Let $X$ and $Y$ be elliptic curves over an algebraically closed field $K$. If the characteristic of $K$ is nonzero, assume both curves are ordinary or both are supersingular. Does it follow that $X$ and $Y$ have equivalent small etale toposes?

  • 3
    $\begingroup$ I like the question! If K is the complex numbers I think the answer is yes, because in that case the small etale site of X depends only on the homeomorphism class of X(C): is it the category of local homeomorphisms U --> X admitting a factorization U --> Y --> X where Y is a compact Riemann surface mapping holomorphically to X and U --> Y is an open inclusion missing only finitely many points. (Key point: the complex structure on such a Y is unique and functorial.) $\endgroup$ Sep 12, 2010 at 16:02
  • 1
    $\begingroup$ Yeah, this sure seems puzzling. The constructible sheaves are intrinsic (noetherian objects in the topos). Suppose $K$ is algebraic closure of its prime field, and descend the elliptic curves to finite extensions $k$ of the prime field. Can do the same for any constructible abelian sheaves on the elliptic curves by increasing $k$ a bit (depending on the sheaf), thereby showing that the cohomology of such sheaves inherits an action of open subgroup of Gal($K/k$) that is well-defined and functorial "near the identity". But is it intrinsic to the topos??? $\endgroup$
    – BCnrd
    Sep 12, 2010 at 23:28
  • 1
    $\begingroup$ A question; given two such elliptic curves, are the absolute Galois groups of their function fields known to be isomorphic? $\endgroup$ Sep 13, 2010 at 14:15
  • $\begingroup$ Here's a simpler question. Given a point x in E(K), is the etale fundamental group of E\{x} independent of E and x? If yes, can it be described explicitely, e.g. as some particular completion of a free group on two enerators (just like over C)? $\endgroup$ Sep 14, 2010 at 14:16
  • $\begingroup$ @André: I would be surprised. The corresponding question for $\mathbb{P}^1$ would ask whether $\mathbb{A}^1$ is simply connected over an algebraically closed field of characteristic p, and the answer is definitely "no". $\endgroup$ Sep 15, 2010 at 13:52

2 Answers 2


In characteristic zero, I believe the answer is "yes". More generally, if $X$ and $Y$ are two smooth proper curves over algebraically closed fields $K$ and $L$ of characteristic zero, then $X$ and $Y$ have equivalent etale topoi if and only if they have the same genus and card($K$) = card($L$). The "only if" claim follows from the fact that the etale topos has both the etale fundamental group and the Zariski site as invariants (the first being determined by the locally finite constant objects and the second being determined by the lattice of subobjects of the final object). This is more of a remark; the question is about the "if" claim.

For that, choose a set S and bijections $f: S \to X(K)$ and $g: S \to Y(L)$. Now, for every finite subset $T$ of $S$, let $$a_T: Spec(K) \to M_{g,T}$$ be the map classifying $f(T)$ inside $X$, and ditto $b_T$ for $g(T)$ inside $Y$. Furthermore let $A_T$ be the Henselization of $a_T$, and ditto $B_T$ of $b_T$. For $T$ inside $T'$ we have maps $A_{T'} \to A_T$ and $B_{T'} \to B_T$, mapping compatibly to $M_{g,T'} \to M_{g,T}$; we consider the inverse limit of the fiber products $$A_T \times_{M_{g,T}} B_T$$ along these maps. Since it is (eventually) a filtered limit of nonempty schemes along affine dominant maps, it has a point over some algebraically closed field $F$. Let $Z$ be the induced curve over $F$ and $h:S \to Z(F)$ be the induced injection.

Then I claim that the étale site of $X$ is equivalent to the subcategory of the étale site of $Z$ consisting of those $U \to Z$ with either $U$ empty or $Z-im(U)$ inside $h(S)$. By symmetry this will prove our desired result.

Here is how that equivalence goes. For $U \to X$ be etale (suppose nonempty), let $X'$ in $X$ be an open subset of $X$ for which the pullback $U' \to X'$ is finite (i.e. proper). Take $T = f^{-1}(X-X')$, and recall we have maps $Spec(K) \to A_T \to M_{g,T}$ where the composite classifies $f(T)$ inside $X$, and similarly $Spec(F) \to A_T \to M_{g,T}$ classifying $h(T)$ inside $Z$. Let $C'$ denote the pullback to $A_T$ of the universal $T$-punctured curve over $M_{g,T}$; it gives $X'$ over $K$ (special fiber) and $Z':=Z-h(T)$ over $F$ (generic fiber). Since $C' \to A_T$ is a smooth map with a smooth proper relative normal crossings compactification, the maps $X' \to C' \leftarrow Z'$ induce equivalences on categories of finite étale covers; thus we can transport $U' \to X'$ uniquely over to a $V' \to Z'$, via some $W' \to C'$.

But in fact I claim that the same is true of $U \to X$. Indeed, let $C$ be the completed curve of $C'$ (over $A_T$), and for $t$ in $T$ with corresponding section $c$ of $C-C'$ consider the completion of $C$ along $c$, then subtract the section $c$ to get $L \to A_T$. On pullback to $K$ this becomes $L_x$, Spec of a Laurent field at $x=f(t)$ in $X$, and ditto for $F$ and $L_z$; furthermore once again the maps $L_x \to L \leftarrow L_z$ induce isomorphisms on categories of etale covers of $L$. On the other hand, the Galois type of the pullback of $U' \to X'$ to $L_x$ completely determines the points at infinity of $U'$ lying above $x$ (and ditto on the $F$ side): they are the orbits of Galois (which is the profinite completion of $\mathbb Z$) over any geometric fiber. Thus we can add in points at infinity on both sides in a consistent and canonical manner.

This finishes the proof, I suppose: functoriality of this one-to-one correspondence (and its independence of $X'$) comes from us having made our constructions compatible with changing $T$ and the fact that smooth proper compactifications of curves over fields are unique and functorial.

Some remarks:

1) This proof is close to the proof I gave in the comments for $K$ the complex numbers: the fact that all elliptic curves are homeomorphic is symptomatic of connected moduli, which was again a key fact in this proof (though in a hidden manner?)

2) However, the real surprise lies in the differences: to get an equivalence of &eactue;tale topoi one needn't choose a homeomorphism -- in fact any bijection will do! The field $F$ whose existence was ensured above guarantees some measure of continuity no matter what the bijection. In fact I imagine that in higher dimensions the notions of being homeomorphic and having equivalent étale topoi diverges. For instance, I'll bet that $E \times \mathbb A^1$ doesn't have the same étale topos as $\mathbb G_m \times \mathbb G_m$. Similarly, I wonder about two varieties occurring in the same proper smooth family having inequivalent étale topoi.

3) The positive characteristic case is a bummer for me because of non-topological (i.e. "wild") phenomena and I don't have much to say.

  • $\begingroup$ Nice. You find a curve over a big field that degenerates to the two given ones, crucially using the connectedness of $M_{g,T}$, which is also nice. I'm not so sure about a few of the details, but I doubt it's too hard to get around them. (For instance, unless I'm misunderstanding you, the base change of $L$ to $F$ is not the full Laurent series field over $F$. Also the Galois type of the pullback of $U'$ to $L_x$ doesn't seem to determine the points of $U$ lying over $x$. For instance, take $U_1=X\coprod X'$ and $U_2=X'\coprod X'$. These two $U$'s have the same Galois structure generically.) $\endgroup$
    – JBorger
    Sep 16, 2010 at 10:10
  • $\begingroup$ Thank you for the corrections! You're right, the base change is not the Laurent series (unless we think of these things as algebras in pro-abelian groups, say), but we still have L_x --> L <-- L_z between the Laurent series with the desired properties (I believe). As for the second point, maybe the issue is that I was being imprecise: I didn't mean to imply that the Galois structure knows what U was, only that it knows how to build the canonical compactification U-bar of U', which lets us build U from U' analogously on both sides (special and generic). $\endgroup$ Sep 16, 2010 at 13:33

Yes, these toposes are equivalent. (see Tyler's comment below)

Let $E$ be your curve. Let $x:=(E \leftarrow E_1 \leftarrow E_2 \ldots)$ be a system of etale covers that is cofinal in the sense that any etale cover of $E$ factors through an $E_n$. The etale fundamental group of $E$ based at $x$ is then given by $$ \pi_1(E,x) = \lim_n Gal(E_n/E) $$ and is a pro-(finite group). The pro-set $X:=\lim_n E_n(K)$ has an action of $G$, and can be thought of as the set of "set of points of the universal cover of $E$".

Key fact: the small etale topos of $E$ only depends on $G$, and on the $G$-pro-set $X$ (and here $E$ doesn't need to be genus 1, and $K$ doesn't need to be algebraically closed).

Here is a description of that topos. Its elements are diagrams $$ > A \rightarrow B \leftarrow X > $$ where $A \to B$ is an inclusion that misses finitely many points, and $B \to X$ is a projection given by modding out by the action of an open subgroup of $G$.

So it all comes down to the following observation:
If two elliptic curves are ordinary, and the base field $K$ is algebraically closed, then their etale fundamental groups are isomorphic (they are a product of $\mathbb Z_q^2$ for every $q$ not equal to the characteristic, and of a $\mathbb Z_p$ if $char(K)=p$), and their "set of points of universal covers" are equivariantly isomorphic (they are free $G$-pro-sets).

In the case of supersingular curves, you get a similar description of $G$, but the factor $\mathbb Z_p$ is now gone.

  • $\begingroup$ @Andre: How does your description of the etale topos recover etale covers of open subsets? $\endgroup$ Sep 13, 2010 at 14:14
  • $\begingroup$ @Tyler: You're right. I completely forgot those. Ok, my whole answer is wrong. $\endgroup$ Sep 13, 2010 at 14:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.