Let $R$ be a complete discrete valuation ring with field of fractions $K$ and algebraically closed residue field $k$, and let $X$ be a proper, smooth, geometrically connected curve over $K$. Take a finite extension $L/K$ and a regular proper model $\widetilde{X}$ of $X$ over the ring of integers of $L$ whose special fibre $\widetilde X_k$ is semi-stable. Let $e$ be the ramification index of $L/K$. The reduction graph of $X$ is a metrised graph obtained as follows. Take a set of intervals of lengh $1/e$ indexed by the singular points of $\widetilde X_k$. For every singular point $x$ of $\widetilde X_k$, label the endpoints of the corresponding edge by the two irreducible components (possibly the same) on which $x$ lies. For every irreducible component $C$ of $\widetilde X_k$, identify all the endpoints labelled $C$. The result (as a metric space) is independent of the choice of $L$.

Question: Suppose we have a regular proper model of $X$ over $R$ whose special fibre $X_k$ is reduced, but not necessarily semi-stable. Suppose furthermore that we know the irreducible components and singular points of $X_k$ and their intersection multiplicities in this model. What can be said about the reduction graph of $X$?

It seems reasonable to ask this question in this generality, but I am actually interested in the modular curves X1(n) over Wp[ζp2], where p is a prime number dividing n exactly twice and Wp is the ring of Witt vectors of an algebraic closure of Fp. In this situation non-semi-stable models as above were found by Katz and Mazur. What I would like specifically is an upper bound on the diameter of the reduction graph for these curves.

  • $\begingroup$ @PB: Interesting question; +1. For the special case of $X_1(n)$, since the special fiber has already been computed by Katz-Mazur, isn't the question an essentially graph-theoretic one rather than an arithmetic-geometric one? Maybe you should add graph-theory as a tag. $\endgroup$ Jun 14, 2010 at 15:03
  • $\begingroup$ @Pete L. Clark: The problem is that the model of Katz and Mazur is not semi-stable, and it is not clear (to me) how closely the special fibre in their model is related to that in a semi-stable model. After taking a suitable extension one has to do a sequence of normalisations and blow-ups to arrive at the regular semi-stable model; is there an a priori bound on the number of new components that this creates? $\endgroup$ Jun 14, 2010 at 15:15
  • $\begingroup$ Sorry, I didn't read your first paragraph carefully, so I missed the fact that by the "reduction graph" you mean the topological graph associated to a base change to semistable reduction. I retract my prvious comment: in general it is difficult to find an explicit description of the semistable model. Are you familiar with work of Coleman and McMurdy on $X_0(p^n)$? $\endgroup$ Jun 14, 2010 at 16:58
  • $\begingroup$ In full generalities, there is no hope that you can say something about the semi-stable reduction over L only in terms of data on $X_k$, especially when $L/K$ is wildly ramified. If you know there exists a tamely ramified $L/K$, then a possible method is to first find a regular proper model X' of X over R such that its special fiber is a normal crossing divisor. Then one can resolve by hand the singularities of $X'\times O_L$ and then get a semi-stable regular model over L after contracting some unnecessary irreducible components. You could have a look at B.Edixhoven's PhD thesis on $X_0(n)$. $\endgroup$
    – Qing Liu
    Jun 14, 2010 at 20:59
  • $\begingroup$ I have no hope of saying anything about the degree of $L/K$, but this does not (yet) preclude the possibility of there being some implication of the following form: if $X$ gets semi-stable reduction over some extension $L$ of $K$, then the number of components of the special fibre of the semi-stable model is at most $c[L:K]$ for some $c$ that can be computed a priori from a (not necessarily semi-stable) model à la Katz-Mazur. This would give an explicit bound on the diameter of the reduction graph. Or does wild ramification cause this to be a hopeless question, too? $\endgroup$ Jun 15, 2010 at 15:23

1 Answer 1


This is a sequel to the above comments.

Consider an elliptic curve $E$ over $K$ with additive reduction over $K$ and multiplicative reduction over some extension $L/K$. Then we can find a quadratic extension $L/K$, and the Kodaira symbole of $E$ over $K$ is $I^*_m$ for some $m$ (see below), and the group of components of $E_L$ is $Z/nZ$, where $n=[L:K]\nu_K(\Delta)$ and $\Delta$ is the minimal discriminant of $E$ over $K$.

Now how to compute $n$ from data over $K$ ? I mean from data that you can read from the special fiber of the minimal regular model over $K$ ? The link between $n$ and $m$ is $n/2=(2+\delta) + (m+4)-1$ (Ogg's formula), and $\delta$ is the Swan conductor of $E$. If the residue characteristic is different from $2$, then $\delta=0$ and $n=2(m+5)$. Perfect.

But if the residue characteristic is 2 (so wild ramification happens), then $\delta$ can be arbitrarily big (if the absolute ramification index of $K$ is big). So $n$ is not a function of data from the special fiber over $K$.

  • $\begingroup$ That mostly answers my question, thank you! I have one remaining question (since you say the Swan conductor can be arbitrarily big if the absolute ramification index of $K$ is big): suppose the residue characteristic is $p$ and the absolute ramification index of $K$ is $p-1$, then does one know an upper bound on the Swan conductor? $\endgroup$ Jun 16, 2010 at 11:32
  • $\begingroup$ Yes, search for Brumer and Kramer in Compositio. $\endgroup$
    – Qing Liu
    Jun 16, 2010 at 12:43
  • $\begingroup$ Other remark: for the minimal extension L/K realizing the semi-stable reduction of X, the degree [L:K] is bounded by a constant which depends only on the genus of X. $\endgroup$
    – Qing Liu
    Jun 17, 2010 at 22:54

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