Mordell-Weil rank of some algebraic surface

Question

Consider an elliptic curve $S:y^2 = x^3 + t^2x + (t^3 + 1)$ over $k(t)$, where char($k$) is 0. How can I calculate the Mordell-Weil rank of the surface, or how to get its Picard number $\rho(S)$ of the Neron-Severi group. I guess $\rho(S) = 2$ and its Mordell-Weil rank is 0, but I cannot prove it.

Answer 1

I assume that $k$ is algebraically closed; if you are interested in non-closed fields then one needs to calculate the Picard group over the algebraic closure first then use Galois theory to try to decide what the Picard group over the ground field is.

Any elliptic surface of the form $y^2 = x^3 + a_4(t)x + a_6(t)$ with $\deg a_i \leq i$ is a rational elliptic surface. So your surface is rational.

Let $S$ be the unique smooth projective relatively minimal surface corresponding to your equation. Then $S$ is the blow-up of $\mathbb{P}^2$ in the base-locus of a pencil of plane cubic curves. It follows that $\rho(S) = 10$.

To calculate the rank of the Mordell-Weil group one then uses the Shioda-Tate formula, which requires calculating the Kodaira symbol of the singular fibres using Tate's algorithm.

Rather than writing out the full details I think this is a good exercise to have a go at yourself. You can read all about these results and this method in the book:

Schutt, Shioda - Elliptic surfaces

Chapter 8 in particular is dedicated solely to rational elliptic surfaces.

Answer 2

I can prove the generic rank over $\overline{\mathbb{Q}}$ is 4, however, I can just show the generic rank over $\mathbb{Q}$ is at most 2. In the following, we will first find the structure of singular fibres and then the group $E(\overline{\mathbb{Q}}(t))$. Secondly, we will search for two $\overline{\mathbb{Q}}(t)$-rational points and calculate their height pairing to show they are independent to each other.

Let $E$ be the elliptic curve defined over $\mathbb{Q}(t)$: \begin{equation*} \hspace{3cm} y^2 = x^3 + t^2 x + (t^3 + 1). \hspace{3cm} (1) \end{equation*} Let $f: S \rightarrow \mathbb{P}^1$ be the associated Kodaira-Neron model, and it is rational. Then $\rho(S) = 10, \chi = 1$.

The discriminant $\Delta = 4t^6 + 27(t^3 + 1)^2$has 6 distinct roots \begin{equation*} t_i = - \sqrt[3]{ \frac{27 + 6\sqrt{3}}{31} } \zeta_3^i,\ t_{3 + i} = - \sqrt[3]{ \frac{27 - 6\sqrt{3}}{31} } \zeta_3^i, i = 1, 2, 3. \end{equation*} Except for $\infty$, it is exactly when $t=t_i$ $S$ has singular fibres.

The singular point $(x,y)$ on the fibre $F_t$ satisfies \begin{equation*} \left\{ \begin{array}{lr} 3x^2 + t^2 = 0 \\ 2y = 0. \end{array} \right. \end{equation*} Then we know the singular point is $(- \frac{3(t_i^3 + 1)}{2t_i^2}, 0)$. Make the transform $x \mapsto x - \frac{3(t^3 + 1)}{2t^2}$, then we get the minimal Weierstrass equation \begin{equation*} y^2 = x^3 - 18(t^3 + 1)x^2 + 4 \Delta x - 8(t^3 + 1) \Delta. \end{equation*} Since $t - t_i \nmid a_2(t) = 18(t^3 + 1)$, $E(\overline{\mathbb{Q}}(t))$has multiplicative reduction at $t-t_i$. We define $v_{t_i}(*)$ the normalized valuation associated with $t-t_i$. According to Tate's Algorithm, since $v_{t_i}(\Delta) = 1$, $F_{t_i}$ is a rational curve with a node. It means $m_{t_i} = 1$.

Now we consider the structure of $F_\infty$. Dividing by $t^6$ on the both sides of equation (1). Make the transform $(x,y,t) \mapsto (x/t^2, y/t^3, 1/t)$, then we get the minimal Weierstrass equation \begin{equation*} \hspace{3cm} y^2 = x^3 + t^2x + t^3(t^3+1), \hspace{3cm} (2) \end{equation*} $\Delta = t^6\big( 4 + 27(t^3+1)^2 \big)$. Denote $v_\infty(*)$ the normalized valuation of $t$. Since $v_\infty(a_2) = v_\infty(0) = \infty > 0$, $S$ has additive reduction at $t = \infty$. since $\mbox{char}\mathbb{Q} \neq 2,3$, we have $m_\infty = v_\infty(\Delta) - 1 = 5$. But as for additive reduction, it is only $I_0^*$ that contains exactly 5 components, i.e. , \begin{equation*} F_\infty = \Theta_{\infty, 0} + \Theta_{\infty,1} + \Theta_{\infty, 2} + \Theta_{\infty, 3} + 2\Theta_{\infty, 4}, \end{equation*} $(\Theta_{\infty, i}.\Theta_{\infty, 4}) = 1, i = 0,1,2,3$. According to the correspondence of reducible singular fibres and Dynkin graphs, $T_\infty$ is isomorphic to $D_4$.

Above all, the trivial lattice of $S$ has the structure \begin{equation*} \mbox{Triv}(S) \cong \begin{pmatrix} -1&1\\ 1&0 \end{pmatrix} \oplus \begin{pmatrix} -2 & 0 & 0 & 1\\ 0 & -2 & 0 & 1\\ 0 & 0 & -2 & 1\\ 1 & 1 & 1 & -2 \end{pmatrix} \end{equation*} According to the classfication of rational elliptic surfaces, $E(\overline{\mathbb{Q}}(t))$ is torsion-free, and Mordell-Weil lattice is isomorphic to $D_4^\vee$, $r(E/ \overline{\mathbb{Q}}(t) ) = 4$.

Consider six points of the elliptic curve $E( \overline{\mathbb{Q}}(t) )$ \begin{equation*} P_i = ( \alpha_i t , -1),\ P_{i + 3} = ( \alpha_i t , 1),\ i = 1,2,3, \end{equation*} where $\alpha_i$ are three distince roots of the equaiton $x^3 + x + 1 = 0$. The equation has two complex roots and one algebraic irrational root of degree 3. Thus $P_i,i = 1,2, ... ,6$ are not in $\mathbb{P}_{\mathbb{Q}(t)}^2$. About above points, we have the relation \begin{equation*} \left\{ \begin{array}{lr} P_1 + P_2 + P_3 = O, \\ P_i + P_{i + 3} = O, i = 1,2,3. \end{array} \right. \end{equation*} Hence the rank of the subgroup generated by $P_i, i = 1,2,\cdots,6$ is at most 2. However, $E(\overline{\mathbb{Q}}(t))$ is torsion-free, so the rank of the subgroup is at least 1. Now we calculate the Gram matrix of height pairing between $P_1$ and $P_2$ to prove $P_1$ is independent to $P_2$. The expression for the height pairing is \begin{equation*} \langle P_1,P_2\rangle = \chi + (P_1.O) + (P_2.O) - (P_1.P_2) - \mbox{contr}_\infty(P_1,P_2). \end{equation*} \begin{equation*} \langle P_k,P_k\rangle = 2\chi + 2(P_i.O) - \mbox{contr}_\infty(P_k, P_k), k = 1,2,3. \end{equation*} Let $(P) \subset S$ be the section of $S$ associated with $P \in E(\overline{\mathbb{Q}}(t))$. Suppose $(P), (Q)$ intersect with $\Theta_{v,i}$ and $\Theta_{v,j}$ respectively, then the local contribution of $P$ and $Q$ on the reducible singular fibre $F_v$ is defined by \begin{equation*} \mbox{contr}_v(P,Q) = \left\{ \begin{array}{lr} (-A_v^{-1})_{i,j},\ \mbox{if}\ i\geq 1, j \geq 1, \\ 0\ \mbox{otherwise}, \end{array} \right. \end{equation*} where $A_v$ is the Gram matrix of $T_v^{-}$. Since $T_\infty \cong D_4$, we have \begin{equation*} -A_\infty^{-1} = \begin{pmatrix} 1 & 1/2 & 1/2 & 1 \\ 1/2 & 1 & 1/2 & 1 \\ 1/2 & 1/2 & 1 & 1 \\ 1 & 1 & 1 & 2 \end{pmatrix} \end{equation*} Since $\mbox{deg}(\alpha_i t) \leq 2$, $(O)$ does not intersect $(P_i)$. On the affice piece $\{Z = 1\}$, $(P_1)$ intersect $(P_2)$ at one point $(0, -1)$, and the multiplicity is 1. In the next, we have to blow up to see the local contribution and intersect number of $P_1$ and $P_2$ on the fibre at $\infty$.

Dividing $t^2$ on the both sides of the equation (2), make the transform \begin{equation*} x = \overline{x}t, \hspace{2cm} y = \overline{y}t. \end{equation*} After the first blow-up, we get \begin{equation*} \hspace{3cm} \overline{y}^2 = t \overline{x}^3 + t\overline{x} + t(t^3+1) \hspace{3cm} (3) \end{equation*} Substitute t with 0, then we get the irreducible component with multiplicity 2 of $F_\infty$. \begin{equation*} \Theta_{\infty, 4} = \{\overline{y} = 0\} \end{equation*} Make the transform $\overline{y} = \overline{\overline{y}}t$, we make the second blow-up, \begin{equation*} \hspace{3cm} \overline{\overline{y}}^2 t = \overline{x}^3 + \overline{x} + (t^3+1) \hspace{3cm} (4) \end{equation*} Substitute $t$ with 0, we get the other irreducible components \begin{equation*} \Theta_{\infty, k} = \{ \overline{x} = \alpha_k \}, k = 1, 2, 3. \end{equation*} In the coordinates of $\overline{x}$ and $\overline{\overline{y}}$, $P_k = (\alpha_k, -t), P_{k+3} = (\alpha_k, t), k = 1,2,3$. Clearly, $(P_k)$ and $(P_{3+k})$ intersect with the same component $\Theta_{\infty, k}, k = 1,2,3$.

\begin{equation*} \langle P_1, P_2 \rangle = 1 + 0 + 0 - 1 - 1/2 = -1/2. \end{equation*} \begin{equation*} \langle P_k, P_k \rangle = 2 + 2\times 0 - 1 = 1,\ k = 1,2,3,4,5,6. \end{equation*} Since $1^2 - (-1/2)^2 = 3/4 \neq 0$, $P_1$ is independent to $P_2$.

Finally, we can examine the following relations by calculation \begin{equation*} \left\{ \begin{array}{lr} P_1 + P_2 + P_3 = O, \\ P_i + P_{i + 3} = O, i = 1,2,3. \end{array} \right. \end{equation*} The height pairing of $P_i$ and $P_j$ are as follows \begin{equation*} \langle P_i, P_j \rangle = 1 + 0 + 0 - 1 - 1/2 = -1/2, \ 1 \leq i < j \leq 3. \end{equation*} \begin{equation*} \langle P_k, P_{3+k} \rangle = 1 + 0 + 0 - 1 - 1 = -1, \ k = 1,2,3. \end{equation*}