Is an elliptic curve canonically isomorphic to its quotient by its own $n$-torsion? 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 $S$?
 A: 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. 
A: 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. 
