TITLE: Determining the Mordell-Weil group of a universal elliptic curve
QUESTION [15 upvotes]: 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.

REPLY [9 votes]: 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.