Introduction Preliminaries Survey on the resolution Proof of Theorem 1 . 5 References . On the almost Gorenstein property of determinantal rings . Naoki Taniguchi Meiji University The 38th Japan Symposium on Commutative Algebra November 19, 2016 . . . . . . Naoki Taniguchi ( Meiji University ) Almost Gorenstein determinantal rings 1 / 26 November 19, 2016
Introduction Preliminaries Survey on the resolution Proof of Theorem 1 . 5 References Introduction 2 ≤ t ≤ m ≤ n integers X = [ X ij ] an m × n matrix of indeterminates over an infinite field k S = k [ X ] = k [ X ij | 1 ≤ i ≤ m , 1 ≤ j ≤ n ] the polynomial ring I t ( X ) the ideal of S generated by the t × t minors of the matrix X R = S / I t ( X ) . Fact 1 ([2, 3]) . R is a Cohen-Macaulay normal domain dim R = mn − ( m − ( t − 1))( n − ( t − 1)) K R = Q n − m ( − ( t − 1) m ) where Q = I t − 1 ( Y ) R and Y = [ X ij ] is an m × ( t − 1) matrix obtained from X by choosing the first t − 1 columns. . . . . . . . Naoki Taniguchi ( Meiji University ) Almost Gorenstein determinantal rings 2 / 26 November 19, 2016
Introduction Preliminaries Survey on the resolution Proof of Theorem 1 . 5 References Therefore R is level, a ( R ) = − ( t − 1) n , and R is Gorenstein ⇐ ⇒ m = n . . Question 1.1 . When do the determinantal rings satisfy almost Gorenstein property? . . . . . . . Naoki Taniguchi ( Meiji University ) Almost Gorenstein determinantal rings 3 / 26 November 19, 2016
Introduction Preliminaries Survey on the resolution Proof of Theorem 1 . 5 References Therefore R is level, a ( R ) = − ( t − 1) n , and R is Gorenstein ⇐ ⇒ m = n . . Question 1.1 . When do the determinantal rings satisfy almost Gorenstein property? . . . . . . . Naoki Taniguchi ( Meiji University ) Almost Gorenstein determinantal rings 3 / 26 November 19, 2016
Introduction Preliminaries Survey on the resolution Proof of Theorem 1 . 5 References . Theorem 1.2 (Goto-Takahashi-T, 2015) . Let R = k [ R 1 ] be a Cohen-Macaulay homogeneous ring with d = dim R > 0 . Suppose that R is not a Gorenstein ring and | k | = ∞ . Then TFAE. (1) R is an almost Gorenstein graded ring and level. (2) Q( R ) is a Gorenstein ring and a ( R ) = 1 − d. . . Corollary 1.3 . R = S / I t ( X ) is an almost Gorenstein graded ring ⇐ ⇒ m = n, or m ̸ = n and . m = t = 2 . . . . . . . Naoki Taniguchi ( Meiji University ) Almost Gorenstein determinantal rings 4 / 26 November 19, 2016
Introduction Preliminaries Survey on the resolution Proof of Theorem 1 . 5 References Set M = R + . Then R = k [ X ] / I t ( X ) : AGG = ⇒ R M = ( k [ X ] / I t ( X )) M : AGL ⇐ ⇒ k [[ X ]] / I t ( X ) : AGL . Question 1.4 . Does the implication R M = ( k [ X ] / I t ( X )) M : AGL = ⇒ R = k [ X ] / I t ( X ) : AGG hold true? . . . . . . . Naoki Taniguchi ( Meiji University ) Almost Gorenstein determinantal rings 5 / 26 November 19, 2016
Introduction Preliminaries Survey on the resolution Proof of Theorem 1 . 5 References . Theorem 1.5 . Suppose that k is a field of characteristic 0 . Then TFAE. (1) R is an almost Gorenstein graded ring. (2) R M is an almost Gorenstein local ring. (3) Either m = n, or m ̸ = n and m = t = 2 . . . . . . . . Naoki Taniguchi ( Meiji University ) Almost Gorenstein determinantal rings 6 / 26 November 19, 2016
Introduction Preliminaries Survey on the resolution Proof of Theorem 1 . 5 References Preliminaries . Setting 2.1 . ( R , m ) a Cohen-Macaulay local ring with d = dim R | R / m | = ∞ ∃ K R the canonical module of R . . Definition 2.2 . We say that R is an almost Gorenstein local ring , if ∃ an exact sequence 0 → R → K R → C → 0 of R -modules such that µ R ( C ) = e 0 m ( C ). . . . . . . . Naoki Taniguchi ( Meiji University ) Almost Gorenstein determinantal rings 7 / 26 November 19, 2016
Introduction Preliminaries Survey on the resolution Proof of Theorem 1 . 5 References Look at an exact sequence 0 → R → K R → C → 0 of R -modules. If C ̸ = (0), then C is Cohen-Macaulay and dim R C = d − 1. Set R = R / [(0) : R C ]. Then ∃ f 1 , f 2 , . . . , f d − 1 ∈ m s.t. ( f 1 , f 2 , . . . , f d − 1 ) R forms a minimal reduction of m = m R . Therefore e 0 m ( C ) = e 0 m ( C ) = ℓ R ( C / ( f 1 , f 2 , . . . , f d − 1 ) C ) ≥ ℓ R ( C / m C ) = µ R ( C ) . Thus µ R ( C ) = e 0 m ( C ) ⇐ ⇒ m C = ( f 1 , f 2 , . . . , f d − 1 ) C . Hence C is a maximally generated maximal Cohen-Macaulay R-module in the sense of B. Ulrich, which is called an Ulrich R-module . . . . . . . Naoki Taniguchi ( Meiji University ) Almost Gorenstein determinantal rings 8 / 26 November 19, 2016
Introduction Preliminaries Survey on the resolution Proof of Theorem 1 . 5 References . Lemma 2.3 . Let R be an almost Gorenstein local ring and choose an exact sequence ϕ 0 → R − → K R → C → 0 of R-modules s.t. µ R ( C ) = e 0 m ( C ) . If φ (1) ∈ m K R , then R is a RLR. Therefore µ R ( C ) = r ( R ) − 1 provided R is not a RLR. . . . . . . . Naoki Taniguchi ( Meiji University ) Almost Gorenstein determinantal rings 9 / 26 November 19, 2016
Introduction Preliminaries Survey on the resolution Proof of Theorem 1 . 5 References . Corollary 2.4 . Let R be an almost Gorenstein local ring but not Gorenstein. Choose an exact sequence ϕ 0 → R − → K R → C → 0 of R-modules s.t. C is an Ulrich R-module. Then 0 → m φ (1) → m K R → m C → 0 is an exact sequence of R-modules. . Hence µ R ( m K R ) ≤ µ R ( m ) + µ R ( m C ) . . . . . . . Naoki Taniguchi ( Meiji University ) Almost Gorenstein determinantal rings 10 / 26 November 19, 2016
Introduction Preliminaries Survey on the resolution Proof of Theorem 1 . 5 References Survey on the resolution of determinantal rings . Setting 3.1 . t ≥ 1, m ≥ n ≥ 1 integers ( S , n ) a Noetherian local ring s.t. Q ⊆ S F , G free S -modules with rank S F = m + t − 1, rank S G = n + t − 1 ϕ = ( r ij ) : F → G a S -linear map s.t. r ij ∈ n . . . . . . . Naoki Taniguchi ( Meiji University ) Almost Gorenstein determinantal rings 11 / 26 November 19, 2016
Introduction Preliminaries Survey on the resolution Proof of Theorem 1 . 5 References Let λ ( m , n ) be the Young tableau consisting of rectangle of n rows of m squares, where the i -th row contains the numbers ( i − 1) m + 1 , ( i − 1) m + 2 , . . . , im in increasing order. 1 2 . . . m m − 1 λ ( m , n ) = . . . 2 m m +1 m +2 2 m − 1 . . . . ... . . . . . . . . . . . mn . . . . . . Naoki Taniguchi ( Meiji University ) Almost Gorenstein determinantal rings 12 / 26 November 19, 2016
Introduction Preliminaries Survey on the resolution Proof of Theorem 1 . 5 References k an integer s.t. 0 ≤ k ≤ mn λ = ( λ 1 , λ 2 , . . . , λ n ) a partition of k s.t. λ 1 ≥ λ 2 ≥ · · · ≥ λ n , ∑ n i =1 λ i = k , and λ i ≤ m for 1 ≤ ∀ i ≤ n . Definition 3.2 . We define the tableaux λ F , λ G as follows. The i -th column of λ F consists of λ i squares which contain the numbers of the ( n − i + 1)-th row of λ ( m , n ) in reverse order. λ G is the tableau derived from λ ( m , n ) by removing the numbers of λ F . . . Example 3.3 . Consider the case where m = 4, n = 3, k = 5, and λ = (3 , 2 , 0). Then λ ( m , n ) = 1 2 3 4 λ F = 12 8 and λ G = 1 2 3 4 , , 5 6 7 8 11 7 5 6 9 10 11 12 10 9 . . . . . . . Naoki Taniguchi ( Meiji University ) Almost Gorenstein determinantal rings 13 / 26 November 19, 2016
Introduction Preliminaries Survey on the resolution Proof of Theorem 1 . 5 References To the square in the ( i , j ) position of λ ( m , n ) we associate: . The square in the ( i , j ) position of λ ( m , n + t − 1) if j − i > m − n . The string of t squares from the ( i , j ) position to the ( i + t − 1 , j ) position if j − i = m − n . The square in the ( i + t − 1 , j ) position if j − i < m − n . . . Example 3.4 . Consider the case where m = 4, n = 3, k = 5, λ = (3 , 2 , 0), and t = 3. Then λ ( m , n + t − 1) = 1 2 3 4 λ ( m , n ) = 1 2 3 4 , . 5 6 7 8 5 6 7 8 9 10 11 12 9 10 11 12 13 14 15 16 17 18 19 20 . . . . . . . Naoki Taniguchi ( Meiji University ) Almost Gorenstein determinantal rings 14 / 26 November 19, 2016
Introduction Preliminaries Survey on the resolution Proof of Theorem 1 . 5 References . Definition 3.5 . We define the tableaux λ F ( t ), λ G ( t ) as follows. λ F ( t ) is the tableau constructed by replacing each square of λ F by the associated square or string of squares of λ ( m , n + t − 1). λ G ( t ) is the tableau obtained from λ ( m , n + t − 1) by removing the squares of λ F ( t ). . . Example 3.6 . Consider the case where m = 4, n = 3, k = 5, λ = (3 , 2 , 0), and t = 3. Then λ F = 12 8 so that λ F ( t ) = 12 8 . Therefore λ G ( t ) = 1 2 3 4 , . 11 7 16 7 5 6 10 20 11 9 10 19 15 13 14 18 17 . . . . . . . Naoki Taniguchi ( Meiji University ) Almost Gorenstein determinantal rings 15 / 26 November 19, 2016
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