Conductors and minimal discriminants of hyperelliptic curves Padmavathi Srinivasan MIT AMS Summer Institute in Algebraic Geometry, Salt Lake City July 31, 2015 Outline 1 Introduction 2 Conductors and minimal discriminants 3 Comparing conductors and discriminants What are conductors and minimal discriminants? Degenerating family of hyperelliptic curves Measures of degeneracy Artin conductor Minimal discriminant How are these related? y 2 = (x − 1)(x 2 − t) y2 = x3 − t Outline 1 Introduction 2 Conductors and minimal discriminants 3 Comparing conductors and discriminants Notation R : complete discrete valuation ring K : fraction field of R k : residue field of R, algebraically closed, char 6= 2 ν : discrete valuation K → Z ∪ {∞} t : a uniformizer of R, i.e., ν(t) = 1. Examples: C[[t]], Zunr p Notation f (x) : monic, squarefree, even degree ≥ 4 polynomial in R[x] X : smooth projective model of the plane curve y 2 = f (x) over K g: genus of X If f (x) = x d + ad−1 x d−1 + . . . + a0 factors as (x − α1 ) . . . (x − αd ) in K [x], then Y disc(f ) := (αi − αj )2 i<j ∈ K [a0 , . . . , ad−1 ] Minimal discriminant of X f (x) = x d + ad−1 x d−1 + . . . + a0 = (x − α1 ) . . . (x − αd ) The naive disciminant of y 2 = f (x) is defined to be Y ∆f := ν(disc(f )) = ν (αi − αj )2 . i<j The minimal discriminant ∆min of X is given by ! f (x) ∈ R[x] such that y 2 = f (x) ∆min = ∆min (X ) := min ∆f is birational to X over K ∆min = 0 ⇐⇒ X has good reduction. Artin conductor X : a hyperelliptic curve over K X : a proper, flat, regular R-scheme with XK ' X XK : geometric generic fiber of X Xk : special fiber of X Fix a prime ` 6= char k. For any curve C over an algebraically closed field of char 6= `, let P i (C , Q ) χ(C ) := 2i=0 (−1)i dim Hét ` δ : Swan conductor for the representation H 1 (XK , Q` ) (integer, ≥ 0, measure of wild ramification). Artin Conductor X : a hyperelliptic curve over K X : a proper, flat, regular R-scheme with XK ' X XK : geometric generic fiber of X Xk : special fiber of X χ : `-adic Euler Poincaré characteristic δ : Swan conductor for the representation H 1 (XK , Q` ) − Art(X /R) = χ(Xk ) − χ(XK ) + δ = (Saito) Deligne discriminant. Properties of the Artin Conductor Art(X /R) is independent of `. − Art(X /R) ≥ 0. − Art(X /R) = 0 ⇐⇒ X is smooth or g = 1 and (Xk )red is smooth. Let n be the number of components of Xk and let be the tame conductor. Then, − Art(X /R) = (n − 1) + + δ ≥ n − 1. When X is regular and semi-stable, − Art(X /R) = # singular points of Xk . Outline 1 Introduction 2 Conductors and minimal discriminants 3 Comparing conductors and discriminants Earlier results (Ogg, Saito, Liu) Let X be the minimal proper regular model of X . If g = 1, then − Art(X /R) = ∆min [Ogg-Saito formula]. This also holds when char k = 2. If g = 2, Liu showed that − Art(X /R) ≤ ∆min . He showed that equality does not always hold. Question: Does − Art(X /R) ≤ ∆min hold for hyperelliptic curves of arbitrary genus? Main result Theorem ( ) Let X be a hyperelliptic curve over K and let X be the minimal proper regular model of X . Assume that (a) the Weierstrass points of X are K -rational, and, (b) that the residue characteristic of K is not 2. Then, − Art(X /R) ≤ ∆min . Explicit construction of a regular model X 0 The first step in the proof is the explicit construction of a regular model X 0 for X (not necessarily minimal). Lemma Let Bl P1R be an arithmetic surface birational to P1R . Let f be an element of the function field of P1R . Assume that the odd multiplicity components of the divisor of f on Bl P1R are disjoint. p Then, the normalization of Bl P1R in K (x, f (x)) is a proper regular model for the hyperelliptic curve given by y 2 = f (x). Proof strategy 1. Explicitly construct a regular model X 0 (not necessarily minimal). P 2. X 0 is strict simple normal crossings, Xk0 = mi Γi and mi ∈ {1, 2}. X X X Γi .Γj . (1 − mi )χ(Γi ) + (mj − 1)Γi .Γj + − Art(X 0 /R) = i j6=i 3. Similarly decompose ∆min into local terms. 4. Prove a local comparison inequality. 5. Add local inequalities. − Art(X /R) ≤ − Art(X 0 /R) ≤ ∆min . i<j Finally . . . Thank you!