Resolution of Singularities I

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1 Resolution of Singularities I The Blow-Up Operation Will Pazner February 12, 2010

2 1. Set up. k := field of characteristic 0 Defn. Algebraic variety k n X := zeros of f 1,..., f k Pol Defn. Multiplicity { of f at a is µ a (f ) := min d : d f x d 1 i1... x d k i k (a) 0 for d d k = d Defn. a Sing(V(f )) µ a (f ) > 1 e.g. Cone = V(f = x 2 y 2 z 2 ) Sing(Cone) = {0} µ 0 (f ) = 2 }

3 Σ For (x, y) X, (x, y) (0, 0) Slope y = y x 0 line[x : y] x-axis

4 2. Enticement. Take any f on manifold M (f is in some fixed category, e.g. polynomials, or analytic functions) Then proper morphism φ : M M (φ is a composition of quadratic maps) f φ is locally a monomial (up to invertible factor) on M Furthermore, even J(φ) (f φ) is locally a monomial

5 3. Blowing-up: key instrument. C M manifolds. U M coordinate chart. C = V(I), ideal I = (f 0,..., f m ) on U. [f ] : U\C x [f 0 (x) :... : f m (x)] P m [ξ] Bl I U := closure graph [f ] U P m σ U id closure graph [f ] = {(x, [ξ]) : ξ j f i (x) = ξ i f j (x)} i,j m So σ : Bl I U U where σ = π BlI U. π U

6 4. U P m = j V j V j := {ξ j 0}. U j := V j Bl I U. Say I := (x 1,..., x m ) where (x 1,..., x n ) are coordinates on U. So C = V(I) = {(x) : x 1 =... = x m = 0}. U j = V j {ξ jx i = ξ i x j } i,j m So coordinates on U j are: y j = x j y i = ξ i /ξ j y i = x i for 1 i m, i j otherwise e.g. on chart U j U j \U i = {y i = 0} and U j σ 1 (C) = {y j = 0} Furthermore σ U j : x j = y j x i = y j y i (1 i m, i j) x i = y i (otherwise)

7 5. Example: f := x n y m = 0, X = {f = 0} k 2 Assume gcd(n, m) = 1, n > m. I = (x, y), C = {0} = Sing(X) Fact: σ 1 (X\C) U x σ U x : (x 0, y 0 ) (x 0, x 0 y 0 ) f σ(x 0, y 0 ) = x n 0 (x 0y 0 ) m = x m 0 (xn m 0 y m 0 ) for (x 0, y 0 ) U x f := x m 0 (f σ) = x n m 0 y m 0, σ 1 (C) = {x 0 = 0}. Summarizing. σ 1 (X) U x = {x 0 = 0} {f = 0} Repeat. By Euclidean Algorithm: Finitely many blowups = X n has no singularities.

8 6. Effect of blowing up. M X := {x: f (x) = 0} a X d := µ a (f ) Linear coordinate change = a = 0 and d f (a) 0 x d n Why: f = α d c α x α = α =d c α x α 0, x ξ for some line ξ Make ξ into the x n -axis. x := (x 1,..., x n 1 ). Near a can write f (x) = c 0 ( x) + c 1 ( x)x n c d 1 ( x)x d 1 n + c d (x)x d n c d (x) 0. Im.F.T. = d 1 f (x) (x n h( x)) =: x n new coord. N := { x : d 1 f x d 1 n x d 1 n } (x) = 0, c i := 1 i! i f x i n, N i < d

9 7. Blowing-up with C = {x 1 =... = x m = x n = 0} Two types of charts: U n and all the others. On U n\ m j=1 U j = {y 1 =... = y m = 0} we show later f 0. In coordinate chart U j for 1 j m. Say j = 1. σ 1 (C) U 1 = {y 1 = 0}. c i := yi d 1 (c i σ) = c i = c i (ỹ), for i < d ỹ := (y 1,..., y n 1 ) f := y d 1 (f σ) = c 0(ỹ) c d 2 (ỹ)yd 2 n + c d (y)yd n c d (y) 0 y σ 1 (C) U 1 since σ n y n (y) = y 1 = 0 In particular µ y (f ) d.

10 8. On U n\ m j=1 U j = {y 1 =... = y m = 0} y f (y) := yn d (f σ)(y) = c 0 (y) + c 1 (y) c d (y), where c i := yi d n (c i σ) µ x (c i ) d i x C, i.e. c i = c iα (x m+1,..., x n 1 ) x α xα m m α d i Recall (x 1,..., x m ) = y n (y 1,..., y m ) = c d } (y) 0 c j (y) = 0 = µ y (f ) = 0 d Corollary. d = µ a (f ) = µ a (f ) µ a (f ) a σ 1 (a). In fact, f (y) 0 for y U n\ j n U j Conclusion: we don t need to consider U n chart.

11 9. Monomial assumption Assumption: (proved later using induction on dimension) Ω Q n 1 such that c i ( x) d!/(d i) = ( x Ω ) d! c i ( x) Where x Ω := x Ω x Ω n 1 n 1, d!ω i N with some ci ( x) 0 x. S (f,d) := {x: µ x (f ) = d} = {x: x n = 0 and µ x ( x Ω ) 1} = J Z J Z J := {x n = 0, x j = 0 j J}, for J s.th. j J Ω j 1 Suffices to consider J s.th. 0 ( j J Ω j ) 1 < Ω k k ( ) Choose C = Z J for any such J. Say J = {1,..., m}.

12 10. Multiplicity decreases. We will show Ω := n 1 k=1 Ω k decreases at points y σ 1 (a) with µ a (f ) = µ a (f ) = d. On U j say j = 1 Reminder: corollary = µ y (f ) d. Either multiplicity decreases (done) or multiplicities are equal. c i (ỹ)d!/(d i) = (y (i d)d!/(d i) 1 )(σ(ỹ) Ω ) d! (c i σ(ỹ)) = (y ( k J Ω k) 1 1 y Ω y Ω n 1 n 1 )d! (c i σ(ỹ)) Recall: f = c 0 (ỹ) c d 2 (ỹ)yd 2 n + c d (y)yd n with c d (y) 0 y

13 11. If µ a (f ) = d then y n = 0 Ω := ( m k=1 Ω k 1, Ω 2,..., Ω n 1 ) Then (ỹ Ω ) d! (c i )d!/(d i) i with some (ci σ)(ỹ) 0 ỹ i.e. f satisfies monomial assumption Ω < Ω by choice of C (see inequality ( )) = (µ a (f ), Ω ) < (µ a (f ), Ω ) (ordered lexicographically) Conclusion: multiplicity d decreases when n+1 k=1 Ω k < 1. After blowing up, Ω Ω 1/d! since d!ω i N = after at most d! Ω blowings-up, multiplicity decreases.

14 12. Induction on dim. = monomial assumption A f ( x) := d 2 i=0 c d!/(d i) i ( x) ( their differences) means include only nonzero factors Desing. in (n 1)-dim = φ = σ 1 σ l such that (A f φ)(ỹ) is locally monomial So, W.L.O.G. may assume { each factor } in the product is locally a monomial on N = d 1 f (x) = 0. x d 1 n Lemma (to come) = exponents are totally ordered = Ω such that ( x Ω ) d! (c i ) d!/(d i)

15 13. Lemma α, β, γ N n a(x), b(x), c(x) invertible a(x)x α b(x)x β = c(x)x γ = { either αi β i i or β i α i i Proof. If α k β k k done. Otherwise k s.t. β k < α k. Write α β k = (α 1,..., α k β k,..., α n ) a(x)x α β k b(x)x β β k = c(x)x γ β k. Evaluate at x k = 0 = c(x)x γ β k = b(x)x β β k 0 = β = γ = a(x)x α = (b(x) + c(x))x β. If j s.t. α j < β j, proceed similarly. Done

16 14. I cheated: used stronger induction than proved I ignored exceptional factors that accumulated before the last drop in multiplicity of f f := yexc(f d σ) (say old exceptional divisors). As a consequence, choices of centres might not have normal crossings with with old exceptional divisors as well. Say E old := {H k : H k old exc.} E old (a) := {H k E old : a H k } s(a) := #E old (a) {g k = 0} := H k = d g k 0

17 15. Choose x n such that both d f g (a) x d 0 and k (a) 0 H n x k E old (a) n Consider besides c j ( x) = j f, 0 j d 2 N all a k = g k N. Now, x j n A f ( x) := c d!/(d j) j ( x) a d! k ( x) ( their differences) Before: (µ a (f ), Ω ) < (µ a (f ), Ω ) for a σ 1 (a) Now: (µ a (f ), s(a ), Ω ) < (µ a (f ), s(a), Ω )

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