Explicit two-cover descent for genus 2 curves

Note that $\operatorname{ev}_{\alpha}$ is well-defined since $f(\alpha)=0$. The first step is to understand how the base $\mathbf{P}^{1}$ embeds into $\mathcal{K}$. Recall that a trope of a quartic surface $\mathcal{K}\subset\mathbf{P}^{3}$ is a tangent plane that intersects the surface at a conic (with multiplicity two). A Kummer surface has exactly $16$ tropes, which are the projective duals of the $16$ nodes [10, §3.7][23, §8-9].

The description of six of the tropes of the Kummer surface given in [10, §7.6] shows that the image of $\mathbf{P}^{1}$ in $\mathcal{K}$ corresponding to the Abel–Jacobi map with base point $(\alpha,0)$ is contained in the trope $T_{\alpha}$ with equation $\alpha^{2}\kappa_{1}-\alpha\kappa_{2}+\kappa_{3}=0$, where $\kappa_{1},\kappa_{2},\kappa_{3},\kappa_{4}$ are the coordinates of the usual embedding of $\mathcal{K}$ as a quartic surface in $\mathbf{P}^{3}$ (as described in [10, §3.1], using the letters $\xi_{i}$ instead of $\kappa_{i}$). (Strictly speaking, $T_{\alpha}$ intersects $\mathcal{K}$ in twice the conic corresponding to the base $\mathbf{P}^{1}$; we also denote this conic by $T_{\alpha}$.)

Instead of constructing $Z_{\delta}$ as the quotient of a genus $17$ curve embedded in $J$, we compute $Z_{\delta}$ using maps between the twisted Kummer surfaces. The preimage of $T_{\alpha}$ under the map $\bar{\pi}_{\delta}\colon\mathcal{Y}_{\delta}\to\mathcal{K}$ also contains some exceptional divisors, which we wish to omit when constructing $Z_{\delta}$, so we instead consider the condition for a general element of $\mathcal{Y}_{\delta}$ to map to $T_{\alpha}$ via the map $\bar{\pi}_{\delta}$.

Let $P\in\mathcal{Y}_{\delta}$ be an arbitrary point not contained in the locus of indeterminacy of the rational map $p_{\delta}\colon A_{\delta}\dashrightarrow\mathcal{Y}_{\delta}$. Let $\xi\in L$ be an arbitrary lift of $P$ from $\mathbf{P}(L)$ to $L$. We will show that $\bar{\pi}_{\delta}(P)\in T_{\alpha}$ if and only if $\xi(\alpha)=0$.

We first treat the untwisted case $\delta=1$. Let $D=((x_{1},y_{1}))+((x_{2},y_{2}))-K_{C}$ such that $p_{1}([D])=[\pm D]=P$. As explained in the paragraph preceding [17, Prop. 4.11], the $x$-coordinates of the points $R_{1}$ and $R_{2}$ such that $2D\sim R_{1}+R_{2}-K_{C}$ are the roots of the quadratic polynomial $H(X)$ corresponding to $\xi^{2}$. (This does not depend on the choice of lift $\xi\in L$ since choosing a different lift multiplies $H$ by an element of $k^{*}$, which does not change the roots.) The condition that $\pm 2D$ is contained in $T_{\alpha}$ is exactly that one of the $x$-coordinates of $R_{1}$ and $R_{2}$ is $\alpha$, i.e., that $\alpha$ is a root of $H$, or equivalently, that $\xi^{2}(\alpha)=0$, which is the case if and only if $\xi(\alpha)=0$.

Now we handle the twisted case. Let $D=((x_{1},y_{1}))+((x_{2},y_{2}))-K_{C}$ such that $p_{\delta}([D])=P$. Let $k^{s}$ be a separable closure of $k$, let $L^{s}=L\otimes_{k}k^{s}$, and let $\varepsilon\in L^{s}$ such that $\varepsilon^{2}=\delta$. Then $\delta\xi^{2}=(\varepsilon\xi)^{2}$, so $\varepsilon\xi\in\mathcal{Y}_{1}$. Let $D^{\prime}\in J$ such that the image of $D^{\prime}$ in $\mathcal{Y}_{1}$ is $\varepsilon\xi$. By [17, §7], we have $\bar{\pi}_{\delta}=[2]\circ g$, where $g$ is defined by multiplication by $\varepsilon$ in $L$. So

and lifting to the Jacobian, the divisor class corresponding to $\bar{\pi}_{\delta}(\xi)$ is equal to $[2D^{\prime}]$, i.e., $\pi_{\delta}([D])=[2D^{\prime}]$. As in the previous paragraph, the roots $r_{1},r_{2}$ of the quadratic polynomial $H$ such that $\delta\xi^{2}\equiv H\pmod{f}$ are the $x$-coordinates of points $R_{1},R_{2}\in C$ such that $(R_{1})+(R_{2})-K_{C}\sim 2D^{\prime}$. Thus, $\bar{\pi}_{\delta}(P)\in T_{\alpha}$ if and only if $\alpha$ is a root of $H$, which is equivalent to $\delta(\alpha)\xi^{2}(\alpha)=0$. Since $\delta\in L^{*}$, we have $\delta(\alpha)\neq 0$, so this is equivalent to $\xi(\alpha)=0$.

Represent $\xi$ in the basis $g_{1},\dots,g_{6}$ as $\xi=\sum_{i=1}^{6}v_{i}(\xi)g_{i}$. Then

so in the basis $v_{1},\dots,v_{6}$, the condition $\xi(\alpha)=0$ becomes

It follows immediately from the definitions of $g_{1},\dots,g_{6}$ that

i.e., $\gamma_{i}=g_{i}(\alpha)$ for each $i\in\{1,\dots,6\}$, so $Z_{\delta}$ is in fact the hyperplane section of $\mathcal{Y}_{\delta}$ whose coefficients in the basis $v_{1},\dots,v_{6}$ are the coefficients of the polynomial $f(x)/(x-\alpha)$. ∎

Let $q_{\delta}\colon A_{\delta}\to\mathcal{K}_{\delta}$ be the quotient map. The ramification divisor of $q_{\delta}$ is $\pi_{\delta}^{-1}(0)$. At each point $P\in\pi_{\delta}^{-1}(0)$, since $q_{\delta}$ has degree $2$, the point $q_{\delta}(P)$ is either nonsingular or a simple node. The map $\mathcal{Y}_{\delta}\to\mathcal{K}_{\delta}$ is given by blowing up at $q_{\delta}(\pi_{\delta}^{-1}(0))$, which desingularizes any simple nodes, so $Z_{\delta}$ (being the proper transform of $q_{\delta}(W_{\delta})$) is smooth.

By Theorem 2.2, the curve $Z_{\delta}$ is a complete intersection of three quadrics in $\mathbf{P}^{4}$, so $Z_{\delta}$ is a canonical curve of genus $5$ (cf. [20, Ch. IV, Ex. 5.5.3]). ∎

Write $q=p^{n}$, where $p$ is an odd prime. By [19, Exposé X, Cor. 3.9], specialization to $\mathbb{F}_{q}$ induces an isomorphism between the prime-to-$p$ parts of the étale fundamental groups of $C$ and the special fiber $\bar{C}/\mathbb{F}_{q}$. Thus $W_{\delta}$, being a degree $16$ étale cover of $C$, also has good reduction. Euler characteristic (and hence also arithmetic genus) are locally constant in proper flat families [28, §5, Cor. 1], so Proposition 2.3 implies that the special fiber $\bar{Z}_{\delta}/\mathbb{F}_{q}$ has arithmetic genus $5$, hence geometric genus at most $5$. Since $p$ is odd, the quotient map $\bar{W}_{\delta}\to\bar{Z}_{\delta}$ is tamely ramified, so the Riemann–Hurwitz formula implies that $\bar{Z}_{\delta}$ has geometric genus exactly $5$ and thus is smooth over $\mathbb{F}_{q}$. ∎

As in [17, §4], let $C_{0}^{(\delta)},\dots,C_{5}^{(\delta)}\in\operatorname{Sym}^{2}(\hat{L})$ be quadratic forms such that $C_{j}^{(\delta)}(z)=p_{j}(\delta z^{2})$ for $z\in L$, where $p_{j}$ gives the coefficient of $X^{j}$. We have

where $T$ is the matrix defined in Definition 2.1, so that in particular

Thus, taking into account that $Q_{j}^{(\delta)}$ vanishes on $\mathcal{Y}_{\delta}$ for $j\in\{0,1,2\}$, we have

Moreover, $C_{3}^{(\delta)}=C_{4}^{(\delta)}=C_{5}^{(\delta)}=0$ on $\mathcal{Y}_{\delta}$.

Let $\xi\in L$ be a lift of $P\in Z_{\delta}\subset\mathbf{P}(L)$. By construction of $\mathcal{Y}_{\delta}$, we have

As explained in the proof of Theorem 2.2, the roots of this quadratic polynomial are the $x$-coordinates of points of the divisor in $J$ corresponding to $\bar{\pi}_{\delta}(P)$. Moreover, since $P\in Z_{\delta}$, one of these roots is $\alpha$. Thus, in the affine patch where the second coordinate of $\mathbf{P}^{1}$ is nonzero, writing $\bar{\pi}_{\delta}(P)=(r:1)$, we have

for some nonzero $c\in k^{s}$. Comparing coefficients, we obtain $C_{2}^{(\delta)}(\xi)=c$ and $C_{1}^{(\delta)}(\xi)=-c(\alpha+r)$, so

This gives the desired formula for $\bar{\pi}_{\delta}(P)$. Finally, we have $\bar{\pi}_{\delta}(P)=(1:0)$ if and only if $C_{2}^{(\delta)}(\xi)=0$, completing the proof. ∎

Observe that $\pi_{\delta}\colon W_{\delta}\to C$ is étale, the branch locus of $i\colon C\to\mathbf{P}^{1}$ is $\Omega$, and the branch locus of $p_{\delta}\colon W_{\delta}\to Z_{\delta}$ is $\bar{\pi}_{\delta}^{-1}(\alpha)$, with all ramification indices in the preimage of the branch locus equal to $2$. Thus, commutativity of diagram (2.1) implies that the branch locus of $\bar{\pi}_{\delta}$ is $\Omega\setminus\{\alpha\}$, and for each $\omega\in\Omega\setminus\{\alpha\}$, the preimage $\bar{\pi}_{\delta}^{-1}(\omega)$ consists of $8$ geometric points of ramification index $2$.

The remaining claim that $\bar{\pi}_{\delta}^{-1}(\omega)$ is the hyperplane section of $Z_{\delta}$ given by intersection with $\mathbf{P}(\ker(\operatorname{ev}_{\omega}))$ follows from the description of $\bar{\pi}_{\delta}$ given in the proofs of Theorems 2.2 and 3.1: For $\xi\in L$ lifting a point $P\in Z_{\delta}$, we have $\bar{\pi}_{\delta}(P)=(\omega:1)$ if and only if the quadratic polynomial defining $\delta\xi^{2}$ has roots $\alpha$ and $\omega$, which is equivalent to the condition $\xi(\alpha)=\xi(\omega)=0$, i.e., $P$ is in the kernel of both the evaluation maps $\operatorname{ev}_{\alpha}$ (which defines $Z_{\delta}$ as a hyperplane section of $\mathcal{Y}_{\delta}$) and $\operatorname{ev}_{\omega}$, as was to be shown. ∎

Since $\operatorname{ev}_{\beta_{j}}$ is a ring homomorphism for each $j$, the quartic form $Y_{\alpha,\omega}$ is multiplicative with respect to $L$, i.e., $Y_{\alpha,\omega}(\xi\eta)=Y_{\alpha,\omega}(\xi)Y_{\alpha,\omega}(\eta)$ for all $\xi,\eta\in L$. As proved in Theorem 3.1, for all $\xi\in L^{s}$ lifting a point $P\in Z_{\delta}(k^{s})$, we have

Putting these together, we obtain

Thus $\varphi(Z_{\delta})\subseteq D_{\delta,\omega}$. Since $\varphi$ is non-constant and $D_{\delta,\omega}$ is an irreducible curve, $\varphi(Z_{\delta})=D_{\delta,\omega}$. Commutativity of the diagram is immediate from the formulas. ∎

Our strategy is to consider universal families of curves and abelian varieties corresponding to the above situation, observe that the properties of interest are deformation-invariant, and deform the problem to a more computationally tractable case.

Let $S=\operatorname{Spec}A$ be the space parametrizing triples $(g,\alpha,\delta)\in k[w]\times k\times k[X]$ such that $g$ is a monic squarefree quintic polynomial with $g(\alpha)\neq 0$, the degree of $\delta$ is at most $5$, and $\delta$ is invertible modulo $(X-\alpha)\cdot g(X)$. Let $P\in A[w]$ be the generic monic quintic polynomial, and let $T=\operatorname{Spec}A[w]/\langle P(w)\rangle$. Let $\mathcal{Z}\to S$ and $\mathcal{D}\to T$ be the relative curves whose fibers above a point $(g,\alpha,\delta)\in S$ are the genus $5$ curve $Z_{\delta}$ and the genus $1$ curve $D_{\delta}$, respectively, that are associated to the twisting parameter $\delta$ for the hyperelliptic curve $y^{2}=(x-\alpha)g(x)$. Let $\mathcal{J}\to S$ be the relative Jacobian variety of $\mathcal{Z}$, and let $\mathcal{A}=\operatorname{Res}_{S}^{T}(\operatorname{Jac}(\mathcal{D}))$, which exists as a scheme since $T\to S$ is étale.