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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 $X$?

edited Nov 14 2011 at 19:55

asked Jun 12 2011 at 11:18

Ariyan Javanpeykar
4,032●5●28

ulrich · Accepted Answer · 2011-06-12 13:50:15Z UTC

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.)

Joe Silverman · Answer 2 · 2011-06-12 13:47:44Z UTC

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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 $\dim(J)=\text{genus}(X)$.

answered Jun 12 2011 at 13:47

Joe Silverman
7,800●24●45

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 2011 at 21:06

The SZ paper: aif.cedram.org/item?id=AIF_1995__45_2_403_0 – Junkie Jun 13 2011 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 2011 at 3:38

For elliptic curves, Kraus explicitly computed the field. springerlink.com/content/bm6555l33861p521 Another SZ work is: sciencedirect.com/science/article/pii/… – Junkie Jun 13 2011 at 5:07

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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 2011 at 8:37

Can we bound the minimal degree of a field extension required to obtain semi-stable reduction

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