# Reference request: Fibre functor for elliptic curves is pro-representable

I am writing a project on étale fundamental groups of elliptic curves and I want to include a proof of a key theorem: the fibre functor on the category of finite étale covers of an elliptic curve is "pro-representable". Firstly, let me just recall the general case:

Given an algebraic variety $X$ over a field $K$ and a geometric point $\bar{x}:\text{Spec}(\bar{K})\to X$ we can consider the category of pointed finite étale covers $Y\to X$. Let's call this category $\mathsf{Cov}(X,\bar{x})$. The fibre functor $\Phi$ on this category associates to each cover $\phi: Y\to X$ the underlying set of the fibre above the geometric point $\bar{x}$ and to each morphism of covers the corresponding function on fibres. The étale fundamental group of $(X,\bar{x})$ is defined as the automorphism group of this functor, and it is a general theorem that it is pro-representable i.e. there's a projective system of étale covers of $X$ that represents $\Phi$.

Now I'm able to find many quite abstract proofs of the above pro-representablity theorem (for example, in Szamuely's book it is done for connected schemes). However in the interests of not investing an impossible amount of energy into my project I am looking for a far more basic version: just the case that $X = E$ is an elliptic curve with basepoint $O\in E(K)$. In this case the category $\mathsf{Cov}(E,O)$ actually becomes a category of covers of $E$ by isogenies of elliptic curves (over extensions of $K$) once a point in the fibre above $O$ is chosen, and the fibre functor actually factors through the category of finite abelian groups because it associates to each isogeny its kernel.

In this case, one of the nice results I have seen mentioned often is that the fibre functor on $\mathsf{Cov}(E,O)$ is pro-represented by the system of multiplication-by-$m$ isogenies $$[m]: E\to E$$ (given a partial order by division) which are étale provided $m$ is a unit in $K$.

I haven't been able to find a proof of this specific result anywhere. It's possible one doesn't exist and that a proof using the fully general construction I mentioned above is required, but I would be very happy if anyone out there can correct me and point me in the direction of a proof this result for elliptic curves. I am trying to avoid "schemey" arguments as much as possible but I would be happy using something that uses results about elliptic curves à la Silverman's AEC.

Many thanks!

As you've already intuited, this is really a statement about the theory of elliptic curves, not about algebraic geometry proper.

The real statement that you're after, in the sense that I assume you want this as the corollary, is the following: let $(E,p)$ be an elliptic curve then

$$\pi_1^{\acute{e}\text{t}}(E,\overline{p})=T(E)$$

where $$T(E)=\prod_\ell T_\ell(E)$$

where, here,

$$T_\ell(E)=\varprojlim E[\ell^n]\left(\overline{k}\right)$$

which, in the process, will give you want. Of course, I assume that you're assuming that $K$ is algebraically closed, else we get a canonical isomorphism

$$\pi_1^{\acute{e}t}(E,\overline{p})=T(E)\rtimes G_K$$

where $G_K=\text{Gal}(K^\text{sep}/K)$.

Note that, in particular, this shows that in characteristic $0$ the étale fundamental group is just $\widehat{\mathbb{Z}}^2$ and in positive characteristic it's $(\mathbb{Z}^{(p)})^2\times A$ (here $\mathbb{Z}^{(p)}$ is the product of $\mathbb{Z}_\ell$ for $\ell\ne p$) and $A$ is either $\mathbb{Z}_p$ or $0$ depending whether $E$ is ordinary or supersingular.

(Slight pedantic remark: if one wants to think about Galois representations, one should really replace the $\overline{k}$ above with $k^\text{sep}$. If $\ell\ne p$, where forever $p=\mathrm{char}(k)$, then there is no difference since, as you pointed out, the finite flat group schemes are étale so pick all their points up over the separable closure. But, for us, we need to actually think about the points over the algebraic closure.)

So, assume that $f:C\to E$ is a finite Galois cover of $E$. Then, we immediately deduce many things about $C$: it's a smooth projective connected curve (these all follow from basic considerations about finite étale maps). I claim that $C$ has genus $1$. But, this follows fairly obviously from the Riemann-Hurwitz formula:

$$2g(C)-2=\deg(f)(2(1)-2)+\sum 0=0$$

and thus $g(C)=0$ as desired.

Thus, for a choice $q\in f^{-1}(p)$ we see that $(C,q)$ is an elliptic curve. Moreover, since $f(q)=p$ (by construction) it follows from the basic theory of elliptic curves (a corollary of the ‘Rigidity Lemma’) that $C\to E$ is actually a group map and, in fact, an isogeny.

But, note that $\ker(f)$ is then a subgroup (scheme) of $C$ of degree $\deg(f)$ and since translation by the points of $\ker(f)$ (of which there are actually $\deg(f)$ since $f$, and thus $\ker(f)$ are étale!) are all in $\text{Aut}(C/E)$ we conclude that

$$\text{Gal}(C/E)=\text{Aut}(C/E)^\text{op}=\ker(f)^\text{op}=\ker(f)$$

in particular, $C\to E$ is necessarily abelian.

In particular, take a $p$-Sylow subgroup $P\subseteq \text{Gal}(C/E)$ and consider the associated tower one gets

$$C\to C/P\to E$$

which allows us to, essentially, divide our efforts amongst the $p$-power case, and the prime-to-$p$ case.

The latter is simple. Since

$$f:C\to E$$

is an isogeny, there then exists by the basic theory of elliptic curves, a dual isogeny

$$\widehat{f}:E\to C$$

such that

$$f\circ \widehat{f}=[\deg(f)]$$

as a morphism on $E$. Since we’re in the prime-to-$p$ setting we can conclude that $[\deg(f)]$ is étale and thus we’ve dominated our Galois cover by one of the form $[m]$.

If we’re in the $p$-power case, it’s slightly more annoying but not too bad. Namely, we know that since

$$C\to E$$

is an isogeny, that $E=C/\ker(f)$ and, as we observed above, that $\text{Gal}(C/E)=\ker(f)$. Since $f$ is étale $\ker(f)$ must be a constant subgroup (scheme) of $C$ and so a quotient of $T_p(C)$. But, as before, we can find a dual isogeny

$$\widehat{f}:E\to C$$

and so by the same token, we see that $C=E/M$ for some subgroup (scheme) of $M$, and thus we can identify $T_p(C)$ with $T_p(E)$—the subgroup $\ker(f)\subseteq C$ can really be thought of as a subgroup of $E$. Conversely, we get the full $T_p(E)$ by considering the tower $\{E/E[p^n]^\circ\xrightarrow{f_n}E\}$, where $f_n$ is quotient by $E[p^n]^{\acute{e}\text{t}}$, as $n$-grows. The same sort of uniformization is at play here since we're tacitly identifying

$$(E/E[p^n]^\circ)/(E[p^n]^{\acute{e}\text{t}})=E/E[p^n]=E$$

in our above tower.

EDIT: Most of what I said above is fairly basic algebraic geometry, except, possibly, the dual isogeny statement. So, for self-containment, let me give an easy way to think about such a morphism.

Let's assume for a second that you know that an elliptic curve $(E,e)/k$ is a group scheme in such a way such that for any $L/k$-the group $E(L)$ is functorially identified with $\mathrm{Pic}^0(L)$.

Now, let $\mathscr{C}$ be the category of smooth projective, geometrically connected curves over $k$ with morphisms finite morphisms. Then, it's well-known (by the valuative criterion) that for any object $C\in\mathscr{C}$ we have that

$$\text{Hom}(C,E)=E(K(C))=\mathrm{Pic}^0(E_{K(C)})$$

in particular, we see that to define a group map $E\to E'$ of elliptic curves, it suffices to define a group map

$$\mathrm{Pic}^0(E(K(C))=E(C)\to E'(C)=\mathrm{Pic}^0(E'(K(C))$$

functorially in $C$.

So, suppose that we're given a morphism $f:E\to E'$ in $\mathscr{C}$ which is, in fact, a group map. We then want to define a map $\widehat{f}:E'\to E$ such that $f\circ \widehat{f}=[\deg(f)]$. To do this, we use the above observation. Namely, for all $C\in\mathscr{C}$ let us define

$$\widehat{f}:\mathrm{Pic}(E'_{K(C))})\to \mathrm{Pic}(E_{K(C)})$$

by

$$\mathscr{L}\mapsto f^\ast(\mathscr{L})$$

it's evident that this is a functorial group map and thus we obtain a group morphism

$$\widehat{f}:E'\to E$$

as desired.

To see that it satisfies the desired property that $f\circ \widehat{f}=[\deg(f)]$ it suffices to see that they agree on $\overline{k}$-points. But, note that if $\mathscr{L}=\mathcal{O}(p-e)\in\mathrm{Pic}^0(E'_{\overline{k}})$, then

$$f^\ast\mathscr{L}=\mathcal{O}(D)$$

where

$$D=(p-e)\times_{E'}E=\sum_{q\in f^{-1}(p)}e_q q+\sum_{q'\in f^{-1}(e))}e_{q'}q'$$

and then so, applying $f$ to this we get

\begin{aligned}f\left(\sum_{q\in f^{-1}(p)}e_q q+\sum_{q'\in f^{-1}(e)}e_{q'}e\right) &=\sum_{q\in f^{-1}(q)}e_q f(q)+\sum_{q'\in f^{-1}(e)}e_{q'}f(q')\\ &=\sum_{q\in f^{-1}(p)}e_q p+\sum_{q'\in f^{-1}(e)}e_{q'}e\\ &= \deg(f) p-\deg(f) e\end{aligned}

which proves the claim.

• Dear Alex, thank you for another fantastic answer! You were right in that its this corollary I was really after in the first place, and representability was just something I wanted to show the similarity to the topological fundamental group but wasn't exactly key.Anyway, I'm interested in non-algebraically closed fields too, so does the same basic argument work in this case? Does the semidirect product just come about because as well as translations each cover $C\to E$ also has a Galois action giving extra automorphisms? – Alex Saad Apr 1 '16 at 14:02
• @AlexSaad This is a general fact about (geometrically connected) varieties: one always has a short exact sequence $$1\to \pi_1^\text{geom}(X)\to \pi_1^\text{arith.}(X)\to G_k\to 1$$ where $\pi_1^\text{geom}(X)=\pi_1(\overline{X})$, $\pi^\text{arith.}(X)=\pi_1(X)$, and, as you know, $G_k$ denotes the absolute Galois group (NB: here $\overline{k}$ really should be $k^\text{sep}$). The rough idea being that every finite Galois cover $Z\to X$ roughly looks like $Z\to X_L\to X$ where $L/k$ is finite separable (i.e. picked up by $G_k$) and $Z$ is geometrically connected over $L$ – Alex Youcis Apr 1 '16 at 15:11
• so that $Z_\overline{L}\to X_{\overline{k}}$ is still a Galois cover. Note htat if $K(Z)$ contains a non-trivial separable extension then $Z_{\overline{k}}$ won't be connected, so no longer Galois! So, this $L$ is, essentially, the separable closure of $k$ in $K(Z)$ which, as is obvious, is finite. This is roughly the above decomposition. – Alex Youcis Apr 1 '16 at 15:15
• The mappings in the sequence come by functoriality. Also, any $k$-point of $X$ gives a splitting of the above sequence (and, roughly, the converse is conjectured). So, since elliptic curves have canonically defined $k$-points, they have canonically defined splittings of their arithmetic/geometric fundamental group sequences. – Alex Youcis Apr 1 '16 at 15:17
• Also, as a good exercise, it's worth deciphering where in the above proof I used that $K=K^\text{sep}$! – Alex Youcis Apr 1 '16 at 15:24