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Let $K$ be a number field, and let $P(t,X)$ be a monic polynomial in $X$ with coefficients in $K(t)$.

I would like to understand the set $T$ consisting of those $t_0 \in K$ such that the polynomial $P(t_0,X)$ is (totally) split over $K$.

Q1. Is there an algorithm to decide whether $T$ is empty, finite or infinite?

Q2. In the case $T$ is infinite, does there exist a non-constant rational function $f \in K(u)$ such that $t_0 \in T$ if and only if $t_0 = f(u_0)$ for some $u_0 \in K$?

Q3. In the case $T$ is infinite, is there an algorithm to find a non-constant rational function $f \in K(u)$ such that $P(f(u),X)$ splits over $K(u)$?

Here are some easy examples.

Example 1. Take $K=\mathbf{Q}$ and $P=X^2-t$. Obviously $T$ is the set of rational squares, and $f(u)=u^2$ satisfies the conditions in Q2 and Q3.

Example 2. Take $K$ arbitrary and $P=X^3-t$. If $t_0 \in K^\times$ and $P(t_0,X)$ splits over $K$, then $K$ must contain the cube roots of unity, and in this case the rational function $f(u)=u^3$ satisfies the conditions in Q2 and Q3.

My motivation for this question is that I'm trying to determine explicit equations for universal elliptic curves. For example, let $E_1(4)$ (resp. $E(4)$) be the universal elliptic curve over the modular curve $Y_1(4)$ (resp. $Y(4)$). A Weierstrass equation for $E_1(4)$ is given by $y^2+xy+ty=x^3+tx^2$ where $t$ is a generator of the function field of $Y_1(4)$ and the universal point of order 4 is given by $(0,0)$. Now determining $E(4)$ amounts to find those $t$ such that the $4$-division polynomial of $E_1(4)$ splits completely over $\mathbf{Q}(i)$.

So here is a concrete question: take $K=\mathbf{Q}(i)$ and $P(t,X)=X^4 + \frac12 X^3 + \frac32 tX^2 + 2t^2 X + t^3$. Can you find an explicit rational function $f \in K(u)$ such that $P(f(u),X)$ splits over $K(u)$?

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Your question 1 is open and is equivalent to the problem of determining the rank of an elliptic curve over a number field. (In most "practical cases" it should be solvable.) In particular, we don't have an algorithm to determine if an elliptic curve $E/K$ has positive rank, if $E : y^2 = x^{3} + Ax + B$, this is asking precisely for the values of $t$ for which $P(t,X) = X^{2} - t^{3} - At - B$ splits. In general, if you let $L$ be the Galois closure of $K(t)[X]/(p(t,X))$ over $K(t)$, then $L$ is the function field of a curve, and you are asking about the $K$-rational points of this curve. Faltings's theorem says that if the genus of this curve is $> 1$, then there are finitely many $K$-rational points. The genus $1$ cases are a bit messier, and the genus zero case should be fairly simple.

Given the explanation above, you can probably see that the answer to question 2 is no. In particular, because $E : y^{2} = x^{3} - 2x$ is an elliptic curve and has positive rank, there are infinitely many $t$ so that $P(t,X) = X^{2} - (t^{3} - 2t)$ splits over $\mathbb{Q}$. However, there is no $f(u)$ so that $P(f(u),X)$ splits over $\mathbb{Q}(u)$. This would imply that there is a rational curve on $E$, which would force $E$ to be rational (which it's not because it has genus $1$).

You might also like to know that you can find a model for the univeral elliptic curve over $X(4)$ here. For me, $Y(4)$ is the modular curve consisting of pairs $(E,i)$, where $i : E[4] \to \mathbb{Z}/4\mathbb{Z} \times \mu_{4}$ is a Galois-equivariant map. To get the $E(4)$ you want, take this curve and consider the base change to $\mathbb{Q}(i)$. [ In my paper with David Zureick-Brown here, we explain how to compute models of these universal elliptic curves. ]

In regard to your last question about finding $f(u)$ so that $P(f(u),X)$ splits, the relevant curve is simply $P(t,X) = 0$. This has genus zero with a non-singular rational point, and thus can be parametrized. This yields $$ f(u) = \frac{u^{3} - 3u^{2} + 4u - 2}{2u^{4}}. $$

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  • $\begingroup$ This is very interesting -- I didn't know this connection. So I guess the assumption in Q2 and Q3 is wrong, I should replace it with $L$ being of genus 0 with a rational point. Thanks a lot also for $E(4)$ and working out my example! $\endgroup$ – François Brunault Mar 15 '17 at 6:20

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