4

Let $E$ be an elliptic curve (say, over a field $K$), and $E[n]$ be its $n$-torsion subgroup-scheme (suppose char $K$ is coprime to $n$). Is $E$ canonically isomorphic to $E/E[n]$? (What is this canonical isomorphism?)

I feel like there should be a slick way to state this using the autoduality of elliptic curves.

EDIT: I suppose the question I should be asking is, if $E$ is an elliptic curve over $S$, where $S$ is a scheme on which $n$ is invertible, then is the map $[n] : E\rightarrow E$ a cokernel for the map $E[n]\hookrightarrow E$ in the category of group schemes over $S$?

oxeimon
  • 12,279

2 Answers2

2

There's no need to invoke duality, although the answer depends on what you mean by the quotient $E/E[n]$. There's a natural sequence

$$E[n] \to E \xrightarrow{n} E$$

where the right map is multiplication by $n$. Under suitable hypotheses (which I'm not sure of, but $K$ a field of characteristic zero is probably fine) this is a short exact sequence of group schemes, the elliptic curve Kummer sequence, and so it exhibits $E$ canonically as the quotient in the sense of group schemes by $E[n]$. Whether this is true at the level of $K$-points is less clear; if $K$ is an algebraically closed field of characteristic zero then things are probably still fine, but otherwise there's no reason to expect multiplication by $n$ to be surjective.

Qiaochu Yuan
  • 419,620
  • Hmm, the link you provided for the elliptic Kummer sequence doesn't seem to do anything at the level of group schemes. I think what I really want to ask is - is $[n] : E\rightarrow E$ a cokernel for the morphism $E[n]\hookrightarrow E$ (in the category of group schemes)? – oxeimon Jan 22 '16 at 06:30
  • (In the above comment I want $E$ to be an elliptic curve over a scheme $S$ with all residue characteristics coprime to $n$, and the question occurs in the category of group schemes over $S$) – oxeimon Jan 22 '16 at 06:38
  • 1
    @oxeimon: with suitable hypotheses (which, again, I'm not sure of), I believe the exactness of a sequence of group schemes over a field $K$ can be checked on $\bar{K}$-points. This is far from my area of expertise, though, so maybe I should let a real arithmetic geometer say something here. I'm very hesitant to say anything over a more general scheme. – Qiaochu Yuan Jan 22 '16 at 07:13
2

I guess it depends on what you mean by a quotient, but in this case, there is a natural one, and this works over an arbitrary base. If $A$ is an abelian scheme over an arbitrary base $S$ and $H$ is a closed subgroup scheme of $A$ which is finite locally free over $S$, then the fppf quotient sheaf $A/H$ is representable by an abelian scheme over $S$.

The map $[n]:E\to E$ is finite locally free, so $E[n]$ is finite locally free over $S$. It is surjective on each fiber, so it is surjective. It is then an fppf covering, and in particular it is fppf surjective, which means that $E/E[n]\simeq E$ are isomorphic as fppf sheaves, and hence as abelian schemes over $S$. As far as I know, in this context, the meaning of the assertion that $0\to E[n]\to E\xrightarrow{[n]} E\to 0$ is exact is precisely that $E[n]\to E$ is the scheme-theoretic kernel of $[n]:E\to E$ and that $[n]:E\to E$ is a surjection of fppf sheaves.

Keenan Kidwell
  • 26,020
  • Hmm I think the fact that $[n] : E\rightarrow E$ is the cokernel of $E[n]\hookrightarrow E$ follows from the fact that the categorical quotient $E/E[n]$ exists. Do you have a nice reference for when this quotient is representable? – oxeimon Jan 23 '16 at 01:55
  • Not exactly. It's proved over a locally Noetherian base as Theorem 4.39(c) in the group scheme quotient chapter of Moonen and van der Geer's book-in-progress on abelian varieties, but I'm sure I've seen elsewhere that it holds without the locally Noetherian hypothesis. And fppf quotients are geometric quotients are categorical quotients, so this seems to work. – Keenan Kidwell Jan 23 '16 at 02:03
  • In any case I guess you can prove directly the desired universal property. Say $G$ is an $S$-group and there is a homomorphism $\pi:E\to G$ which kills $E[n]$ (i.e. $En\to E(T)\to G(T)$ is zero for all $S$-schemes $T$). Then there is a unique map of fppf sheaves $f:E\to G$ such that $f\circ [n]=\pi$. But everything is representable so by Yoneda this is a map of group schemes over $S$. Does that seem okay? And you can do this just knowing that $[n]:E\to E$ is the fppf cokernel of $E[n]\to E$, i.e. you don't need to know in advance that fppf quotients exist in this setup...it seems to me. – Keenan Kidwell Jan 23 '16 at 02:08
  • How are you arguing that $[n] : E\rightarrow E$ is the fppf cokernel? (It seems to me that fppf cokernels of $E[n]\hookrightarrow E$ is the same as the fppf quotient of $E$ by the action of $E[n]$ by translation) – oxeimon Jan 23 '16 at 04:02
  • Dear @oxeimon, Is it the fppf surjectivity that's the problem? If we accept the fppf surjectivity of $[n]:E\to E$, then this is an epimorphism in the abelian category of fppf sheaves on $S$, so it is the cokernel of its kernel $E[n]\to E$. – Keenan Kidwell Jan 23 '16 at 13:44