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I know this should be pretty simple, but right now the only way I can see how to prove it is to sit down and write out explicit formulae for the group law, and see that everything works out. What's the geometric or abstract-nonsense reason why the abelian group structure of elliptic curves behaves nicely under homomorphisms?

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This follows by a rigidity property of certain morphisms. It is important to note that elliptic curves are complete, that is, proper and integral schemes. Then we have the following "Rigidity Lemma" (see Mumford's Abelian Varieties, the beginning of chapter II (page 43 of the old edition), for example):

Let $X$ be a complete variety, $Y$ and $Z$ any varieties, and $f:X\times Y\rightarrow Z$ a morphism such that for some $y_0\in Y$, $f(X\times\{y_0\})$ is a single point $z_0$ of $Z$. Then there is a morphism $g:Y\rightarrow Z$ such that if $p_2:X\times Y\rightarrow Y$ is the projection, $f=g\circ p_2$.

How does this help? Well, elliptic curves have a distinguished point (the origin), and a morphism of elliptic curves is a morphism of their underlying schemes taking one distinguished point to the other. We'd like to show that any such morphism is actually compatible with the group law. So consider a morphism of elliptic curves $f:E_1\rightarrow E_2$, and let $\Phi:E_1\times E_1\rightarrow E_2$ be defined by $\Phi(x,y)=f(x+y)-f(x)-f(y)$. Then $\Phi(E_1\times\{0\})=0$ so by the lemma I quoted, there is a morphism $g:E_1\rightarrow E_2$ such that $\Phi=g\circ p_2$. And since $\Phi(\{0\}\times E_1)=0$, as well, $g$ must be zero. But then $\Phi$ itself must be zero.

None of this relied on $E_1$ and $E_2$ being elliptic curves; it works exactly the same way for abelian varieties.

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The same rigidity theorem implies that an abelian variety (or an elliptic curve) is a COMMUTATIVE group variety. –  VA. Jan 24 '10 at 5:33
    
The Rigidity Lemma as stated above is valid over algebraically closed base field and $Z$ must be separated (OK this depends on what you call a variety). For general base field, one supposes that $Y, Z$ are geometrically integral, and the points $y_0, z_0$ must be rational over $k$. –  Qing Liu Jan 24 '10 at 16:18
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There is also a simple analytic proof which is enlightening in a different way.

Let $f : E \to E'$ be a morphism of elliptic curves. Both E and E' have $\mathbb C$ as their universal covering space. The composition $\mathbb C \to E \to E'$ lifts uniquely to a continuous map $\overline f$ from $\mathbb C$ into the the universal covering of $E'$ (since the source is simply connected), and $\overline f$ is holomorphic because the map $\mathbb C \to E'$ is locally biholomorphic.

Let $E = \mathbb{C}/\Lambda$, $E' = \mathbb{C}/\Lambda'$. We know that for $\lambda \in \Lambda$ we have $\overline f(z+\lambda) - \overline f(z) \in \Lambda'$, hence $\overline f(z+\lambda) - \overline f(z)$ is constant as a function of z. Differentiating, the derivative of $\overline f$ is a doubly periodic entire function, so it is a constant. Since $f$ should preserve the basepoint on the elliptic curves, which is the image of 0, we can assume that $\overline f (0) = 0$ and so $\overline f (z) = cz$ for some constant $c$. But the group laws on E and E' are induced by the group law on $\mathbb C$, so the fact that f preserves the group structure reduces to the distributive law!

The same proof holds also for any complex torus.

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@dan. To be more precise: why is the derivative of $\tilde f$ doubly periodic? –  Anweshi Jan 24 '10 at 13:33
    
@dan. Thanks. But the lift of a doubly periodic function need not be doubly periodic. Your argument however works if you use the composition $\mathbb{C} \rightarrow \mathbb{C} \rightarrow E^\prime$. –  Anweshi Jan 24 '10 at 15:38
    
Since we don't yet know that $\overline{f}$ is a homomorphism, we should state the equivariance as more than just $\overline{f}(\Lambda)\subseteq\Lambda'$, something like $\overline{f}(z+\Lambda)\subseteq\overline{f}(z)+\Lambda'$. –  Tom Church Jan 24 '10 at 17:09
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This answer is an elaboration on some of the earlier ones. Throughout we suppose that $\phi:E\_1 \rightarrow E\_2$ is a morphism of elliptic curves (so in particular it preserves the origins, i.e. $\phi(O) = O$).


Here is a concrete form of the rigidy argument:

If $P$ is in $E\_1$, then we define a map $\phi\_P:E\_1 \rightarrow E\_2$ via $\phi\_P(X) = \phi(X + P) - \phi(P)$. Note that $\phi\_O = \phi.$ (Here we use the fact that $\phi(O) = O$.) Thus $\phi\_P$ is a family of map $E\_1 \rightarrow E\_2$, parameterized by $E\_1$, passing through $\phi$, with the property that $\phi\_P(O) = O.$

Now $\phi\_P$ induces a corresponding local map on local rings $\phi_P^*: \mathcal O_{E\_2,O} \rightarrow \mathcal O_{E\_1,O},$ and hence also on the finite length quotients $(\phi_P^*)_n: \mathcal O_{E\_2,O}/\mathfrak m^n \rightarrow \mathcal O_{E\_1,O}/ \mathfrak m^n$. (Here we use $\mathfrak m$ to denote the maximal ideal in each of the local rings.)

If $k$ is our ground field, then the source and target of $(\phi_P^*)_n$ are just finite-dimensional vector spaces, and so $P \mapsto (\phi\_P^*)_n$ is a morphism (of varieties) from $E$ to a finite dimensional (and in particular, affine!) space of matrices. Since $E$ is projective and connected, it must be constant.

Thus $(\phi_P^*)_n = (\phi_O^*)_n$ for all $n$, and so passing to the limit we find that $\phi_P^* = \phi_O^*$. Thus $\phi_P$ and $\phi_O$ induce the same map on local rings at $O$, and since $E$ is irreducible, they therefore coincide.
(If you like, passing to fraction fields, we see that they induce the same maps $K(E\_2)\rightarrow K(E\_1)$, and hence coincide as morphisms of curves.)

Thus $\phi\_P = \phi\_O = \phi.$ Unwinding the definition of $\phi\_P$, we find that $\phi(P + X) = \phi(P) + \phi(X)$ for all $X$, and thus that $\phi$ is a group homomorphism.


Here is a concrete version of the argument with Picard and divisors:

To show that $\phi$ is a homomorphism, it is easy to see that it suffices to show that $P + Q + R = O$ implies that $\phi(P) + \phi(Q) + \phi(R) = O.$ (This uses the fact that $\phi(O) = O$; hopefully no confusion is caused by using $O$ to denote the origin on both $E$ and $F$.)

Now $P + Q + R = O$ (in the group law of $E$) if and only if there is a rational function $f$ such that div $f$ = P + Q + R - 3 O (where now the right hand side is a divisor on $E$); i.e. if and only if $P + Q + R - 3 O$ is a principal divisor. (Concretely, if $E$ is given by a cubic Weierstrass equation in ${\mathbb P}^2$ with homogeneous coordinates $X$, $Y$, and $Z$, then $P + Q + R = O$ when $P$, $Q$, and $R$ are collinear, and then if $\ell(X,Y,Z)$ is the equation of the line passing through them, the function $f$ can be taken to be $\ell(X,Y,Z)/Z^3$.)

Similarly $\phi(P) + \phi(Q) + \phi(R) = O$ in the group law on $F$ if and only if the divisor $\phi(P) + \phi(Q) + \phi(R) - 3 O$ is principal.

So we are reduced to showing that $\phi$ takes principal divisors to principal divisors.

This is a general property of maps of smooth projective curves: if $\phi$ is constant, there is nothing to show. Otherwise, if $\phi:C \rightarrow D$ is non-constant, then it induces a finite extension of function fields $K(D) \hookrightarrow K(C)$, and we have the corresponding norm map (in the usual sense of field theory) $K(C) \rightarrow K(D)$. It turns out that for any function $f \in K(C)$, we have div$(\text{norm of }f) = \phi(\text{div }f)$. (This is an exercise whose difficulty will vary with your comfort level with the material; if you understand the basics of rational functions and divisors on curves well, then it is not too hard.)

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have a look at lemma 4.9 and exercise 2.6 in chapter IV of hartshorne. i think it comes down to the fact that the group of an elliptic curve is isomorphic to its picard group--the group of invertible sheaves where the group operation is given by tensor product--and the fact that a morphism $f:X\to Y$ induces a homomorphism $f_*:Pic\ Y \to Pic\ X$.

(the isomorphism sends a point P to the sheaf of functions with a simple zero at P and a simple pole at infinity.)

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maxmoo -- that's right, but an elliptic curve is the connected component of it's Picard. –  algori Jan 24 '10 at 5:28
    
Probably better to call it Pic^0, the degree 0 part. –  Kevin H. Lin Jan 24 '10 at 5:36
    
I don't know if this works. It seems that the map $f^\ast : Pic^0 Y \to Pic^0 X$ would have to be in the other direction for this to have a chance of working. –  Kevin H. Lin Jan 24 '10 at 5:45
    
This is on the right track but not exactly right: your morphism would be better denoted $f^*$, since it is contravariant. We want a covariant morphism on the Picard groups, so we should be taking the Albanese morphism $f_*: Alb^0 X \rightarrow Alb^0 Y$ instead. –  Pete L. Clark Jan 24 '10 at 5:47
    
...It's on the right track, because $f_*$ and $f^*$ are duals of each other, using the autoduality of elliptic curves [or more generally, Jacobians of curves]. –  Pete L. Clark Jan 24 '10 at 5:49
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You don't need explicit formulas. In a more general setting, let $A$, $B$ be abelian varieties and let $f:A\to B$ be a morphism of varieties sending $0$ to $0$. Then $f$ is automatically a morphism of abelian varieties : it preserves the group structure.

See for example Mumford's book on Abelian Varieties, or Milne's online notes. On p. 8, Milne writes

Corollary 1.2. Every regular map $\alpha:A\to B$ of abelian varieties is a composite of a homomorphism with a translation.

This is a consequence of the "rigidity theorem".

You might also consult his earlier expository article 1986b.

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