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.
