Example (cont d): Function fields in one variable...

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1 Example (cont d): Function fields in one variable... Garrett Practice: consider K a finite extension of k = C(X), and O the integral closure in K of o = C[X]. K = C(X, Y ) for some Y, and can renormalize so Y C[X, Y ] O. O, so For example, for hyperelliptic curves Y 2 = P (X) with P (X) C[X] square-free, have O = C[X, Y ] exactly. Puiseux expansions and field extensions of C((X z)). Introduction to Newton polygons.

2 Garrett Hensel s lemma: With monic F (T ) C[[X]][T ], given α 1 C[[X z]] with F (α 1 ) = 0 mod X z, F (α 1 ) 0 mod X z, the recursion α n+1 = α n F (α n) F (α n ) mod (X z)n+1 gives α = lim n α n C[[X z]] with F (α ) = 0 in C[[X z]], and α is the unique solution congruent to α 1 mod X z. Theorem: All finite field extensions of C((X z)) are by adjoining solutions to Y e = X z for e = 2, 3, 4,.... [Proof below.] This approximates the assertion that, locally, Riemann surfaces are combinations of coverings of the z-plane and power maps w e = z. The proof invites extending Hensel s lemma to cover factorization.

3 Garrett Hensel s Lemma II: Let R be a UFD, and π a prime element in R. Given a R, suppose b 1, c 1 R such that a = b 1 c 1 mod π and Rb 1 + Rc 1 + Rπ = R Then there are b, c in the π-adic completion R π = lim n R/π n such that b = b 1 mod π, c = c 1 mod π, and a = b c (in lim n R/π n = R π ) Remark: Apply this to R = C[[X z]][t ] or R = C[X, T ] and π = X z to talk about field extensions of C((X z)).

4 Garrett Proof: With a = b 1 c 1 mod π, try to adjust b 1, c 1 by multiples of π to make the equation hold mod π 2 : require a = ( b 1 + xπ ) (c 1 + yπ ) mod π 2 Simplify: the π 2 term π 2 xy disappears, and a b 1 c 1 π = xc 1 + yb 1 mod π By hypothesis, expressions xc 1 + yb 1 + zπ with x, y, z R give R, so there exist (non-unique!) x, y to make the equation hold.

5 Garrett Thus, the genuine induction step involves a = b n c n mod π n, and trying to solve for x, y in which gives a = ( b n + xπ n) (c n + yπ n) mod π n+1 a b n c n π n = xc n + yb n mod π Inductively, c n = c 1 mod π and b n = b 1 mod π, so Rb n + Rc n + Rπ = Rc 1 + Rb 1 + Rπ = R and there are x, y satisfying the condition. Induction succeeds. ///

6 Garrett Caution: By Gauss lemma, polynomial rings o[x] over UFDs o are UFDs, but what about o[[x]]? We don t really need the more general case, since we only care about C[[X]] = lim n C[X]/X n, which is completely analogous to Z p, where we recall that the ideals in Z p are just p l Z p. Many fewer than in Z, and all coming from Z. Thus, C[[X]] is a PID, with a unique non-zero prime ideal X C[[X]], and all ideals are of the form X n C[[X]]. Even though o[[x]] is much bigger than o[x], it has many more units, for example. At the same time, UFDs like Z[x, y] are not PIDs, so we have to be careful what we imagine... Maybe proving Z[[X]] and C[[X]][T ] are UFDs is a good exercise.

7 Corollary: (Now z = 0 and X z = X.) Consider Garrett f(x, T ) = T n + a n 1 (X)T n a 1 (X)T + a o (X) with a j (X) C[[X]] and such that the equation f(0, Y ) = (Y w 1 ) ν 1 (Y w 2 ) ν 2... (Y w m ) ν m with w i w j for i j. Then f(x, T ) factors in C[[X]][T ] into m monic-in-t factors, of degrees ν j in T : T n + a n 1 (X)T n a o (X) = f 1 (X, T )... f m (X, T ) with f j (0, T ) = (T w j ) ν j That is, f j (X, T ) = (T w j ) ν j mod X

8 Proof: In Hensel II, take R = C[[X]][T ], π = X, and Garrett b 1 = (T w 1 ) ν 1 c 1 = (T w 2 ) ν 2... (T w m ) ν m An equality of polynomials g(x) = h(x) mod X is equality of complex numbers g(0) = h(0). Since w 1 is distinct from w 2,..., w m, there are r 1, r 2 in the PID C[T ] such that r 1 b 1 + r 2 c 1 = 1, so certainly Rb 1 + Rc 1 + Rπ = R. By Hensel II, f(x, T ) = g(x, T ) h(t, X) (in C[[X]][T ]) and g(x, T ) = (T w 1 ) ν 1 mod X h(x, T ) = (T w 2 ) ν 2... (T w m ) ν m mod X Since 1 + c 1 X +... C[[X]], we can make g, h monic in T. Induction on m. ///

9 Garrett Corollary: Unless f(0, w) = 0 has just a single (distinct) root in C, f(x, T ) has a proper factor in C[[X]][T ]. /// That is, over scalars C[[X]], the irreducible factors of f(x, T ) are (factors of) the groupings-by-distinct-factors mod X. Now consider w 1 = 0, and f(x, T ) = T n mod X. That is, f(x, T ) is of the form f(x, T ) = T n + X a n 1 (X) T n X a o (X) In the simplest case a o (0) 0, Eisenstein s criterion in C[[X]][T ] gives irreducibility of f(x, T ). Let s consider this case.

10 Garrett Extend C[[X]] by adjoining X 1/n. Replacing T by X 1/n T, the polynomial becomes X T n n X n an 1 (X) T n X 1+ 1 n a1 (X) T + Xa o (X) Taking out the common factor of X gives T n + (X 1/n ) n 1 a n 1 (X) T n X 1/n a 1 (X) T + a o (X) Mod X 1/n, this is T n a o (0) = T n + a o (0) mod X 1/n For a o (0) 0, w n + a o (0) = 0 has distinct linear factors in C. By the Hensel paraphrase, f(x, X 1/n T ) factors into linear factors in C[[X 1/n ]][T ]. We re done in this case: the field extension is C((X))(Y ) = C((X 1/n ))

11 Garrett Example: To warm up to Newton polygons and the general case, consider (T X 1/3 ) 3 (T X 1/2 ) 2. Write ord(x a/b ) = a/b. The symmetric functions of roots have ords ord σ 1 = ord(3x 1/3 + 2X 1/2 ) = 1 3 ord σ 2 = ord(3x X X ) = 2 3 ord σ 3 = ord(x X X ) = 1 ord σ 4 = ord(2x X ) = 3 2 ord σ 5 = ord(x ) = 2 That is, the increments in ord σ l are 1 3, 1 3, 1 3, 1 2, 1 2.

12 Garrett Variant: Varying the example, take f(x, T ) = (T z 1 X 1 3 )(T z2 X 1 3 (T z3 X 1 3 (T z4 X 1 2 )(T z5 X 1 2 ) with non-zero z i C. Now we mostly have inequalities for ords: ord σ 1 = ord((z 1 + z 2 + z 3 )X 1/3 + (z 4 + z 5 )X 1/2 ) 1 3 ord σ 2 = ord((z 1 z )X (...)X z 4 z 5 X ) 2 3 ord σ 3 = ord(z 1 z 2 z 3 X (...)X X ) = 1 ord σ 4 = ord(z 1 z 2 z 3 (z 4 + z 5 )X (...)X ) 3 2 ord σ 5 = ord(z 1 z 2 z 3 z 4 z 5 X ) = 2

13 Garrett A stark example of the latter is f(x, T ) = T 5 XT 2 + X 2 The crucial mechanism is that the smallest ord is 1/3, and replacing T by X 1/3 T will distinguish the two sizes of roots: f(x, X 1/3 T ) = X 5 3 T 5 X 5 3 T 2 + X 2 Dividing through by X 5/3 gives T 5 T 2 + X 1 3 Mod X 1 3, this has 3 non-zero factors, and 2 zero factors, so by Hensel II factors properly into cubic and quadratic.

14 Garrett More generally, consider f(x, T ) = (T X 1/e 1 ) ν 1... (T X 1/e m ) ν m (with 1 e e m ) By the ultrametric inequality, ord(σ l ) ord ( sum of ords of the l smallest-ord zeros ) l 1 e 1 for 1 l ν 1 ν 1 e 1 + (l ν 1 ) 1 e 2 for ν 1 l ν 1 + ν 2 ν 1 e 1 + ν 2 e 2 + (l ν 1 ν 2 ) 1 e 3 for ν 1 + ν 2 l ν 1 + ν 2 + ν with equality at l = 0, ν 1, ν 1 + ν 2,..., ν ν m.

15 Garrett Since 1 e e m, the convex hull (downward) of the points (l, ord σ l ) has boundary the polygon of lines connecting the points in R 2 (0, 0) (ν 1, ν 1 e 1 ) = (ν 1, ord σ ν1 ) (ν 1 + ν 2, ν 1 e 1 + ν 2 e 2 ) = (ν 1 + ν 2, ord σ ν1 +ν 2 )... (ν ν m, ν 1 e ν m em ) = (ν ν m, ord σ ν1 +...ν m ) This convex hull is the Newton polygon of the polynomial. For f(x, T ) C[[X]][T ], the ords are in Z. Eisenstein s criterion is the case ν 1 = n, and ord σ n = 1, and all the exponents are 1/n.

16 Garrett The general case was reduced to f(x, T ) = T n a o (X) with an n-fold multiple zero w o at X = 0. Replacing T by T + w o, without loss of generality, this root is 0, so a j (0) = 0 for all j. Replace T by X ρ T with ρ the slope of the first segment from (0, 0) to (l, ord σ l ) on the Newton polygon. That is, disregard any (l, ord σ l ) with l < l lying above that segment. Replacing T by X ρ T and dividing through by X nρ gives T n a n l(x) X lρ T n l +... The Newton polygon says the ord of the coefficient of T j for n j > n l is non-negative, at T n l the ord is 0, and for n l > j it is strictly positive.

17 Garrett That is, mod X, f(0, T ) = T n b n l (0) T n l }{{} non zero Thus, f(0, w) = 0 has l non-zero complex roots, and n l roots 0. Hensel II says that there are degree l factor and degree n l factors in C[[X ρ ]][T ]. Note that C[[X ρ ]] C[[X]], so the argument can be repeated. Induction on degree. ///