# Determining the Mordell-Weil group of a universal elliptic curve

Let $K$ be a number field and let $K(a,b)$ be the field of rational functions with two indeterminates over $K$. Consider the elliptic curve $E$ over $K(a,b)$ defined by the Weierstrass equation \begin{equation*} E : y^2=x^3+ax+b. \end{equation*} What is the torsion subgroup and the rank of the Mordell-Weil group of $E$ over $K(a,b)$?

In the case $K=\mathbf{Q}$, this Mordell-Weil group is trivial because $E$ admits specializations $a,b \in \mathbf{Q}$ such that $E_{a,b}$ has trivial Mordell-Weil group. In the general case, I would like to show similarly the existence of $a,b \in K$ such that $E_{a,b}$ has trivial Mordell-Weil group over $K$, but I don't know how to proceed.

Since $K(a,b)$ is a finitely generated field, the Lang-Néron theorem tells us that the Mordell-Weil group of $E$ is finitely generated, but I'm not familiar enough to tell whether the proof actually gives us a way to compute this Mordell-Weil group.

## 2 Answers

Specialize $a,b$ to functions giving the universal elliptic curve over the modular curve $X_0(N)$. These are known to have rank zero over the function field of the modular curve with coefficients over $\mathbb{C}$ even. They can have torsion but, by varying $N$, you can show that the torsion is trivial too.

T. Shioda, On elliptic modular surfaces, J. Math. Soc. Japan, 24, (1972) 20-59.

• Oh, so the Mordell-Weil group of $E$ over $\mathbb{C}(a,b)$ is trivial, that's good to know. Thanks for your answer and the reference. Using the modular curves $X(N)$ and $X(N')$ for two coprime integers $N,N' \geq 3$ and Shioda's Thm 5.5, we see that the torsion is indeed trivial. – François Brunault Oct 10 '17 at 11:08

If you want to do this directly, you could "partially specialize" to, say $y^2 = x^3 + Ax + T$ with $A\in\mathbb C$. Then I don't think it's very hard to show, via a standard descent, that as an elliptic curve over $\mathbb C(T)$, the rank is $0$, and that for most $A$, the torsion is trivial, too. Actually, for the rank $0$ part, maybe it's easier to show that $y^2=x(x-A)(x-T)$ has rank 0 over $\mathbb C(T)$, since you can more easily do a 2-descent. And you also get that there is at most 2-torsion for most $A$. But it's pretty clear that your original curve over $\mathbb C(a,b)$ has no 2-torsion.

• That $y^2=x^3+Ax+T$ has trivial torsion for any nonzero $A$ follows from the analog of the Nagell-Lutz theorem over $\mathbb C[T]$ instead of $\mathbb Z$ (essentially the same proof). – Peter Mueller Oct 10 '17 at 15:21