Preliminary Exam Stephen McKean (Duke) March 17 th , 2020 Overview - - PowerPoint PPT Presentation

Preliminary Exam

Stephen McKean (Duke) March 17th, 2020

Overview

• Study self-duality of complete intersections
• Scheja-Storch: construct explicit isomorphism Θ for self-duality
• Goal: describe Scheja-Storch’s isomorphism
• Context: A a commutative ring, B a finitely generated projective

A-algebra

1

Notation

• A a commutative ring, B a finitely generated projective A-algebra
• χ : B ⊗A B → HomA(HomA(B, A), B) defined by

χ(b ⊗ c) = (ϕ → ϕ(b)c)

• Presentation of B as a complete intersection over A:
• A[x] denotes either A[x1, ..., xn] or A[[x1, ..., xn]]
• ρ : A[x] → B with ker ρ = (t1, ..., tn)
• K¨

ahler differentials dtj = tj ⊗ 1 − 1 ⊗ tj = n

i=1 aij(xi ⊗ 1 − 1 ⊗ xi)

• µA[x] : A[x] ⊗A A[x] → A[x] given by b ⊗ c → bc
• ker µA[x] = (xi ⊗ 1 − 1 ⊗ xi : 1 ≤ i ≤ n)
• tj ⊗ 1 − 1 ⊗ tj ∈ ker µA[x]
• ∆ = ρ ⊗ ρ(det(aij)) ∈ B ⊗A B
• Remark: µB(∆) = ρ(det(∂tj/∂xi)) := J

2

Main Theorem

• Scheja-Storch: construct explicit isomorphism between B and B∗

Theorem The homomorphism Θ := χ(∆) : HomA(B, A) → B is a B-linear isomorphism.

• Proof uses the following lemmas:
• Lemma 1: χ induces a B-linear isomorphism

AnnB⊗AB(ker µB) → HomB(HomA(B, A), B).

• Lemma 2: (B ⊗A B) · ∆ = AnnB⊗AB(ker µB) and

ker µB = AnnB⊗AB((B ⊗A B) · ∆).

• Lemma 3: HomA(B, A) is a projective A-module.
• Lemma 4: HomA(B, A) is a free B-module.

3

Proof of Main Theorem

Theorem: Θ := χ(∆) : HomA(B, A) → B is a B-linear isomorphism.

• (B ⊗A B) · ∆ = ((B ⊗A B)/ AnnB⊗AB ∆) · ∆
• Lemma 2 =

⇒ AnnB⊗AB ∆ = ker µB

• (B ⊗A B)/ AnnB⊗AB ∆ = (B ⊗A B)/ ker µB
• First iso. theorem for modules =

⇒ (B ⊗A B)/ ker µB ∼ = B

• Thus (B ⊗A B) · ∆ ∼

= B · ∆

• Lemma 2 =

⇒ AnnB⊗AB ker µB = (B ⊗A B) · ∆ = B · ∆

• AnnB⊗AB ker µB is a free B-module with basis ∆
• Lemma 1 =

⇒ χ(AnnB⊗AB ker µB) is a free B-module with basis χ(∆)

• Lemma 1 =

⇒ HomB(HomA(B, A), B) is a free B-module with basis χ(∆)

• Lemmas 3,4 =

⇒ HomA(B, A) is a free B-module of rank 1

• χ(∆) : HomA(B, A) → B is B-linear homomorphism of rank 1 free

B-modules

• Any HomA(B, A) → B is a B-multiple of χ(∆), so χ(∆) is an

isomorphism

4

Proof of Lemma 1

Lemma 1: χ induces AnnB⊗AB ker µB ∼ = HomB(HomA(B, A), B).

• χ : B ⊗A B → HomA(HomA(B, A), B) is a bijection
• Look at two B-module structures compatible with χ
• On B ⊗A B, let a · (b ⊗ c) = ab ⊗ c and a ∗ (b ⊗ c) = b ⊗ ac
• ker µB = (a ⊗ 1 − 1 ⊗ a : a ∈ B)
• AnnB⊗AB ker µB = {x ∈ B ⊗A B : (a ⊗ 1 − 1 ⊗ a)x = 0 ∀a ∈ B}
• AnnB⊗AB ker µB = {x ∈ B ⊗A B : a · x = a ∗ x ∀a ∈ B}
• =

⇒ AnnB⊗AB ker µB is the largest submodule where · and ∗ agree

• On HomA(HomA(B, A), B), let a · ϕ = (ψ → ϕ(aψ)) and

a ∗ ϕ = (ψ → aϕ(ψ))

• a · ϕ = a ∗ ϕ for all a ∈ B iff ϕ is B-linear
• =

⇒ HomB(HomA(B, A), B) is the largest submodule where · and ∗ agree

• χ(a · (b ⊗ c)) = a · χ(b ⊗ c) and χ(a ∗ (b ⊗ c)) = a ∗ χ(b ⊗ c)

5

Proof of Lemma 2

Lemma 2: (B ⊗A B) · ∆ = AnnB⊗AB ker µB and ker µB = AnnB⊗AB((B ⊗A B) · ∆)

• Reduce to the following lemma:

Lemma 5: Let A be a commutative ring. Assume:

• Ideals (g1, ..., gn) = b ⊆ a = (f1, ..., fn)
• gj = n

i=1 aijfi and ∆ = det(aij)

• a′ = a/b in A′ = A/b and ∆′ image of ∆ in A′
• f1, ..., fn and g1, ..., gn are prime sequences in Ap for all p ⊇ a

Then: (a) ∆′ is independent of choice of aij (b) ∆′A′ = FittA′(a′) (c) ∆′A′ = AnnA′(a′) and a′ = AnnA′(∆′A′)

6

Proof of Lemma 2

Lemma 2: (B ⊗A B) · ∆ = AnnB⊗AB ker µB and ker µB = AnnB⊗AB((B ⊗A B) · ∆) A[x] ⊗A A[x] A[x] B ⊗A A[x] B ⊗A B B

µA[x] ρ⊗id ρ id ⊗ρ µB

• For A[x] = A[[x1, ..., xn]], consider id ⊗ρ : B ⊗A A[x] → B ⊗A B
• ker(id ⊗ρ) = (−1 ⊗ t1, ..., −1 ⊗ tn) plays the role of b
• (ρ(x1) ⊗ 1 − 1 ⊗ x1, ..., ρ(xn) ⊗ 1 − 1 ⊗ xn) plays the role of a
• For A[x] = A[x1, ..., xn], reduce to the local case in two steps:
• Locally in B, i.e. over Bn ⊗A Bn
• Locally in B ⊗A B, i.e. over (B ⊗A B)N
• Clear denominators to go from locally in B to locally in B ⊗A B

7

Proof of Lemmas 3 and 4

Lemma 3: HomA(B, A) is a projective A-module.

• B finitely generated =

⇒ π : Am ։ B

• 0 → ker π → Am → B → 0 and B projective =

⇒ B ⊕ ker π ∼ = Am

• Am ∼

= HomA(Am, A) ∼ = HomA(B ⊕ ker π, A) ∼ = HomA(B, A) ⊕ HomA(ker π, A)

Lemma 4: HomA(B, A) is a free B-module.

• If M finitely generated B-module, M projective A-module, and

HomB(M, B) projective (free) B-module, then M is a projective (free) B-module.

8

Proof of Lemma 5

Lemma 5: Let A be a commutative ring. Assume:

• Ideals (g1, ..., gn) = b ⊆ a = (f1, ..., fn)
• gj = n

i=1 aijfi and ∆ = det(aij)

• a′ = a/b in A/b and ∆′ image of ∆ in A′
• f1, ..., fn and g1, ..., gn are prime sequences in Ap for all p ⊇ a

Then: (a) ∆′ is independent of choice of aij (b) ∆′A′ = FittA′(a′) (c) ∆′A′ = AnnA′(a′) and a′ = AnnA′(∆′A′)

• A prime sequence is a regular sequence f1, ..., fn such that (f1, ..., fi)

is a prime ideal for each 1 ≤ i ≤ n

• The Fitting ideal FittA′(a′) is generated by the determinants of all

n × n minors det(bij) of the collection of syzygies n

i=1 bijfi = 0,

where f1, ..., fn generate a′ over A′

9

Proof of Lemma 5

Lemma 5: Let A be a commutative ring. Assume:

• Ideals (g1, ..., gn) = b ⊆ a = (f1, ..., fn)
• gj = n

i=1 aijfi and ∆ = det(aij)

• a′ = a/b in A/b and ∆′ image of ∆ in A′
• f1, ..., fn and g1, ..., gn are prime sequences in Ap for all p ⊇ a

Then: (a) ∆′ is independent of choice of aij (b) ∆′A′ = FittA′(a′) (c) ∆′A′ = AnnA′(a′) and a′ = AnnA′(∆′A′)

• (a), (b), (c) are local in A′
• If a′ = A′, then a′ is one-dimensional over A′
• Any n × n minor contains linearly dependent rows, so ∆′ = 0
• May assume A′ is local, {fi} and {gi} regular sequences contained in

unique max ideal

10

Proof of Lemma 5

(a) ∆′ is independent of choice of aij.

• Suppose gj = n

i=1 aijfi = n i=1 bijfi

• Need to show det(aij) − det(bij) ∈ b
• WLOG, assume first n − 1 rows of (aij) and (bij) agree
• Set cij = aij = bij for 1 ≤ j ≤ n − 1 and cin = ain − bin
• det(cij) = det(aij) − det(bij)
• Cramer’s rule =

⇒ det(cij) · fi = det(replace ith column of (cij) with (g1, ..., gn−1, 0)T)

• =

⇒ det(cij) · a ⊆ (g1, ..., gn−1)

• g1, ..., gn prime sequence =

⇒ (g1, ..., gn−1) a prime ideal

• g1, ..., gn a regular sequence in a =

⇒ a ⊆ (g1, ..., gn−1)

• =

⇒ det(cij) ∈ (g1, ..., gn−1) ⊆ b

11

Proof of Lemma 5

(b) ∆′A′ = FittA′(a′).

• FittA′(a′) is the image of FittA(a) in A′
• a′ ∼

= An/U

• U is generated by (a1j, ..., anj) and (0, ..., fq, ..., −fp, ...0)
• One of these minors is ∆
• Using Cramer’s rule as in (a), all other minors are in b
• Fitting ideal does not depend on choice of generators or relations

12

Proof of Lemma 5

(c) ∆′A′ = AnnA′(a′) and a′ = AnnA′(∆′A′).

• Induct on n
• n = 0:
• ∆′ = det(∅) =

∅ λ = 1 and a′ = 0

• Trivially, AnnA′(0) = A′ and 0 = AnnA′(A′)
• n ≥ 1:
• (b) =

⇒ ∆′A′ is a Fitting ideal

• =

⇒ ∆′A′ does not depend on choice of f1, ..., fn

• Pick f1, ..., fn so that f1 is prime to (g2, ..., gn)
• Set B = A/(g2, ..., gn), bij = aij for j ≥ 2, b11 = 1 and bi1 = 0 for

i ≥ 2, ∆b = det(bij)

• Inductive hypothesis on f2, ..., fn and g2, ..., gn in A/f1A =

⇒ :

(i) (B/f1B) · ∆b = AnnB/f1B(aB/f1B) = (f1B : aB)/f1B (ii) aB/f1B = AnnB/f1B((B/f1B) · ∆b) = (f1B : ∆bB)/f1B

• (ii) =

⇒ aB = (f1B : ∆bB)

• Cramer’s rule =

⇒ ∆f1, ∆bg1 ∈ b = ⇒ ∆bg1 ≡ ∆f1 mod b

13

Proof of Lemma 5

(c) ∆′A′ = AnnA′(a′) and a′ = AnnA′(∆′A′).

• Inductive hypothesis on f2, ..., fn and g2, ..., gn in A/f1A =

⇒ :

(i) (B/f1B) · ∆b = AnnB/f1B(aB/f1B) = (f1B : aB)/f1B (ii) aB/f1B = AnnB/f1B((B/f1B) · ∆b) = (f1B : ∆bB)/f1B

• (ii) =

⇒ aB = (f1B : ∆bB)

• Cramer’s rule =

⇒ ∆f1, ∆bg1 ∈ b = ⇒ ∆bg1 ≡ ∆f1 mod b

• =

⇒ (g1B : ∆B) = (f1B : ∆bB) = ⇒ (g1B : ∆B) = aB

• =

⇒ aB/g1B = (g1B : ∆B)/g1B = AnnB/g1B((B/g1B)∆)

• Canonical isomorphism (f1B : aB)/f1B ∼

= (g1B : aB)/g1B sends image of ∆b in B/f1B to image of ∆ in B/g1B

• (i) =

⇒ (B/g1B)∆ = (g1B : aB)/g1B = AnnB/g1B(aB/g1B)

• Conclude by noting A′ = B/g1B, a′ = aB/g1B, ∆′A′ = ∆(B/g1B)

14

Corollaries

• Main Theorem =

⇒ η := Θ−1(1) generates HomA(B, A)

• Θ(TrB/A) = J = µ(∆) and TrB/A = J · η
• Corollary: Let A be reduced with

(x1, ..., xn)m ⊆ (t1, ..., tn) ⊆ (x1, ..., xn) for some m.

• Write tj = n

i=1 eijxi and set E ≡ det(eij) mod (t1, ..., tn). Then:

• J = (dimA B) · E
• Get bilinear form βη : B × B → A defined by βη(x, y) = η(xy)
• Eisenbud-Levine: if φ : B → A has φ(E) = 1, then βφ ∼

= βη

• Kass-Wickelgren: βη is the local A1-degree

15

Questions?

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