## The Dual Isogeny

Today I want to talk about morphisms between curves and the special properties they have. This is sort of a broad-based topic, given that the study of morphisms and their properties is essentially the whole of algebraic geometry. I decided that perhaps the best way to narrow my focus is to push towards proving that if $\phi: E_1 \to E_2$ is a non-constant isogeny of elliptic curves over a field $k$ with a kernel of order $n$ then there exists a dual isogeny $\hat{\phi}:E_2 \to E_1$  such that $\hat{\phi} \circ \phi = [n]$, the map sending each point $P$ to $P \oplus P \oplus \dots \oplus P = nP$. The end result is that the path I present towards the dual isogeny is brisk but perhaps not optimal in that there are lots of facts I use which deserve a more unified treatment.

First things first: We recall from last time that an elliptic curves is an abstract curve with a distinguished rational point. An isogeny between elliptic curves $E_1 \to E_2$ is a morphism of curves which sends the distinguished point $O_{E_1}$ of $E_1$ to the distinguished point $O_{E_2}$ of $E_2$.

Implicit in this business of a dual isogeny is that of a finite kernel, which implies that an isogeny must be finite and a homomorphism on the groups of points. We recall some facts about morphisms of curves:

Fact 1: If $f: C \to C'$ is a morphism of irreducible curves, $f(C)$ is either all of $C'$ or just a point.

The proof of this fact hinges entirely on the fact that $f$ is continuous and $C$ is irreducible. Being a morphism implies that it is continuous and the image of an irreducible set under a continuous map is irreducible. Then we are reduced to dimension-counting. Since $C'$ has dimension 1, $f(C)$ has dimension 0 or 1. The only 0-dimensional irreducible set is a point and the only 1-dimensional irreducible subset of $C'$ is itself.

In the latter case, we have an injection of function fields $k(C') \hookrightarrow k(C)$. To see this, let $r$ be a rational function on $C'$ defined on the open set $U$. Then $latex f^*r = r\circ f$ is a rational function on $C$ defined on the open set  $f^{-1}(U)$. This map is a ring homomorphism, as $(r +r')\circ f = r\circ f + r'\circ f$ and $rr'\circ f = (r\circ f)(r'\circ f)$ on the intersection of the domains of $r$ and $r'$. Since it is a ring homomorphism between fields, it is injective, so $k(C')$ is isomorphic to a subfield of $k(C)$ which we call $f^*k(C')= \{f^*r: r\in k(C')\}$.

Fact 2: $k(C)$ is a finite extension of $k(C')$ up to isomorphism.

We already know that $k(C') \subset k(C)$. However, recall that we define the dimension of an irreducible variety to be the transcendence degree of its function field over $k$. Since $k(C)$ and $k(C')$ are extensions of transcendence degree 1 over $k$,  $k(C)$ is an algebraic extension of $k(C')$. Since both are finitely generated over $k$, it is a finite algebraic extension.

We define the degree (likewise the separable and inseparable degrees) of a morphism $f$ to be the degree of the extension $k(C)/f^*k(C')$.

Inseparable morphisms are frequently encountered in arithmetic geometry and cannot be ignored, but it’s very easy to get bogged down in details when one studies them. We will revisit them in a future post and for now restrict to the case when $f$ is a separable morphism ($k(C)/f^*k(C')$ is a finite separable extension) and $C,C'$ are elliptic curves.

Fact 3: Isogenies of elliptic curves induce group homomorphisms on $Pic^0$ and thus they ARE homomorphisms in the chord and tangent group law.

Recall first that if $\phi:E_1 \to E_2$ is an isogeny with a single point as its image, that single point must be $O_{E_2}$, so $\phi = [0]$.

So consider $\phi$ to be a nonconstant isogeny. Since it maps points to points, we can extend it  linearly to divisors ($\phi(\sum n_P P) = \sum n_P\phi(P)$). And if $N$ is the norm from $k(E_1)$ down to $\phi^*E_2$ then $div(f)\mapsto div(Nf)$. Hence we have an induced map $\phi_*:Pic^0(E_1) \to Pic^0(E_2)$.

The kernel of this map is then $\phi^{-1}(O_{E_2})$, a finite subgroup of $E_1$ of size $\deg(\phi)$(in the algebraic closure).

Fact 4: There is an injective map(really a group isomorphism, but we don’t need that) $\ker\phi \to Aut(\bar{k}(C)/\phi^*\bar{k}(C'))$ , thus $\bar{k}(C)$ is a Galois extension since $\phi$ is separable.

Let $P$ be a point in $\ker\phi$ and $\tau_P: Q \mapsto Q \oplus P$ be the automorphism of $E_1$ be induced by translation. This is a nonconstant morphism $E_1 \to E_1$ and thus induces an automorphism $\tau_P^*$ of $k(E_1)$(This identification $P \mapsto \tau_P^*$ is injective by checking at $O$). For each $P\in \ker\phi$, $\tau_P^*$ fixes every element of $\phi^*k(E_2)$ because for all points $Q$, $\phi \circ \tau_P(Q) = \phi(Q)+\phi(P) = \phi(Q) + O_{E_2} = \phi(Q)$ and $\tau^*(\phi^* f) = (\phi^*f)\circ \tau_P = f\circ\phi\circ \tau_P = f \circ \phi = \phi^* f$ .

Fact 5: If $\phi_2:E_1 \to E_2$ is separable and $\phi_3:E_1\to E_3$ has $\ker \phi_2 \subset \ker\phi_3$ then there is a unique isogeny $\psi$ such that $\psi \circ \phi_2 = \phi_3$.

We know $k(E_1)$ is Galois over $\phi_2^*k(E_2)$. It may or may not be Galois over $\phi_3^*k(E_3)$ but we know the automorphism group of $k(E_1)$ over $\phi_3^*k(E_3)$ contains the Galois group above. Hence every element of $Gal(k(E_1)/\phi_2^*k(E_2))$ fixes $\phi_3^*k(E_3)$ and we have $\phi_3^*k(E_3)\subset \phi_2^*k(E_2) \subset k(E_1)$. This inclusion gives us an injection $i:k(E_3) \hookrightarrow k(E_2)$, which gives a unique morphism $\psi:E_2 \to E_3$ such that $i = \psi^*$ and thus $\phi_2^*\psi^*k(E_3) = \phi_3^*k(E_3)$. Thus $\psi\circ\phi_2 = \phi_3$.

Fact 6: For any natural number $n$, $\ [n]$ described above is a nonconstant isogeny

By the group law, the function $\ [n] O = O$, and addition is continuous so we have a morphism. If we can show $\ [2]$ is nonconstant, then because $\ [nm] = [n]\circ [m]$, we need only prove that $\ [m]$ for $m = 2k+1$ is nonconstant. To do this, we find a nontrivial 2-torsion point $P$, so that $\ [m]P = [2k +1]P = [2k]P +P = P \ne O$.

Thus it suffices to find nontrivial 2-torsion. If we work over a field of characteristic not 2, we can use coordinates and the transformation in a previous post on elliptic curves to write our elliptic curve as $y^2 = 4x^3 + b_2x^2 + 2*b_4x + b_6 =:g(x)$ for $b_i \in k$ that we can calculate in terms of $a_1, \dots, a_6$. The nontrivial 2-torsion elements of our elliptic curve correspond to the places where the above equation has a vertical tangent line, thus the roots of the polynomial $g(x)$. Since there are exactly 3 such points in the algebraic closure of $k$, the morphism is nonconstant. If $k$ is a field of characteristic 2, we have to get our hands dirty and compute a duplication formula(or else just look it up in Silverman’s book, under theorem II.2.3).

If $P = (x,y)$ then the $x$-coordinate of $\ [2]P$ is a rational function with denominator $g(x)$ described above. Thus once more, the 2-torsion occurs when $g(x) =0$, which only occurs at finitely many points. The trouble is that in characteristic 2, $g(x) = b_2x^2 + b_6$, which is not a separable polynomial so there’s reason to suspect that $\ [2]$ is inseparable and there is no nontrivial 2-torsion(which is indeed the case as we’ll see next time). In this case we calculate a triplication formula, which will yield a similar polynomial formula for the denominator. This tells us that $\ [3]$ is a nonconstant morphism and there are nontrivial 3-torsion points, i.e. points such that $P \ne O, 2P \ne O$. A similar argument shows that the integers coprime to 6(they must be 1 or 2 mod 3) give nonconstant morphisms, completing the proof.

Now we can state the following:

If $\phi:E_1 \to E_2$ is a nonconstant separable isogeny with kernel of size $n$, \$ there is a unique isogeny $\hat{\phi}:E_2 \to E_1$ such that $\hat{\phi}\circ \phi = [n]$.

The proof follows from each of the facts given. In fact if $\phi= [0]$, then $\ [0]$ would serve as a suitable dual. Next time, The Dual Isogeny 2: Inseparable Isogenies.

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### 5 Responses to The Dual Isogeny

1. notedscholar says:

Nice try.

2. Zygmund says:

This works in general for abelian varieties, right? Even though the proof would have to be different.

3. rmb says:

This does generalize to abelian varieties, but it’s subtle. Elliptic curves are special because they are canonically isomorphic to their duals (that is, varieties that parametrize degree zero line bundles). For an isogeny f:X–>Y of abelian varieties, there is an induced isogeny f^:Y^–>X^ between the duals, but unless X and Y are elliptic curves, this is not the same as an isogeny Y–>X.

4. Marc says:

I have a question: if A is an abelian variety over some field K, and N:A->A is the multiplication-by-N isogeny, is it true that the dual of N is the multiplication-by-N isogeny on the dual of A? In particular, is this true also if N is not prime with char(K)?

5. Jim Stankewicz says:

Well if we want to talk about abelian varieties, it is a little bit better to talk about line bundles $L$ on the abelian variety $A$.

Then what you’re really asking is that $[n]^*L \cong L^n$ if $L\in Pic^0(A)$. This result is found in section 8 of Mumford’s Abelian Varieties, which is about the dual abelian variety in characteristic zero. Title aside, he proves it using the Seesaw Theorem, which is true regardless of characteristic.