Preliminary Exam Stephen McKean (Duke) March 17 th , 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 → Hom A (Hom A ( 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 [ x 1 , ..., x n ] or A [[ x 1 , ..., x n ]] • ρ : A [ x ] → B with ker ρ = ( t 1 , ..., t n ) ahler differentials dt j = t j ⊗ 1 − 1 ⊗ t j = � n • K¨ i =1 a ij ( x i ⊗ 1 − 1 ⊗ x i ) • µ A [ x ] : A [ x ] ⊗ A A [ x ] → A [ x ] given by b ⊗ c �→ bc • ker µ A [ x ] = ( x i ⊗ 1 − 1 ⊗ x i : 1 ≤ i ≤ n ) • t j ⊗ 1 − 1 ⊗ t j ∈ ker µ A [ x ] • ∆ = ρ ⊗ ρ (det( a ij )) ∈ B ⊗ A B • Remark: µ B (∆) = ρ (det( ∂ t j /∂ x i )) := J 2
Main Theorem • Scheja-Storch: construct explicit isomorphism between B and B ∗ Theorem The homomorphism Θ := χ (∆) : Hom A ( B , A ) → B is a B -linear isomorphism. • Proof uses the following lemmas: • Lemma 1: χ induces a B -linear isomorphism Ann B ⊗ A B (ker µ B ) → Hom B (Hom A ( B , A ) , B ). • Lemma 2: ( B ⊗ A B ) · ∆ = Ann B ⊗ A B (ker µ B ) and ker µ B = Ann B ⊗ A B (( B ⊗ A B ) · ∆). • Lemma 3: Hom A ( B , A ) is a projective A -module. • Lemma 4: Hom A ( B , A ) is a free B -module. 3
Proof of Main Theorem Theorem: Θ := χ (∆) : Hom A ( B , A ) → B is a B -linear isomorphism. • ( B ⊗ A B ) · ∆ = (( B ⊗ A B ) / Ann B ⊗ A B ∆) · ∆ • Lemma 2 = ⇒ Ann B ⊗ A B ∆ = ker µ B • ( B ⊗ A B ) / Ann B ⊗ A B ∆ = ( B ⊗ A B ) / ker µ B ⇒ ( B ⊗ A B ) / ker µ B ∼ • First iso. theorem for modules = = B • Thus ( B ⊗ A B ) · ∆ ∼ = B · ∆ • Lemma 2 = ⇒ Ann B ⊗ A B ker µ B = ( B ⊗ A B ) · ∆ = B · ∆ • Ann B ⊗ A B ker µ B is a free B -module with basis ∆ • Lemma 1 = ⇒ χ (Ann B ⊗ A B ker µ B ) is a free B -module with basis χ (∆) • Lemma 1 = ⇒ Hom B (Hom A ( B , A ) , B ) is a free B -module with basis χ (∆) • Lemmas 3,4 = ⇒ Hom A ( B , A ) is a free B -module of rank 1 • χ (∆) : Hom A ( B , A ) → B is B -linear homomorphism of rank 1 free B -modules • Any Hom A ( B , A ) → B is a B -multiple of χ (∆), so χ (∆) is an isomorphism 4
Proof of Lemma 1 Lemma 1: χ induces Ann B ⊗ A B ker µ B ∼ = Hom B (Hom A ( B , A ) , B ). • χ : B ⊗ A B → Hom A (Hom A ( 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 µ B = ( a ⊗ 1 − 1 ⊗ a : a ∈ B ) • Ann B ⊗ A B ker µ B = { x ∈ B ⊗ A B : ( a ⊗ 1 − 1 ⊗ a ) x = 0 ∀ a ∈ B } • Ann B ⊗ A B ker µ B = { x ∈ B ⊗ A B : a · x = a ∗ x ∀ a ∈ B } • = ⇒ Ann B ⊗ A B ker µ B is the largest submodule where · and ∗ agree • On Hom A (Hom A ( B , A ) , B ), let a · ϕ = ( ψ �→ ϕ ( a ψ )) and a ∗ ϕ = ( ψ �→ a ϕ ( ψ )) • a · ϕ = a ∗ ϕ for all a ∈ B iff ϕ is B -linear • ⇒ Hom B (Hom A ( 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 ) · ∆ = Ann B ⊗ A B ker µ B and ker µ B = Ann B ⊗ A B (( B ⊗ A B ) · ∆) • Reduce to the following lemma: Lemma 5: Let A be a commutative ring. Assume: • Ideals ( g 1 , ..., g n ) = b ⊆ a = ( f 1 , ..., f n ) • g j = � n i =1 a ij f i and ∆ = det( a ij ) • a ′ = a / b in A ′ = A / b and ∆ ′ image of ∆ in A ′ • f 1 , ..., f n and g 1 , ..., g n are prime sequences in A p for all p ⊇ a Then: (a) ∆ ′ is independent of choice of a ij (b) ∆ ′ A ′ = Fitt A ′ ( a ′ ) (c) ∆ ′ A ′ = Ann A ′ ( a ′ ) and a ′ = Ann A ′ (∆ ′ A ′ ) 6
Proof of Lemma 2 Lemma 2: ( B ⊗ A B ) · ∆ = Ann B ⊗ A B ker µ B and ker µ B = Ann B ⊗ A B (( B ⊗ A B ) · ∆) µ A [ x ] A [ x ] ⊗ A A [ x ] A [ x ] ρ ⊗ id B ⊗ A A [ x ] ρ id ⊗ ρ µ B B ⊗ A B B • For A [ x ] = A [[ x 1 , ..., x n ]], consider id ⊗ ρ : B ⊗ A A [ x ] → B ⊗ A B • ker(id ⊗ ρ ) = ( − 1 ⊗ t 1 , ..., − 1 ⊗ t n ) plays the role of b • ( ρ ( x 1 ) ⊗ 1 − 1 ⊗ x 1 , ..., ρ ( x n ) ⊗ 1 − 1 ⊗ x n ) plays the role of a • For A [ x ] = A [ x 1 , ..., x n ], reduce to the local case in two steps: • Locally in B , i.e. over B n ⊗ A B n • 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: Hom A ( B , A ) is a projective A -module. ⇒ π : A m ։ B • B finitely generated = • 0 → ker π → A m → B → 0 and B projective = ⇒ B ⊕ ker π ∼ = A m • A m ∼ = Hom A ( A m , A ) ∼ = Hom A ( B ⊕ ker π, A ) ∼ = Hom A ( B , A ) ⊕ Hom A (ker π, A ) Lemma 4: Hom A ( B , A ) is a free B -module. • If M finitely generated B -module, M projective A -module, and Hom B ( 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 ( g 1 , ..., g n ) = b ⊆ a = ( f 1 , ..., f n ) • g j = � n i =1 a ij f i and ∆ = det( a ij ) • a ′ = a / b in A / b and ∆ ′ image of ∆ in A ′ • f 1 , ..., f n and g 1 , ..., g n are prime sequences in A p for all p ⊇ a Then: (a) ∆ ′ is independent of choice of a ij (b) ∆ ′ A ′ = Fitt A ′ ( a ′ ) (c) ∆ ′ A ′ = Ann A ′ ( a ′ ) and a ′ = Ann A ′ (∆ ′ A ′ ) • A prime sequence is a regular sequence f 1 , ..., f n such that ( f 1 , ..., f i ) is a prime ideal for each 1 ≤ i ≤ n • The Fitting ideal Fitt A ′ ( a ′ ) is generated by the determinants of all n × n minors det( b ij ) of the collection of syzygies � n i =1 b ij f i = 0, where f 1 , ..., f n generate a ′ over A ′ 9
Proof of Lemma 5 Lemma 5: Let A be a commutative ring. Assume: • Ideals ( g 1 , ..., g n ) = b ⊆ a = ( f 1 , ..., f n ) • g j = � n i =1 a ij f i and ∆ = det( a ij ) • a ′ = a / b in A / b and ∆ ′ image of ∆ in A ′ • f 1 , ..., f n and g 1 , ..., g n are prime sequences in A p for all p ⊇ a Then: (a) ∆ ′ is independent of choice of a ij (b) ∆ ′ A ′ = Fitt A ′ ( a ′ ) (c) ∆ ′ A ′ = Ann A ′ ( a ′ ) and a ′ = Ann A ′ (∆ ′ 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, { f i } and { g i } regular sequences contained in unique max ideal 10
Proof of Lemma 5 (a) ∆ ′ is independent of choice of a ij . • Suppose g j = � n i =1 a ij f i = � n i =1 b ij f i • Need to show det( a ij ) − det( b ij ) ∈ b • WLOG, assume first n − 1 rows of ( a ij ) and ( b ij ) agree • Set c ij = a ij = b ij for 1 ≤ j ≤ n − 1 and c in = a in − b in • det( c ij ) = det( a ij ) − det( b ij ) • Cramer’s rule = ⇒ det( c ij ) · f i = det(replace i th column of ( c ij ) with ( g 1 , ..., g n − 1 , 0) T ) • = ⇒ det( c ij ) · a ⊆ ( g 1 , ..., g n − 1 ) • g 1 , ..., g n prime sequence = ⇒ ( g 1 , ..., g n − 1 ) a prime ideal • g 1 , ..., g n a regular sequence in a = ⇒ a �⊆ ( g 1 , ..., g n − 1 ) • = ⇒ det( c ij ) ∈ ( g 1 , ..., g n − 1 ) ⊆ b 11
Proof of Lemma 5 (b) ∆ ′ A ′ = Fitt A ′ ( a ′ ). • Fitt A ′ ( a ′ ) is the image of Fitt A ( a ) in A ′ • a ′ ∼ = A n / U • U is generated by ( a 1 j , ..., a nj ) and (0 , ..., f q , ..., − f p , ... 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 ′ = Ann A ′ ( a ′ ) and a ′ = Ann A ′ (∆ ′ A ′ ). • Induct on n • n = 0: • ∆ ′ = det( ∅ ) = � ∅ λ = 1 and a ′ = 0 • Trivially, Ann A ′ (0) = A ′ and 0 = Ann A ′ ( A ′ ) • n ≥ 1: ⇒ ∆ ′ A ′ is a Fitting ideal • (b) = ⇒ ∆ ′ A ′ does not depend on choice of f 1 , ..., f n • = • Pick f 1 , ..., f n so that f 1 is prime to ( g 2 , ..., g n ) • Set B = A / ( g 2 , ..., g n ), b ij = a ij for j ≥ 2, b 11 = 1 and b i 1 = 0 for i ≥ 2, ∆ b = det( b ij ) • Inductive hypothesis on f 2 , ..., f n and g 2 , ..., g n in A / f 1 A = ⇒ : (i) ( B / f 1 B ) · ∆ b = Ann B / f 1 B ( a B / f 1 B ) = ( f 1 B : a B ) / f 1 B (ii) a B / f 1 B = Ann B / f 1 B (( B / f 1 B ) · ∆ b ) = ( f 1 B : ∆ b B ) / f 1 B • (ii) = ⇒ a B = ( f 1 B : ∆ b B ) • Cramer’s rule = ⇒ ∆ f 1 , ∆ b g 1 ∈ b = ⇒ ∆ b g 1 ≡ ∆ f 1 mod b 13
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