9
$\begingroup$

Let $K$ be a number field and let $g$ be a positive integer. Does there exist a smooth projective geometrically connected curve $X/K$ of genus $g$ such that $X$ does not have semi-stable reduction over $K$?

I can write down curves over certain number fields without semi-stable reduction, but I can't do it for a general number field.

$\endgroup$

1 Answer 1

15
$\begingroup$

Yes. Take $f\in O_K$ a uniformizing element of some prime $\mathfrak p$. Consider the hyperelliptic curve defined by the equation $$y^2=x^{2g+1}+f.$$ Then this curve doesn't have semi-stable reduction at $\mathfrak p$. In fact, this equation defines a proper regular model of the curve over the localization $O_{K, \mathfrak p}$ and this model is minimal because its closed fiber is irreducible (defined by $y^2=x^{2g+1}$), and it is not semi-stable. If the curve had semi-stable reduction, the minimal regular model would also be semi-stable. Further, this curve is far from being semi-stable because its Jacobian has purely additive reduction at $\mathfrak p$.

Add a more elementary explanation on why the curve doesn't have semi-stable reduction at $\mathfrak p$. Suppose for simplicity that $\mathfrak p$ is prime to $2(2g+1)$. As saw in the comments, the curve has potentially good reduction above $\mathfrak p$. So if it had already semi-stable reduction over $K$, then it would already have good reduction over $K$. This implies (as for elliptic curves) that after a suitable homographic transformation on $x$, we will get a new polynomial with discriminant invertible in $O_{K, \mathfrak p}$. But such transformation changes the valuation of the discriminant by a multiple of $2(2g+1)$ while the initial discrimiant has valuation $2g$. Impossible.

$\endgroup$
2
  • $\begingroup$ Perfect. Can one also write down a number field $L/K$ such that $X_L$ has semi-stable reduction over $L$? I'm guessing $L$ should contain some $n$-th root of $f$, where $n$ depends only on $g$... $\endgroup$
    – Shaye
    Oct 12, 2011 at 22:42
  • 4
    $\begingroup$ If $\pi$ is a $2(2g+1)$-th root of $f$, then you have $(y/\pi^{2g+1})^2=(x/\pi^2)^{2g+1}+1$ which defines a curves having good reduction everywhere except at places dividing $2(2g+1)$. At the small primes $\mathfrak p$, the problem is more complicated. If $\mathfrak p$ is prime to $2$, a quadratic extension of the decomposition field of $X^{2g+1}+1$ will realize the semi-stable reduction. If $\mathfrak p$ divides $2$, change variables to $z=2+2(y/\pi^{2g+1})$ and $w=(x/\pi^2)/4^{1/(2g+1)}$. Then $z^2+z=w^{2g+1}$ which is an equation of good reduction. $\endgroup$
    – Qing Liu
    Oct 13, 2011 at 7:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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