mathbb{Q}})=g(\mathscr{C})$

thus proving that the genus of $\mathscr{C}=0$ (in which case the rest is done as in the proof of Theorem 3).

This proof uses the fact that if $\mathscr{X}$ is such a scheme, then for any $\ell$ the representation $\rho:G_\mathbb{Q}\to H^i_{\acute{e}\text{t}}(X_{\overline{\mathbb{Q}}},\mathbb{Q}_\ell)$ is unramified at every $p\ne\ell$ and crystalline at $\ell$ and deduces strong restrictions on the cohomology of $X$ as a result (the Hodge numbers coming into play via the $C_{\text{dR}}$-conjecture—which is a theorem).

In the direction of what happens if one allows just very mild bad reduction we have the following theorem of Abrashkin:

Theorem 7(Abrashkin): Let $X/\mathbb{Q}$ be a smooth projective variety such that $X$ has good reduction away from $3$ and bad semi-stable reduction at $3$. Then, $h^{1,1}(X)=b_2(X)$.

Here, $b_2(X)$ denotes the second Betti number of $X$ whose definition can be given, for example, as $\dim_\mathbb{Q} H^2(X_\mathbb{C},\mathbb{Q})$.

As an example of this, let $X/\mathbb{Q}$ be an abelian variety of dimension $g$ at least $2$. Then, note that

$h^{1,1}(X)=H^1(X,\Omega^1_{X/\mathbb{Q}})=H^1(X,\mathcal{O}_X)=H^0(X,\Omega^1_{X/\mathbb{Q}})=H^0(X,\mathcal{O}_X)=1$

that said $\displaystyle b_2(X)={2g \choose 2}>1$. Thus, it can’t be the case that $X$ has good reduction outside of $3$ and semi-stable reduction at $3$.

Analogies and intuitions

One thing that one might be interested in is why one would expect Fontaine’s theorem to hold. In particular, one might try and analogize the statement to other situations and prove the result holds there. This is what we try and do in this section.

In particular, we are interested in studying abelian schemes over $\mathbb{A}^1_{\mathbb{F}_q}$, $\mathbb{A}^1_{\overline{\mathbb{F}_q}}$, and $\mathbb{A}^1_\mathbb{C}$. The first of these is a reasonable quest due to the classic analogy between $\text{Spec}(\mathbb{Z})$ and $\mathbb{A}^1_{\mathbb{F}_q}$ (or, if you prefer, the number field/function field analogy). The case $\mathbb{A}^1_{\overline{\mathbb{F}_q}}$ is a simpler first step to solving the case over $\mathbb{A}^1_{\mathbb{F}_q}$, and $\mathbb{A}^1_\mathbb{C}$ is a toy case where things should be easier.

But, before we start, we need to decide what it is that we’re trying to show exactly. Namely, what we’d like to suss out is whether in all three of these cases all curves/abelian schemes have to be ‘isotrivial’. By this, we mean a ‘constant family’ of such objects. Rigorously, a family $\mathscr{X}/\mathbb{A}^1_F$ of curves/abelian schemes is isotrivial if there exists an object $X/F$ such that $\mathscr{X}=X\times_F \mathbb{A}^1_F$—it’s a constant family with fiber $X$.

This seems like the right generalization in the sense that such a result, and Fontaine’s theorem, say that the only curves/abelian schemes are the obvious ones. For $\mathbb{Z}$ there are none, and over $\mathbb{A}^1_F$ there are just the isotrivial ones. Of course, one could say ‘something something $\mathbb{F}_1$ something something’, but I am not that man. And, as the man himself says

The case over the complex line

Let us start with what is the most intuitive case. What are the abelian schemes over $\mathbb{A}^1_\mathbb{C}$? In particular, do there exist any which are not isotrivial?

One might start with the case of elliptic schemes where, again, one might proceed as in Tate’s proof over $\mathbb{Z}$. Namely, since $\mathbb{A}^1_\mathbb{C}$ has only trivial vector bundles we can write such an elliptic scheme $\mathscr{E}$ as a Weierstrass form. Moreover, since we’re over characteristic $0$ we can even write it in the usual edulcoration

$\mathscr{E}:y^2=x^3+A(t)x+B(t)$

where $A(t),B(t)\in\mathbb{C}[t]$. Then, one can try and show that

$-16(4A(t)^3+27B(t)^2)=c$

has no solution where $c\in\mathbb{C}^\times$. That said, again, this lacks the conceptual air to be satisfying and it also lacks any obvious recourse to higher-dimensions.

That said, one might have an intuition that it would be hard to make a non-isotrivial family of curves/abelian varieties because there can be no ‘twisting’. Namely, one might have the following topological intuition. Let

$f:\mathscr{A}\to\mathbb{A}^1_\mathbb{C}$

be an abelian scheme and consider its analytification

$f^\text{an}:\mathscr{A}^\text{an}\to\mathbb{C}$

Then, since $f^\text{an}$ is a smooth and proper map we know, by Ehresmann’s lemma, that $f^\text{an}$ is a locally trivial fiber bundle. But, locally trivial fiber bundles on the complex line are, well, trivial. Thus, topologically the family is trivial. But, abelian varieties are essentially composed of two parts: topology+Hodge filtration (see the end of this post for details). We’ve now shown that the topological portion of the family is trivial, and so one might then imagine that this can be leveraged to show isotriviality in any case where we have a simply connected base.

Of course, there is a snag. Just because the topological aspect of the family (the local system of singular homology groups) is trivial there is no, a priori, reason to believe that the filtration aspect must also be trivial. The precise issue will be made clear in the following proof:

Theorem 8: Let $V/\mathbb{C}$ be a smooth connected algebraic variety such that $V^\text{an}$ is simply connected. Then, every family of abelian varieties $f:\mathscr{A}\to V$ is isotrivial.

Proof: Let $f:\mathscr{A}\to V$ be such an abelian variety. By 4.4.3 of Deligne’s Theorie de Hodge II there is an equivalence of categories between abelian schemes and polarizable $\mathbb{Z}$-variations of Hodge structure. The map is given by

$\mathscr{A}\mapsto (R^1f_\ast^\text{an}\underline{\mathbb{Z}})^\vee$

In particular, since $V^\text{an}$ is simply connected then the $\mathbb{Z}$-local system $(R^1f_\ast^\text{an}\underline{\mathbb{Z}})^\vee$ is a constant local system but, a priori, one might be worried that the Hodge filtration is not constant.

That said, here is a way one might remedy this. Since $V^\text{an}$ is compactifiable by Hironaka’s theorem and by the discussion following Theorem 11 of this article we may conclude that $(R^1f_\ast^\text{an}\underline{\mathbb{Q}})^\vee$ is constant and thus, by Deligne’s theorem, that $\mathscr{A}$ is isogenous to a constant family $\underline{A}$. But, the kernel of some isogeny $\underline{A}\to \mathscr{A}$ must be a finite group constant group scheme $\underline{\mathbb{Z}/n\mathbb{Z}}$. Thus, $\mathscr{A}\cong \underline{A/(\mathbb{Z}/n\mathbb{Z})}$ as desired. $\blacksquare$

Remark: Using the results mentioned in this mathoverflow post, namely the cited rigidity result of Strauch, one can use the above cited result of Peters and Steenbrink to show that not only must a $\mathbb{Q}$-VHS on a compactifiable complex manifold be constant (as a VHS) but also that the same is true for integral Hodge structures. Thus, one can bypass the above argument dealing with the isogeny category.

In the special case of the affine line, the above cited mathoverflow post also contains a ‘more elementary’ proof.

In particular, we derive the following two corollaries:

Corollary 9: Let $C/\mathbb{A}^1_\mathbb{C}$ be a smooth proper curve. Then, $C$ is isotrivial.

Proof: The proof is as in the proof of Theorem 3 or Theorem 4 for the positive genus case, and for the genus $0$ case it follows, again, from the fact that $\mathrm{Br}(\mathbb{A}^1_\mathbb{C})=0$. $\blacksquare$

Remark: It should be noted that the above argument doesn’t totally emphasize the key property that $V$ is algebraic. It was used in two places. First, Deligne’s result about classifying abelian schemes in terms of $\mathbb{Z}$-VHS (of the right type) required algebraicity of $V$. That said, the veracity of the statement where $V$ is an arbitrary smooth connected complex manifold is not obvious to me. The second place we used it was in the context of the Peters and Steenbrink paper to know that $V$ was compactifiable.

Now, these seem like silly remarks, but as soon as one leaves the algebraic category Theorem 8 becomes wildly false. For example consider the following family of elliptic curves over $\mathfrak{h}$, the upper half-plane:

$\mathscr{E}:=\left\{[X:Y:Z:t]\in\mathbb{P}^2_\mathbb{C}\times\mathfrak{h}:Y^2Z=4X^3-g_4(t)XZ^2-g_6(t)Z^3\right\}$

where $g_4$ and $g_6$ are the usual modular forms of level $\Gamma(1)$ and weights $4$ and $6$. This is, of course, not an isotrivial family even though $\mathfrak{h}$ is a simply connected complex manifold. That said, this not a contradiction since $\mathfrak{h}$ is not algebraic!

The case of the line over a finite field

We’d now like to deal with the case of the line $\mathbb{A}^1_{\mathbb{F}_q}$. Namely, we’d like to suss out whether or not there are non-isotrivial curves/abelian varieties. To spoil the drama, let us remark that we will not be able to prove anything substantive for genus greater than $1$ or abelian varieties of dimension greater than $1$.

So, let us begin with the obvious case: genus $0$ curves. We then have the following obvious extension of the above ideas:

Theorem 10: Let $C/\mathbb{A}^1_F$ be a smooth proper curve of genus $0$ where $F=\mathbb{F}_q$ or $F=\overline{\mathbb{F}_q}$. Then, $C\cong \mathbb{P}^1_{\mathbb{A}^1_F}$.

The proof, again, is just the fact that

$\mathrm{Br}(\mathbb{A}^1_{\mathbb{F}_q})=\mathrm{Br}(\mathbb{A}^1_{\overline{\mathbb{F}_q}})=0$

and, of course, this extends to show that any curve of genus 0 over $\mathbb{A}^1_F$ is just $\mathbb{P}^1_{\mathbb{A}^1_F}$ as long as $\mathrm{Br}(F)=0$. This follows from the fact that

$\mathrm{Br}(\mathbb{A}^1_F)=\mathrm{Br}(F)$

as is fairly easily computed.

Let us now move on to the slightly more interesting case of genus $1$ curves. We start, not shockingly, with the case of elliptic schemes. Namely, is every elliptic scheme $\mathscr{E}/\mathbb{A}^1_F$, where $F\in\{\mathbb{F}_q,\overline{\mathbb{F}_q}\}$, isotrivial?

Here is where we want to use the case of the complex line as spiritual guidance. Namely, there we saw that the operative thing about $\mathbb{A}^1_\mathbb{C}$ was that it was simply connected and that it forced the constancy of the local system $(R^1 f_\ast^\text{an}\underline{\mathbb{Z}})^\vee$ (and the Hodge filtration). Of course, $\mathbb{A}^1_{\overline{\mathbb{F}_q}}$ is not simply connected, we can’t access something like integral cohomology, and we have no such thing as a Hodge filtration.

So, at first glance, it seems like it’s literally impossible to adapt the proof in any meaningful way. That said, we do have ways of working around each of these difficulties. Namely, the line over $\overline{\mathbb{F}_q}$ is not (étale) simply connected but it is ‘prime-to-p’ simply connected—its fundamental group has no prime-to-p quotients. We cannot access the integral cohomology but we can access the $\mathbb{Z}/n\mathbb{Z}$-cohomology as the sheaf $(R^1f_\ast\underline{\mathbb{Z}/n\mathbb{Z}})^\vee$. And, what we lack in our ability to deal with Hodge theory we’ll make up for in an understanding of the moduli space of elliptic curves with a fixed cohomology class.

NB: Let’s assume that $q$ is not a power of $2$ or $3$ for the sake of convenience.

In particular, let us state the following rigorously:

Theorem 11: Let $\mathscr{E}/\mathbb{A}^1_{\overline{\mathbb{F}_q}}$ be an elliptic scheme. Then, $\mathscr{E}$ is isotrivial.

Proof: Let us begin by noting that for all $N$ coprime to $p$ we have that $\mathscr{E}[N]$ is a finite étale group scheme over $\mathbb{A}^1_{\overline{\mathbb{F}_q}}$ with (geometric fiber) isomorphic to $\underline{(\mathbb{Z}/N\mathbb{Z})^2}$. We claim that for arbitrarily large $N$ it is, in fact, constant.

To see this, note that it is constant if and only if the monodromy action $G:=\pi_1(\mathbb{A}^1_{\overline{\mathbb{F}_q}},\overline{x})$ is trivial on the fiber $\mathcal{E}[N]_{\overline{x}}$ for some/any geometric point $\overline{x}$. Since the actual geometric point is irrelevant we suppress it from the notation. But, note that the action of $G$ on the fiber of $\mathcal{E}[N]$ corresponds to an element of the group

$\text{Hom}_\text{cont.}(G,\text{GL}_2(\mathbb{Z}/N\mathbb{Z}))$

but for any $N\not\equiv 0,1,2\mod p$ we have that

$|\text{GL}_2(\mathbb{Z}/N\mathbb{Z})|=(N^2-1)(N^2-N)$

has prime-to-$p$ order and thus, since $G$ has no prime-to-p quotients, this map must be trivial.

Thus, we see that for arbitrarily large $N$ the group scheme $\mathscr{E}[N]$ is constant. Thus, we can choose, for arbitrarily large $N$, trivializations

$\alpha:\mathscr{E}[N]\xrightarrow{\approx}\underline{(\mathbb{Z}/N\mathbb{Z})^2}$

The pair $(\mathscr{E},\alpha)$ then defines a map $\mathbb{A}^1_{\overline{\mathbb{F}_q}}\to Y(N)_{\overline{\mathbb{F}_q}}$ where, here, $Y(N)/\mathbb{Z}[\frac{1}{N}]$ denotes the open modular curve of level $\Gamma(N)$. This then extends to a map $\mathbb{P}^1_{\overline{\mathbb{F}_q}}\to X(N)_{\overline{\mathbb{F}_q}}$ (where $X(N)$ is the canonical compactification of $Y(N)$). But, for $N\gg 0$ the genus $X(N)$ is positive, and so any map from the projective line is constant.

This says precisely that the map $\mathbb{A}^1_{\overline{\mathbb{F}_q}}\to Y(N)_{\overline{\mathbb{F}_q}}$ factors through a point which says precisely that the family $(\mathscr{E},\alpha)$ and thus, in particular $\mathscr{E}$, is constant. $\blacksquare$

We would now like to extend this to the case over $\mathbb{F}_q$ which will largely just be a calculation of twists:

Theorem 12: Let $\mathscr{E}/\mathbb{A}^1_{\mathbb{F}_q}$ be an abelian scheme. Then, $\mathscr{E}$ is isotrivial.

Proof: By Theorem 11 we know that $\mathscr{E}_{\mathbb{A}^1_{\overline{\mathbb{F}_q}}}$ is isotrivial, let’s say isomorphic to $E\times_{\overline{\mathbb{F}_q}}\mathbb{A}^1_{\overline{\mathbb{F}_q}}$ for $E/\overline{\mathbb{F}_q}$ an elliptic curve. We claim that $E$ is actually defined over $\mathbb{F}_q$. To see this, we merely note that

$j(\mathscr{E})=j(\mathscr{E}_{\mathbb{A}^1_{\overline{\mathbb{F}_q}}})=j(E\times_{\overline{\mathbb{F}_q}}\mathbb{A}^1_{\overline{\mathbb{F}_q}})=j(E)$

and thus this common $j$-invariant lies in $\mathbb{F}_q(T)\cap\overline{\mathbb{F}_q}$ and thus in $\mathbb{F}_q$. So, we may assume without loss of generality that $E=E'_{\overline{\mathbb{F}_q}}$ for $E'/\mathbb{F}_q$.

So, we know now that $\mathscr{E}$ is a $\mathbb{A}^1_{\overline{\mathbb{F}_q}}/\mathbb{A}^1_{\mathbb{F}_q}$-twist of $E\times_{\mathbb{F}_q}\mathbb{A}^1_{\mathbb{F}_q}$. So, it suffices to show that all such twists are isotrivial.

That said, note that such twists are classified by

$H^1(G_{\mathbb{F}_q},\text{Aut}(E\times_{\mathbb{A}^1_{\mathbb{F}_q}}\mathbb{A}^1_{\overline{\mathbb{F}_q}}))$

where these automorphisms are as elliptic schemes over $\mathbb{A}^1_{\overline{\mathbb{F}_q}}$. But, it’s easy to see that all such automorphisms are just automorphisms of $E_{\overline{\mathbb{F}_q}}$ and thus all the twists classes are represented by twists of $E$, and thus are isotrivial. $\blacksquare$

One should be able to extend Theorem 12 to all genus $1$ curves, by analyzing the Jacobian, but I have not checked this entirely.

One is very hopeful after all of this. We might be able to extend the above to prove that every curve/abelian scheme over $\mathbb{A}^1_{\mathbb{F}_q}$, or at least $\mathbb{A}^1_{\overline{\mathbb{F}_q}}$, is isotrivial. Unfortunately, our world comes crashing down—it’s false. See this mathoverflow post. Thus, this is an example where the number field/fucntion field analogy breaks down a little bit.

Remark: It’s somewhat interesting to note the following. One might be able to summarize the above proof’s idea as: leverage the moduli space of elliptic curves (or, rather, some decorated elliptic curves) to show that all elliptic schemes are isotrivial. Thus, the question of isotriviality becomes one of pure geometry: is the geometry of the moduli space such that it can support interesting functions from the affine line/

This is interesting because in this (already mentioned) mathoverflow post Donu Arapura gives a proof that $\mathbb{A}^1_\mathbb{C}$ has no abelian schemes using, again, the geometry of some moduli space (or, rather, a cover of it). That said, his approach is to use the complex differential geometry of a moduli space to show that the affine line cannot support non-isotrivial abelian schemes (a hyperbolicity result). This cannot be extended to some sort of interesting algebraic condition on any of the locally symmetric varieties that are quotients of this moduli space else, in theory, this should translate to the positive characteristic world where, as mentioned above, the result is false for higher-dimensional abelian varieties!

1. “Namely, it follows from the Néron-Ogg-Shafarevich that A has good reduction at p if and only if the \ell-adic representation

\rho:G_{\mathbb{Q}}\to \text{GL}(V_\ell(A))

is unramified at p for \ell\ne p and crystalline when \ell=p.”

When \ell = p, this is actually a much more recent theorem of Coleman-Iovita. Of course crystalline Galois representations hadn’t been defined at the time the original NOS theorem was proved. 🙂

2. “Namely, it’s a classic fact that every abelian variety over a field is isogenous to a Jacobian.” This is wildly false. The correct statement (I believe) is that abelian varieties over reasonable fields are always isogenous to *quotients* of Jacobians.

Having said that, I really enjoyed this post!

1. Dear Arithmetica,

1. You’re totally right–the misattribution there is sloppy and, I think, the result of last minute changes to the post. As I’ve stated it now it’s also way weaker than need be (i.e. good reduction if and only if for all \ell, p blah blah if and only if for some \ell,p blah blah).

I’ll fix it right now.

2. You are absolutely correct! I’ll fix this mistake immediately.

Thanks again 🙂

2. By the way, it’s an OPEN PROBLEM to decide whether there’s a smooth proper scheme over Z whose geometric etale cohomology isn’t a direct sum of Q_ell(i)’s!!

1. Dear Arithmetica,

This is not *overly* shocking to me given that a) the easiest objects to even compute cohomology of explicitly are curves and abelian varieties and b) the above says that those are essentially non-existent.

That said, I have a question you might have some knowledge on. Do you know what the state-of-the-art is on what the maximal unramified-away-from-$p$ extension of $\mathbb{Q}$ looks like? Certainly in Fontaine’s original work showing the non-existence of abelian schemes he leverages that their Galois representations are unramified way-from-$p$ and crystalline at $p$, but can we say anything about these representations.

Pointedly my question is this: you made a comment about it not being known that there exists smooth proper $\mathbb{Z}$-schemes with cohomology not a direct sum of Tate twists. It seems like such a problem can be broken down into three parts:

1) Classifying $G_\mathbb{Q}$ representations unramified away-from-$p$ and crystalline at $p$.
2) Understanding when these come from $\mathbb{Q}$-schemes (i.e. partially solving the FM conjecture).
3) Understanding some sort of NOS type theorem for such $\mathbb{Q}$-schemes (i.e. when the nice properties of the cohomology propagates to the scheme itself).

I know that the answer to 3) is ‘not always’ (i.e. we needn’t have perfect propagation unless you are, say, a curve or abelian scheme) but I also don’t know partial results.

Can you say anything about 1), 2), or 3)?

Thanks again!

1. Dear Alex,
Here are my (possibly idiosyncratic or just plain dumb) opinions on your points 1.-3.

1. – This seems impossible outside of the cases dealt with by Fontaine. The Langlands program of course gives a conjectural description of these in terms of everywhere-unramified algebraic automorphic representations of GL_n(A_Q), but these are essentially impossible to count/find/anything. (Open problem: Rigorously show the existence of a non-essentially-self-dual cuspidal automorphic representation of GL_n(A_Q) for some n > 4.)

2. – As far as I’m concerned, this is solved by the FM conjecture. 🙂

3. – This seems like a fascinating area to explore. As you know, this already gets interesting for curves (lower central series quotients of pi_1 and the like); maybe that picture propagates to more general hyperbolic varieties?

BTW, if I had to guess, I would presume that there IS a smooth proper Z-scheme with interesting cohomology. There’s some weak evidence for this (e.g., there are smooth proper DM stacks over Z, like overline{M_g,n}), and the nonexistence just seems too good to be true – after all (as de Jong pointed out to me), how do we know that some random hypersurface in P^29 won’t do the trick? 🙂 It would be interesting to come up with some kind of heuristic involving “Z-points on Hilbert schemes” to get a handle on this…

Cheers,
Dave

3. I thought about Hilbert schemes for a second before noticing your discussions. But the only Hilb smooth over \mathbb{Z} I can find out is flag varieties… Do we even know a non-rational smooth proper variety over \mathbb{Z}? As for rational varieties, probably one could just use combinatorial data to construct some toric variety whose smoothness doesn’t depend on characteristic (I’m not sure on this…)?

1. Hey Mayorliat,

I certainly do not. In fact, I spent a few hours trying to think of interesting non-trivial examples of smooth proper $\mathbb{Z}$-schemes and had difficulty coming up with anything. If toric varieties are key then, unfrotunately, I am currently not the person to ask–I know very little about them (especially over not $\mathbb{C}$–does the theory work in any generality?).

1. Actually my name goes like mayor li at math :).

The toric construction is associate ring to some combinatorial lattice and glue along some combinatorial data, one can change coefficients casually (in the category of fields). And I thought smoothness is just a combinatorial data but I’m not sure if that result is in char 0 or not. As for trying Hilbert functor, it’s worth to notice Vakil’s Murphy law, so probably it wouldn’t give us any good example other than flag varieties.