i ntroduction i ntroduction Invariants of Hilbert series numerical - PowerPoint PPT Presentation

l inear inequalities for the h ilbert depth of graded modules over polynomial rings Julio Jos´ e Moyano Fern´ andez July 8th, 2016 Universitat Jaume I de Castell´ on

i ntroduction

i ntroduction • Invariants of Hilbert series − → numerical semigroups • New interpretation of some characterization already explained in Vila-Real and Cortona • This talk is based on a series of common works with * Lukas Katth¨ an, Goethe-Universit¨ at Frankfurt am Main * Jan Uliczka, Universit¨ at Osnabr¨ uck All available on the arXiv. 3

t he setting

Let K be a field. Let R := K [ X 1 , . . . , X n ] be a polynomial ring endowed with a grading, typically ⋄ standard- Z -grading, i.e., deg X i = 1 ⋄ nonstandard- Z -grading ⋄ ( Z r -grading) Let 0 � = M = � ℓ M ℓ be a finitely generated graded R -module, with Hilbert series (dim K M ℓ ) t ℓ ∈ Z [ � ][ t − 1 ] H M ( t ) = [ t ] ℓ ∈ Z r Series without negative coefficients: nonnegative series. 5

p revious results

h ilbert depth For the moment, let us restrict ourselves to Z -gradings Set d i := deg X i ∈ N for all i = 1 , . . . n . Definition [Hilbert depth] Hdep ( M ) := max { depth N | N a f.g. gr. module with H N = H M } . This is a well-defined but opaque quantity! Characterizations? 7

Theorem [—, Uliczka 13] A formal Laurent series H with denominator � i (1 − t d i ) is the Hilbert series of a f.g. graded R -module M if and only if Q I ( t ) � H ( t ) = j ∈ I (1 − t d j ) � I ⊆{ 1 ,..., n } with nonnegative Q I ( t ). Definition [Decomposition Hilbert depth] � � � H M admits a decompos. as above � decHdep ( M ) := max r ∈ N . � � with Q I = 0 ∀ I such th. | I | < r � 8

c ase of two variables Let R = K [ X , Y ] be with α := deg X , β := deg Y coprime. Set Γ := � α, β � the numerical semigroup generated by α and β . Theorem [—, Uliczka 13] Let M be a finitely generated graded R –module. Then n h n t n satisfies the Hdep ( M ) > 0 if and only if H M ( t ) = � condition � � h i + n ≤ h j + n i ∈ I j ∈ J for all n ∈ Z and all “fundamental couples” [ I , J ] . 9

(I) What is a “fundamental couple” [ I , J ]? Let L be the set of gaps of � α, β � . An ( α, β ) –fundamental couple [ I , J ] consists of two integer sequences I = ( i k ) m k =0 and J = ( j k ) m k =0 , such that (0) i 0 = 0. (1) i 1 , . . . , i m , j 1 , . . . , j m − 1 ∈ L and j 0 , j m ≤ αβ . (2) i k ≡ j k mod α and i k < j k for k = 0 , . . . , m ; j k ≡ i k +1 mod β and j k > i k +1 for k = 0 , . . . , m − 1; j m ≡ i 0 mod β and j m ≥ i 0 . (3) | i k − i ℓ | ∈ L for 1 ≤ k < ℓ ≤ m . 10

(II) What is a “fundamental couple” [ I , J ]? • I consists of minimal generators of “relative ideals” = “semimodules” ∆ of Γ. • J contains “small shifts” of I -sets which turn out to generate a sort of syzygy Syz ∆ . Syzygy in the sense that any element in Syz ∆ admits more than one presentation in the form i + x with i ∈ I and x ∈ Γ. 11

In the special case Γ = � 3 , 5 � the criterion is given by the inequalities h n +0 ≤ h n +15 , h n +0 + h n +1 ≤ h n +6 + h n +10 , h n +0 + h n +2 ≤ h n +12 + h n +5 , h n +0 + h n +4 ≤ h n +9 + h n +10 , h n +0 + h n +7 ≤ h n +12 + h n +10 , h n +0 + h n +1 + h n +2 ≤ h n +5 + h n +6 + h n +7 , h n +0 + h n +2 + h n +4 ≤ h n +5 + h n +7 + h n +9 12

Lattice paths for Γ = � 5 , 7 � 2 9 4 16 11 6 1 23 18 13 8 3 J = [15 , 13 , 16 , 14]. 13

Lattice paths for Γ = � 5 , 7 � 2 9 4 16 11 6 1 23 18 13 8 3 I = [0 , 8 , 6 , 9] J = [15 , 13 , 16 , 14]. 14

Lattice paths for Γ = � 5 , 7 � 2 9 4 16 11 6 1 23 18 13 8 3 I = [0 , 8 , 6 , 9] J = [15 , 13 , 16 , 14]. 15

Lattice paths for Γ = � 5 , 7 � (0) 2 (7) 9 4 (14) 16 11 6 1 (21) 23 18 13 8 3 (28) (35) (30) (25) (20) (15) (10) (5) (0) I = [0 , 8 , 6 , 9] and J = [15 , 13 , 16 , 14]. 16

n ew results

A deep algebraic meaning of the inequalities � i ∈ I h i + n ≤ � j ∈ J h j + n remained rather hidden. New insights appeared when considering the Z r -grading. The starting question arose by looking at the decomposition theorem (already mentioned): i (1 − t d i ) is the A formal Laurent series H with denominator � Hilbert series of a f.g. graded R -module M iff Q I ( t ) � H ( t ) = with nonnegative Q I . � j ∈ I (1 − t d j ) I ⊆{ 1 ,..., n } Question: Is the condition of the Thm satisfied by every rational function with the given denominator and nonnegative coefficients? 18

[excursus] Question: Which formal Laurent series arise as Hilbert series of R -modules (in a certain class)? Conditions: The series must... • ... have nonnegative coefficients. i (1 − t deg X i ). • ... be rational function with denominator � • ... Related work: • Macaulay, 1927: cyclic modules, standard Z -grading. • Boij & Smith, 2015: modules generated in degree 0, standard Z -grading + technical details 19

Theorem [Katth¨ an, —, Uliczka 2016] ][ t − 1 ] be a formal Laurent series, which is the Hilbert Let H ∈ Z [ [ t ] series of some finitely generated graded R -module M . Let further S := R / ( X β − Y α ). Then the following statements are equivalent: (a) Hdep ( M ) > 0 (b) For any finitely generated torsionfree S -module N , it holds that H · H N ≥ 0 . H R (c) Condition (b) holds for any finitely generated torsionfree S -module of rank 1. i h i t i satisfies (d) For all n ∈ Z , [ I , J ] fundamental couple, H = � � � h i + n ≤ h j + n ( ⋆ ) i ∈ I j ∈ J 20

We need the following result about the structure of fundamental couples. Lemma Let [ I = ( i k ) , J = ( j k )] be a fundamental couple of length m . Then there exist two integer sequences β > a 0 > a 1 > · · · > a m = 0 , and 0 = b 0 < b 1 < · · · < b m < α such that i k = αβ − a k − 1 α − b k β for 1 ≤ k ≤ m , and j k = αβ − a k α − b k β for 0 ≤ k ≤ m 21

(c) ⇒ (d): Let [ I , J ] be a fundamental couple. Recall that S = K [ t α , t β ] is the monoid algebra of Γ. Let N ⊆ K [ t ] be the S -module generated by t αβ − j 0 , . . . , t αβ − j m . This module is torsionfree, hence H M H N ≥ 0 by assumption. H R To see that this inequality implies ( ⋆ ), we need to compute H N . Let ( a k ) m k =0 , ( b k ) m k =0 be the sequences as in Lemma and let ˜ N := ( X a 0 Y b 0 , . . . , X a m Y b m ) . It is easy to see that ˜ N is the preimage of N under the projection R → S . In particular, note that X β − Y α ∈ ˜ N , because X a 0 , Y b m ∈ ˜ N . N / ( X β − Y α ) and thus H N = H ˜ Hence N ∼ = ˜ N − t αβ H R . 22

By considering the minimal free resolution of ˜ N , one sees that its syzygies are generated in the degrees a k − 1 α + b k β for 1 ≤ k ≤ m . Therefore m m N − t αβ H R = H ˜ H N t a k α + b k β − t a k − 1 α + b k β − t αβ � � = H R H R k =0 k =1   m m t − j − t αβ − j k − t αβ − i k − t αβ − i 0 = t αβ � � � � t − i =  k =0 k =1 j ∈ J i ∈ I Then we obtain   0 ≤ H · H N t − j − � h n t n ) t αβ � � t − i = (  H R n ∈ Z j ∈ J i ∈ I   = t αβ � � �  , t n h n + j − h n + i n ∈ Z j ∈ J i ∈ I and ( ⋆ ) is satisfied for [ I , J ]. 23