Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $K$ be a number field and let $X$ be a smooth projective geometrically connected curve over $K$.

There exists a finite field extension $L/K$ such that $X_L=X\otimes_K L$ has semi-stable reduction, i.e., there exists a semi-stable arithmetic surface $\mathcal{X}$ over the ring of integers $O_L$ with generic fibre $L$-isomorphic to $X_L$. Let $L_m$ be such an extension of minimal degree over $K$.

Question 1. Can we bound $[L_m:K]$ in terms of data depending only on $X$?

share|improve this question

2 Answers 2

up vote 6 down vote accepted

I think the answer to Question 1 is yes. One may use the fact that a curve has semistable reduction iff its Jacobian does and apply Grothendieck's theorem which says that an abelian variety has semistable reduction (over a local field) iff the representation of the intertia group on the Tate module is unipotent (it is always quasi-unipotent). One may ensure the unipotence by requiring the Jacobian variety have level $n$ structure for some $n> 2$; to do this one it suffices to make an extension of the base field of degree at most $|Sp(2g,\mathbb{Z}/n)|$, where $g$ is the dimension of the abelian variety, so the genus of the curve in your case.

(Note that a regular scheme over the ring of integers of a number field remains regular when it is base changed to the ring of integers of its completion at any finite prime, so semistability of a curve is not affected by completion.)

share|improve this answer
A reference for the fact that a curve has semi-stable reduction iff the Jacobian does is the book "Neron models" by Bosch, Raynaud and Lutkebohmert or see Theorem 2.4 in Deligne, Mumford: The irreducibility of the space of curves of a given genus. –  ulrich Jun 12 '11 at 14:02
Sorry to bring back such an old question, but do I understand correctly that one can take $n=3$ above? Thus, for any curve $X$ of genus $g$ over $K$, there exists a number field $L$ (depending on $X$ of course) whose degree is bounded by $\vert \mathrm{Sp}(2g, \mathbf{Z}/3)$ such that $X_L$ has semi-stable reduction over $L$? –  Ariyan Javanpeykar Oct 24 '11 at 17:57
As pointed out in the comments to Joe Silverman's answer, $3$-torsion is not enough since this does not guarantee semi-stable reduction at primes above $3$. Therefore one must take $n$ to be divisible by at least two primes, so $n=12$ or $n=15$ would do. Also, $Sp$ should be replaced with $GSp$ since the Weil pairing takes values in $\mu_n$ (so one also needs to have the $n$'th roots of unity in the field). –  ulrich Oct 25 '11 at 5:16

For an abelian variety $A/K$, Serre-Tate says that you get semistable reduction if you adjoin enough torsion to $K$. For example, adjoining all of the 3-torsion will suffice. It seems plausible (but I don't know for certain) that if the Jacobian $J$ of your curve $X$ has semistable reduction, then so does $X$. If that's the case, then you can take $L$ to be $K(J[3])$, whose degree is bounded by a function of $\dim(J)=\text{genus}(X)$.

share|improve this answer
Are you sure that 3-torsion is sufficient here? Once I read a paper by Zarhin and Silverberg; as far as I remember they stated that 3-torsion is not sufficient, though 5-torsion is ok (already). –  Mikhail Bondarko Jun 12 '11 at 21:06
The SZ paper: aif.cedram.org/item?id=AIF_1995__45_2_403_0 –  Junkie Jun 13 '11 at 1:07
Okay, good point, I forgot about reduction at primes dividing $n$. One has to be a bit more careful. For elliptic curves, I'm pretty sure that $3$-torsion is enough. Of course, for the original question, if one just wants a bound that depends on the genus of the curve $X$, one could adjoin, say, all of the 15-torsion, which would certainly suffice at all primes. –  Joe Silverman Jun 13 '11 at 3:38
For elliptic curves, Kraus explicitly computed the field. springerlink.com/content/bm6555l33861p521 Another SZ work is: sciencedirect.com/science/article/pii/S0022404998800261 –  Junkie Jun 13 '11 at 5:07
Actually, neither 3-torsion nor 5-torsion is sufficient by itself. E.g. E/Q: y^2+y=x^3 (27a3) has a 3-torsion point, so it acquires full 3-torsion over an extension of degree at most 3*2=6 (the image of Galois is inside $\begin{pmatrix}1&*\cr 0&*\end{pmatrix}$), but $v_3(\Delta)=3$ so it needs at least an extension of degree multiple of 4 (=12/3) to get good reduction at 3. Similarly E=50b1 has a 5-torsion point (|Galois| divides 20) and $v_5(\Delta)=2$ (need degree 6). As Joe says, 15-torsion is always enough though. I think Silverberg and Zarhin also deal with reduction "away from p". –  Tim Dokchitser Jun 13 '11 at 8:37

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.