MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $ X \to \mathbb{P}^1 $ is an elliptic surface with section, with Weierstrass model defined over $ \mathbb{Q} $. If $ \sigma: \mathbb{P}^1 \to X $ is a torsion section with order $n$, then for generic $ t \in \mathbb{P}^1 $, $ \sigma(t) $ is a point of order $n$ in the fiber $ X_t $. Now it seems to me that there will be many rational $t$ that are generic, and hence by Mazur's theorem, $ n \leq 12 $. Now for my question(s):

1) Is this correct?

2) Will $ \sigma(t) $ have order $n$ for every $t$ such that $ X_t $ is smooth, or is it possible that $ \sigma(t) $ has order $ m | n $?

I'm hoping that the answer to (2) is yes, since I can quickly compute the torsion of a particular elliptic curve over $ \mathbb{Q} $.

share|cite|improve this question
up vote 4 down vote accepted

1): Suppose we work over an arbitrary field $K$ and suppose $\pi: X\to \mathbb{P}^1$ is an elliptic fibration with a torsion section of order $n$. Then you obtain a natural classification map $\mathbb{P}^1 \to X_1(n)$.

If this map is dominant then $g(X_1(n))=0$ and hence $n\leq 12$. (You do not need Mazur's theorem at all for this. The genus of $X_1(n)$ was known before Mazur's theorem.)

If the above classifying map is not dominant then $j(\pi^{-1}(t))$ is independent of $t$. The worst case here is that $X$ is a product $E\times \mathbb{P}^1$. In this case the torsion group of the fibration is just the torsion group of $E$.

If $X$ is (not birational to) a product, but has constant $j$-invariant then you can use the structure of the singular fibers to limit the possibilities of the torsion group. This is described in the final chapter of Miranda's book on elliptic surfaces.

If you are only interested in sections that are defined over $\mathbb{Q}$ then things are easier: there exists infinitely many $t$ such that the specialization maps is injective. Now applying Mazur's theorem shows that the order of the section is at most 12.

2): Consider the divisor $[n]^{-1}(\sigma_0)$ the inverse inverse image of the zero section under multiplication by $n$. (I.e., in each smooth fiber you take the union of all the points of order dividing $n$, and then you take the closure in X.) Now restrict the fibration $\pi$ to X[n] then you obtain a degree $n^2$ cover of $\mathbb{P}^1$ (which might have several irreducible components). Since each smooth fiber has (geometrically) $n^2$ points of order dividing $n$, we obtain that this covering is unramified over every points in $\mathbb{P}^1$ that does not belong to the discriminant of the fibration.

If you would have a fiber where the order of the specialization of the section is strictly smaller then the order of section then this would yield a ramification point of $\pi|_{X[n]}:X[n]\to \mathbb{P}^1$. Hence this cannot happen in smooth fibers.

share|cite|improve this answer
Perfect, thanks so much. In practice for the surfaces I'm considering, the types of singular fibers present give easy and small bounds for the size of the torsion subgroup, but it's good to know that even if this weren't the case, I would still be in good shape. – user4192 Apr 12 '11 at 18:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.