ag.algebraic geometry - Why does the group law commute with morphisms of elliptic curves?

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. $phi(O) = O$).

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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.

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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.)