Basic proof that Isogeny = group homomorphism without Riemann Roch?

Question

It is a well known fact that an isogeny $f:E_1\rightarrow E_2$ between elliptic curves (i.e., where $f(O)=O$) is a group homomorphism under the natural group structure of the elliptic curves.

Every proof that I have found (usually based on Silverman's book) uses Riemann-Roch theorem. Of course, it is a very important theorem, so why not.

But I was wondering if it is possible to prove it directly, without first going to $\operatorname{Pic}^0(E_i)$. Does anyone know of any such proof? (The simple the better, of course).

Cheers.

Answer 1

This is proven for abelian varieties (of any dimension, the 1-dimensional case being elliptic curves) as Corollary 1 in section 4 of Mumford's Abelian Varieties. It is an immediate corollary of the "rigidity lemma" stated immediately prior:

Rigidity Lemma (Form I). Let $X$ be a complete variety, $Y$ and $Z$ any varieties, and $f\colon X \times Y \to Z$ 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 $g\colon Y \to Z$ such that if $p_2\colon X \times Y \to Y$ is the projection, $f = g \circ p_2$.

In other words, if $f$ is constant on one fiber of the projection map $p_2$, then $f$ is constant on every fiber of $p_2$.

The proof is brief, so I'll just reproduce it here:

Proof. Choose any point $x_0 \in X$, and define $g\colon Y \to Z$ by $g(y) = f(x_0, y)$. Since $X \times Y$ is a variety, to show that $f = g \circ p_2$, it is sufficient to show that these morphisms coincide on some open subset of $X \times Y$. Let $U$ be an affine open neighbourhood of $z_0$ in $Z$, $F = Z - U$, and $G = p_2(f^{-1}(F))$; then $G$ is closed in $Y$ since $X$ is complete and hence $p_2$ is a closed map. Further $y_0 \notin G$ since $f(X \times \{y_0\}) = \{z_0\}$. Therefore $Y - G = V$ is a non-empty open subset of $Y$. For each $y \in V$, the complete variety $X \times \{y\}$ gets mapped by $f$ into the affine variety $U$, and hence to a single point of $U$. But this means that for any $x \in X$, $y \in V$, $f(x, y) = f(x_0, y) = g \circ p_2(x, y)$, and this proves our assertion.

In particular, if $f \colon X \to Y$ is a morphism of abelian varieties such that $f(0_X) = 0_Y$, then the morphism $\phi\colon X \times X \to Y$ defined for all $x_1, x_2 \in X$ by $$\phi(x_1, x_2) = f(x_1 + x_2) - f(x_1) - f(x_2)$$ is constant with value $0_Y$ on $X \times \{0_X\}$ and $\{0_X\} \times X$. Applying the rigidity lemma, $\phi$ is constant everywhere on $X \times X$, which means that $f$ is a homomorphism.

The analogous relative statement for abelian schemes (over any base scheme) is Corollary 6.4 in Mumford, Fogarty, and Kirwan's Geometric Invariant Theory. The general idea of that proof is similar.

References:

• Mumford, D. Abelian Varieties. Corrected reprint of the second (1974) edition. Tata Institute of Fundamental Research Studies in Mathematics, vol. 5. Hindustan Book Agency, New Delhi, 2008. ISBN: 978-81-85931-86-9. [MR 2514037]
• Mumford, D., Fogarty, J., and Kirwan, F. Geometric Invariant Theory. Third edition. Ergebnisse der Mathematik und ihrer Grenzgebiete, vol. 34. Springer-Verlag, Berlin, 1994. ISBN: 3-540-56963-4. [MR 1304906]