# Mordell–Weil rank of some elliptic $K3$ surface

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.

• It seems that generically these elliptic K3 surfaces have Mordell-Weil rank only 1, coming from the base change of your section of the rational surface over the $s$-line, and thus Picard number only 18. Taking $q=19$ and $b=5$ (the square of $9 \bmod 19$, but not a cube mod $19$), Magma tells me that the characteristic polynomial of Frobenius has a factor $X^4 + 6 X^3 - 437 X^2 + 2166 X + 130321$ whose roots are not of the form $19$ times a root of unity, so the transcendental part of $H^2$ has dimension $4$.  How does this family of K3 surfaces arise in your work? Mar 30, 2021 at 2:06
• It arises in my article eprint.iacr.org/2020/1070. Apr 1, 2021 at 7:24
• Apr 5, 2021 at 0:50

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 $$-3, -2^2 3, -3^2 3, -5^2 3, -7^2 3$$.

• Thank you very much for your detailed response! Apr 10, 2021 at 13:52