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Let $ \phi: A \rightarrow B$ be a separable isogeny between two abelian varieties over a field $k$. One knows that there is a dual isogeny $ \hat {\phi} : B \rightarrow A$ such that $ \hat{\phi} \circ \phi = $ multiplication by $ \mathrm{deg}(\phi)$.

When I studied elliptic curves and abelian varieties, most of the references deal with a base field which is perfect. In this case, the proof for the existence of the dual isogeny is as following:

One makes a base change to work on the algebraic closure $\overline{k}$ of $k$. One considers $\mathrm{ker} (\phi)$, the closed points of the fiber of $\phi$ at the origin of $B$. It's a finite group which acts on $A$ and we have $\mathrm{ ker } (\phi) \subset \mathrm{ker} ( \times \mathrm{deg} (\phi))$. Then using the results about the quotien of a scheme by a finite group, one get the dual isogeny $\hat{\phi}$. Finally, one uses the action of $\mathrm{Gal}(\overline{k}/k)$ to get the result on the original base filed $k$.

Now if $k$ is not perfect, I didn't figure out how to do this. I have a feeling that the reason we work on the algebraic closure $\overline{k}$ is to get a finite group action. Of course, one can think about the action of $\mathrm{ker} (\phi)(k)$, but I think it doesn't work, because it's too small (We need all its closed points).

I would like to know if the dual isogeny exists in a general base field. If yes, what's the idea to see it, (and some reference)

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For elliptic curves, one can treat the inseparable case (with the base field $k$ being perfect or not.) The idea is as following:

(1) For every morphism $\psi : C_1 \rightarrow C_2$ of smooth curves (geometrically integral, complete) over a filed of char($k$) $=p > 0$, it factors as $C_1 \ \xrightarrow{\phi} \ C_1^{(q)} \ \xrightarrow{\lambda} \ C_2$, here $\phi$ is the $q^{\mathrm{th}}$-power Frobenius morphism, $q=$ inseparable degree of $\psi$, and $\lambda$ is separable.

(2) Using (1), one only need to treat with separable case and Frobenius morphism case. Furthermore, one only need to treat with the $p^{\mathrm{th}}$-power Frobenius morphism $\phi$ only. In elliptic curves case, one shows that the morphism $\times p$ is not separable, hence using (1), its factroization must contains $\phi$, i.e $\times p = \lambda \circ \phi^r$ and we are done.

Does this method can be used to higher dimensional abelian varieties? I have some difficulities to di this.

(a) The factorization in (1): In elliptic curves case, the degree of the $p^{\mathrm{th}}$-power Frobenius morphism $\phi$ is $p$, (not a power of $p$), which make it possible to construct this factorization. (Silverman, The Arithmetic of Elliptic Curves, p.30, Corollary 2.12). But for an abelian varieties of dimension $g$, the degree is $p^g$. In order to construct this factorization, one should have a result "the inseparable degree of an isogeny is a power of $p^g$", which I don't know if this is correct.

(b) One need to show that $\times p$ is not separable, which is ok.

So for (a), is it possible to do this for higher dimensional abelian varieties?

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I have learned the quotient of a scheme by a finite group scheme action, but still can't give a proof for this for purely inseparable case. Let $K$ be the kernel of $\phi$, and consider the action $ m: K \times_k A \rightarrow A$ which is induced by the multiplication on $A$ and let $\pi : K \times_k A \rightarrow A$ be the natural projection. Let $\psi : A \rightarrow A$ be the multiplication by $\mathrm{deg}(\phi)$ map. Then one needs to show that $\psi \circ m = \psi \circ \pi$ to get the desired induced morphism, but I can't get a proof for this. If $k$ is perfect, then by working on the function fields of $A$ and $B$, one only need to deal with $\phi =$ Frobenius morphism. (But still need to show that the multiplication by $p$ factors through by the Frobeinus morphism.) I need help to give a proof for this case ( $k$ not necessary being perfect), i.e the equality $\psi \circ m = \psi \circ \pi$. I think one just need to show that the induced morphism of $\psi$ on $K$ factors throguth the structure morphism $K \rightarrow k$ followed by the constant morphism $ e: k \rightarrow K$.

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Thank Qing Liu for the reference. Today I thought I had a way to prove this. Suppose $\phi : A \rightarrow B$ has degree $p^r$. Let $\psi := \times p^r : A \rightarrow A $. In order to show that the existence of $ \hat{ \phi} $, one only need to show that $\psi^* (K(A))$ is contained in $\phi^* (K(B))$, here $K(A)$ and $K(B)$ are the function fields of $A$ and $B$. (Then we have a rational map from $B$ to $A$ and extends to a morphism $ \hat{ \phi }$.) But notice that for each element $g \in K(A)$, $g^{p^r} \in \phi^* (K(B))$. If one can show that for each $f \in K(A)$, there exists $g \in K(A)$ such that $\psi^* (f) = g^{p^r}$, then we are done.

Since the multiplication by $p$ morphism $ [p] : A \rightarrow A$ is the composition of the Frobenius morphism $F: A \rightarrow A^{(p)}$ with the Verschiebung morphism $V : A^{(p)} \rightarrow A$, one sees that for any $f \in K(A)$, $[p]^* (f) = g^p$ for some $g \in K(A)$, if $k$ is perfect. In particular, this holds for an algebraically closed field $k$. So in this case, the statement in the end of the previous paragraph is true.

Finally, for $k$ not necessary being algebraically closed, we take a base change and work on $\overline{k}$ first. We then get $\overline{\psi} = \Phi \circ \overline{\phi}$, here $\Phi : \overline{B} \rightarrow \overline{A}$ and the $ \ \overline{ \ }$ means the schemes and the morphisms obtained from the base change. For any $f \in K(A)$, if one can show that $\Phi^* (f) \in \phi^* (K(B))$, then one completes the proof. But we have $\Phi^* (f) \in K(A) \cap \overline{\phi}^* (\overline{K}(B)) = K(A) \cap \phi^* (K(B)) \otimes_k \overline{k}$. Consider the subfield extension $\phi^* (K(B)) (\Phi^* (f))$ of $K(A)/\phi^* (K(B))$, then $\Phi^* (f) \in K(A) \cap \overline{\phi}^* (\overline{K}(B))$ gives $\phi^* (K(B)) (\Phi^* (f)) = \phi^* (K(B))$, and we are done.

Just a remark about the factorization of the morphism $[p] = V \circ F$. This is somehow different with our $\phi$ and $\hat{ \phi }$, since $\mathrm{deg}(F) = p^g$, not $p$, so it's a stronger result. Also one can show that the kernel of $F$ is of this form (via a $k$-isomorphism) $k[X_1, \cdots, X_n]/(X_1^p, \cdots, X_n^p)$. For a commutative group scheme of this form, I think one can show that the morphism $[p]$ on it is the zero morphism $0$ by dealing with symmetric functions. (I am not sure about this, but I saw somewhere that the Verschiebung morphism $V$ also uses the symmetric functions.) I like this way of approach since it uses the description of $[p]$ by the Frobenius morphism $F$, and the kernel $K$ of $F$ is explicit (if one chooses a coordinate), also the factorization $[p] = V \circ F$ seems can be proved without further knowledge. However, in Mumford's book, there is a proof that $[p] = 0$ for height one commutative group scheme by using the Lie-algebra. It would be appreciated for any comment on this.

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The dual isogeny should be $\vee{\phi} : B^{\vee} \to A^{\vee}$. I think the map $\phi \mapsto \hat{\phi}$ you want to define will not be well-behaved, even in characteristic 0. For example it won't be involutive if the dim is >1 (look at the degrees of the isogenies which are involved). –  François Brunault Jul 4 '11 at 9:58
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If the isogeny is separable, then the kernel of $\phi$ is split after passing to the separable closure of $k$. Now you can form the group quotient, etc., over the separable closure. And you can perform Galois descent for the extension of $k$ to the separable closure to get $\phi^\vee$ defined over $k$. Possibly the references are imposing "perfect" as a hypothesis because they want a result that applies to all isogenies, not just separable isogenies. –  Jason Starr Jul 4 '11 at 10:06
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If the characteristic of the base field divides $\deg(\phi)$ then you need to work with group schemes even if the field is perfect. In general, one can define the quotient of an abelian variety by any finite subgroup scheme, so the same proof works over any field. See for example, SGA 3, Expose VI_A. (As pointed out by Francois Brunault, the map you want is not what is usually called the dual isogeny (in higher dimensions) but it does exist.) –  ulrich Jul 4 '11 at 10:14
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As I have already mentioned, you can get a proof for abelian varieties by working with group schemes. One cannot use just Frobenius in this case; for example, by taking products of elliptic curves it is easy to find inseparable isogenies of degree $p$ between abelian varieties of dimension $g$ for any $g \geq 1$. –  ulrich Jul 5 '11 at 10:14
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It's been a while since I looked (and I don't have a copy handy), but doesn't Mumford give a construction of the dual isogeny over arbitrary fields in his book on Abelian Varieties? Probably Lang does, too, in his *Abelian Varieties" book, but that's in the older Weil-style language, so probably harder to read. –  Joe Silverman Jul 26 '11 at 0:40
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1 Answer

up vote 5 down vote accepted

You can find the statement in the general case (any isogeny over arbitrary field) and the proof in van der Geer and B. Moonen's book (draft) on abelian varieties. More precisely it is in Chapter 5, Prop. 5.12. The quotient by finite group scheme can be found in Chapter 4, §4.

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