# Can we always find a curve which doesn't have semi-stable reduction

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.

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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.
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 $g$... – Shaye Oct 12 '11 at 22:42
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. – Qing Liu Oct 13 '11 at 7:13