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An isogeny of elliptic curves over $k$ is a nonconstant morphism $\varphi :E_1\to E_2$ defined over $k$ and preserving the identity. Such a morphism is a group homomorphism: apply the Mumford rigidity lemma to $\varphi (P+Q)-\varphi (P)-\varphi (Q)$, whose restriction to either coordinate axis is zero. A nonconstant morphism of smooth projective curves is finite and surjective, so its degree of an isogeny is the positive integerDegrees multiply under composition. A separable isogeny has degree equal to the number of geometric points in its kernel of an isogeny. In positive characteristic of a field, inseparable isogenies also occur, for example the Frobenius isogeny; their total degree includes inseparable multiplicity, so counting geometric kernel points alone is insufficient. Every isogeny of degree $d$ has a dual isogeny $\hat{\varphi }$ satisfying $\hat{\varphi }\varphi =[d]$ and $\varphi \hat{\varphi }=[d]$.

Pointwise addition and negation make the homomorphism group of elliptic curves $H={\mathrm{Hom}}_k(E_1,E_2)$ an abelian group. Its zero is the constant identity map; every other element is an isogeny, because its nonconstant image is the whole target curve. Define $q(0)=0$ and $q(\varphi )=\mathrm{deg}\varphi$ otherwise. We prove the positive definiteness of degree on elliptic-curve homomorphisms, including arbitrary characteristic.

Let $L={\mathcal{O}}_{E_2}(O)$, a line bundle associated to a divisor of degree one. On $E_2\times E_2$ write $m(P,Q)=P+Q$, $d(P,Q)=P-Q$ and let ${\pi }_i$ be the projections. The divisor proof of the degree parallelogram law givesHere is a direct justification. The degree-two Weierstrass coordinate $x$ has a double pole at $O$, and $x(P)=x(Q)$ precisely when $Q=P$ or $Q=-P$. Consequentlywhere $D_{+}$ is the diagonal and $D_{-}$ the graph of negation. Since $D_{+}=d^{-1}(O)$ and $D_{-}=m^{-1}(O)$, this principal divisor gives the displayed line bundle identity. This remains valid in characteristic two: the general Weierstrass equation of an elliptic curve of a smooth algebraic curve has a nontrivial generic involution $(x,y)\mapsto (x,-y-a_{1}x-a_3)$, so the two graphs still occur. This identity is also the symmetric-line-bundle form of the Theorem of the square.

Pull back the bundle identity along $(\varphi ,\psi ):E_1\to E_2\times E_2$ and take degrees of line bundles. For a nonzero map, $\mathrm{deg}({\varphi }^{\ast }L)=q(\varphi )$; for a zero map its pullback is trivial and has degree zero. Thus the degree parallelogram law isUsing bundles rather than pulling back the rational function itself is essential when $\varphi =\pm \psi$, where that function can vanish identically. Negation preserves degree, so $q(-\varphi )=q(\varphi )$. Applying the parallelogram identity to $n\varphi$ and $\varphi$ gives the quadratic degree recursion for elliptic multiplication, and induction from $q(0)=0$ yields $q(n\varphi )=n^{2}q(\varphi )$ for every integer $n$.

Define the bilinear degree pairing for elliptic-curve homomorphisms byIt is symmetric and odd in each argument. Two applications of the parallelogram law giveAdding proves $B(x+z,y)=B(x,y)+B(z,y)$, and symmetry proves additivity in the other argument. Also $B(\varphi ,\varphi )=q(\varphi )$. Therefore $q$ is an integer-valued quadratic form. The integral polarization is $2B(\varphi ,\psi )=q(\varphi +\psi )-q(\varphi )-q(\psi )$; $B$ itself may be half-integral.

Finally $q(\varphi )>0$ for every nonzero $\varphi$, since an isogeny has positive finite degree. If $n\varphi =0$ with $n\ne 0$, the identity $q(n\varphi )=n^{2}q(\varphi )$ forces $\varphi =0$, so $H$ is a torsion-free abelian group. On any finite-dimensional rational span in $H\otimes \mathbb{Q}$, the pairing has a rational matrix and is positive on every nonzero rational vector. By density it is nonnegative on real vectors; if that rational matrix were singular, its kernel would contain a nonzero rational vector, contradicting strict positivity. Thus its real extension is a positive-definite quadratic form. This establishes the required positivity as well as the quadratic identity. For example, on the subgroup of multiplication maps $[n]$, $q([n])=n^2$; on the endomorphism ring of an elliptic curve with complex multiplication it is the imaginary-quadratic field norm, explaining the lattice geometry behind isogeny degrees.

The Mumford rigidity lemma says that if $A$ is complete, $B$ is connected, and a morphism $F:A\times B\to Z$ maps the fiber over some $b_0\in B$ to a point, then $F$ is constant on every $A$-fiber and factors through $B$.

Put $y=f(e_X)$ and normalizeso $g(e_X)=e_G$. DefineWhen the first coordinate is $e_X$, this is constantly $e_G$. Apply rigidity with the second copy of the complete variety $X$ as the complete factor. It follows that $F(x,z)$ is independent of $z$, and evaluation at $z=e_X$ gives $F(x,z)=g(x)$. Thereforeso $g$ is a homomorphism of group varieties and $f=T_y\circ g$.

For an abelian variety $X$, consider the commutator morphismIt is the identity whenever either coordinate is the identity. The Mumford rigidity lemma applied successively to the two complete connected factors makes $c$ constant everywhere; its value at $(e,e)$ is $e$. Therefore every pair of points commutes, so the group law is commutative.

One form of the Mumford rigidity lemma says that if $X$ is a complete variety, $S$ is connected, and a morphism $F:X\times S\to Z$ maps $X\times \{s_0\}$ to one point, then $F$ factors through the projection to $S$. In particular, if $F$ also maps $\{x_0\}\times S$ to that point, then $F$ is constant.

Choose $s_0\in S(k)$ and put $g=f{|}_{X\times \{s_0\}}$. To see that the pointed morphism $g:X\to Y$ is a homomorphism, apply rigidity toIt vanishes on $X\times \{e\}$, so it factors through the second projection; it also vanishes on $\{e\}\times X$, so it is identically zero. Now defineThen $F(x,s_0)=e$ for every $x$ and $F(e,s)=e$ for every $s$. Rigidity forces $F$ to be identically $e$, so

A useful form of the Mumford rigidity lemma says that if $X$ is complete and connected and $h:X\times Y\to Z$ is constant on one fiber $X\times \{y_0\}$, then under the pointed separated hypotheses it factors through $Y$.

For $f:X\to G$ with $f(e_X)=e_G$, defineAt $x=e_X$ this is constantly $e_G$. Applying rigidity with the complete factor $X$ in the $y$ variable shows that $h(x,y)$ is independent of $y$. At $y=e_X$ its value is $f(x)$, soThus $f$ is a homomorphism of group varieties.

Completeness is essential. On the additive group variety ${\mathbb{G}}_a$ over a field of characteristic different from two, the morphism $f(t)=t^2$ fixes zero but is not additive.