Mordell–Weil rank of some elliptic $K3$ surface

Question

Consider a finite field $\mathbb{F}_q$ such that $q \equiv 1 \pmod3$ (i.e., $\omega \mathrel{:=} \sqrt[3]{1} \in \mathbb{F}_q$ for $\omega \neq 1$) and an element $b \in (\mathbb{F}_q^*)^2 \setminus (\mathbb{F}_q^*)^3$ (a quadratic residue and cubic non-residue in $\mathbb{F}_q$). Further, I am interested in the elliptic $\mathbb{F}_q$-surface (i.e., elliptic $\mathbb{F}_q(t)$-curve) $$ E\!: \begin{cases} y_1^2 - b = b(y_0^2 - b)t^3,\\ y_2^2 - b = b^2(y_0^2 - b)t^3 \end{cases} \quad \subset \quad \mathbb{A}^4_{(y_0,y_1,y_2,t)}. $$ It is readily checked that $E$ has a Weierstrass form $W\!: y^2 = x^3 + a_4x + a_6$ with \begin{align*} a_4 \mathrel{:=}{} & -3b^4s^4 + 3b^2(b + 1)s^3 - 3(b^2 - b + 1)s^2, \\ a_6 \mathrel{:=}{} & -2b^6s^6 + 3b^4(b + 1)s^5 + 3b^2(b^2 - 4b + 1)s^4 - (2b^3 - 3b^2 - 3b + 2)s^3, \end{align*} where $s \mathrel{:=} t^3$. Note that $W$ has the infinite order $\mathbb{F}_q$-section (i.e., $\mathbb{F}_q(t)$-point) $$ x \mathrel{:=} \big(b(2bs - 1) - (3bs - 2) \big)s, \qquad y \mathrel{:=} 3\sqrt{b}(b - 1)(2\omega + 1)s^2(bs - 1). $$ Besides, the Mordell–Weil rank of $W$ over $\overline{\mathbb{F}_q}(s)$ (and hence over $\mathbb{F}_q(s)$) is equal to $1$. This immediately follows if we look at the degenerate fibers of $W$ as a rational elliptic surface. What about the Mordell–Weil rank of $W$ over $\overline{\mathbb{F}_q}(t)$ and $\mathbb{F}_q(t)$?

Let us identify $W$ with the corresponding Kodaira–Néron model, which is clearly a $K3$ surface. Again, analyzing the degenerate cases of $W$, we see that $\operatorname{rk}W\big( \mathbb{F}_q(t) \big) \leqslant \operatorname{rk}W\big( \overline{\mathbb{F}_q}(t) \big) \leqslant 5$ (even $\leqslant 3$ if $W$ is non-supersingular), because, as is known, the Picard $\overline{\mathbb{F}_q}$-number $\rho(W) \leqslant 22$ (resp., $\leqslant 20$ if $W$ is non-supersingular).

It seems that $W$ is a singular $K3$ surface (i.e., $\rho(W) \mathrel{:=} 20$) and $\operatorname{rk}W\big( \mathbb{F}_q(t) \big) = 1$. At least, for some small $q$ and $b$ (satisfying the restrictions) I checked these conjectures by means of the CAS Magma.

Answer 1

Let $E_0$ be the elliptic curve $y^2 = x^3 + 1$, and choose $\beta$ in $k = {\bf F}_q$ so that $b = \beta^2$. Then $W = W_b$ has $\rho=18$ unless the elliptic curve $$ E_\beta : Y^2 = X^3 + \beta \, (3X + (\beta+1)^2)^2 $$ is isogenous to $E_0$, in which case $\rho(W)$ is $20$ ("singular") or $22$ ("supersingular") depending on whether $E_0$ is ordinary or supersingular, that is, depending on whether the characteristic of $k$ is $1 \bmod 3$ or $-1 \bmod 3$. (This does not depend on the choice of square root $\beta$ of $b$, because the curve $E_\beta$ and $E_{-\beta}$ are $3$-isogenous over $\bar k$.)

The recipe for $\rho(W_b)$ holds whether or not $b$ is a cube, and over any field of characteristic not $2$ or $3$ (the supersingular case does not occur in characteristic zero).

In particular if $b \in {\cal B} := \{-1, 1/4, 4, 4/5, 5/4, 27/28, 28/27\}$ then $E_0$ is isogenous with $E$, so $W$ is supersingular in characteristic $-1 \bmod 3$, and singular otherwise. The fact that ${\cal B}$ is symmetrical under $b \leftrightarrow 1/b$ reflects an isomorphism between $W_b$ and $W_{1/b}$.

The elliptic fibration over the $t$-line is $y^2 = {\rm cubic}(x)$ with a cubic that factors completely, so instead of narrow Weierstrass form we write it as $$ y^2 = x \, (x - b^2 t^4 + t) \, (x - b^2 t^4 + b t). $$ This works over any field where $6 \neq 0$, and becomes equivalent with the OP's formula once we scale $x,y$ by powers of $\sqrt{-3}$. The isomorphism between $E_b$ and $E_{1/b}$ is $(x,y,t,b) \leftrightarrow (x,y,bt,1/b)$. The nontorsion section is $(x,y) = (b^2 t^4, b^{3/2} t^3)$, which has canonical height $3/2$; this section, together with the reducible fibers and $2$-torsion sections, gives a sublattice of ${\rm NS}(W_b)$ of rank $18$ and discriminant $-144$.

We next move to another elliptic fibration with parameter $u$ where $$ x = u t (b t^3 - 1). $$ This yields an even nicer equation $$ Y^2 = X^3 + b (u-1)^2 u^3 (u-b)^4. $$ Its singular fibers at $u=0,\infty$ (type I${}_0^*$), $u=1$ (type IV), and $u=b$ (type IV${}^*$) account for the same sublattice of ${\rm NS}(W_b)$.

This new elliptic fibration tells us that $W_b$ is the quotient of $E_0 \times C_b$ by a $6$-cycle, where $E_0: y^2 = x^3 + 1$ as above and $C_b$ is the curve $$ C_b: v^6 = b (u-1)^2 u^3 (u-b)^4. $$ The $6$-cycle is generated by the product of the automorphisms of $E_0$ on $C_b$ that take $(x,y)$ to $(\zeta^2 x, \zeta^3 y)$ and $(u,v)$ to $(u, \zeta^{-1} v)$ where $\zeta$ is a primitive sixth root of unity (so $\zeta^3 = -1$).

Now $C_b$ has genus $2$, with hyperelliptic involution $(u,v) \leftrightarrow (u,-v)$ because the quotient by this involution is the rational curve $w^3 = b (u-1)^2 u^3 (u-b)^4$ (with $w=v^2$). An explicit parametrization is $$ w = u (u-1) (u-b) z, \quad u = \frac{z^3-b^2}{z^3-b}, $$ and we find that $C_b$ is birational with the curve $\eta^2 = b (z^3 - b) (z^3-b^2)$. Recalling that $b = \beta^2$, we write $z = \beta(1+T)/(1-T)$ and find the equivalent model $$ U^2 = \beta \left( (\beta+1)^2 T^6 - 3(\beta^2-10\beta+1)T^4 + 3(\beta^2+10\beta+1)T^2 - (\beta-1)^2 \right) $$ of $C_b$. The quotients of $C_b$ by the non-hyperelliptic involutions $(T,U) \leftrightarrow (-T,U)$ and $(T,U) \leftrightarrow (-T,-U)$ then give quadratic twists of the elliptic curves $E_{\pm\beta}$. We conclude that the Jacobian of $C_b$ is isogenous over $\bar k$ with $E_\beta \times E_{-\beta}$, and thus with $E_\beta^2$; we soon recover from this our recipe for the Picard number of $W_b$.

The list $\cal B$ of rational values of $b$ for which $W_b$ is singular (or supersingular in characteristic $-1 \bmod 3$) is then obtained by setting the $j$-invariant of $E_\beta$ equal to the rational or quadratic $j$-invariants of the CM elliptic curves with discriminant $-3, -2^2 3, -3^2 3, -5^2 3, -7^2 3$.