Hilbert function, Betti numbers 1 Daniel Gromada
References 2 David Eisenbud: Commutative Algebra with a View Toward Algebraic Geometry 19, 110 David Eisenbud: The Geometry of Syzygies 1A, 1B My own notes available on my webpage http://wwwmathuni-sbde/ag/speicher/gromadahtml
Outline 3 1 Hilbert basis theorem 2 Graded rings, Graded Modules 3 Graded free resolutions 4 Hilbert polynomial 5 Minimal free resolutions, Betti numbers
Hilbert basis theorem A ring R is Noetherian if every ideal of R is finitely generated Theorem (Hilbert basis theorem): If a ring R is Noetherian, then the polynomial ring R[x 1,, x n ] is Noetherian An R-module M is Noetherian if every submodule of N is finitely generated Proposition: If R is a Noetherian ring and M is a finitely generated R-module, then M is Noetherian 4
Graded rings A ring R is (N-)graded if it can be, as an additive abelian group, decomposed into a direct sum of groups 5+ R = i N R i in a way such that R i R j R i+j for i, j N
Graded rings A ring R is (N-)graded if it can be, as an additive abelian group, decomposed into a direct sum of groups 5+ R = i N R i in a way such that R i R j R i+j for i, j N Example: The ring of polynomials S = k[x 1,, x n ] is graded by degree S = i N S i, S i = {homogeneous polynomials of degree i}
Graded rings A ring R is (N-)graded if it can be, as an additive abelian group, decomposed into a direct sum of groups 5 R = i N R i in a way such that R i R j R i+j for i, j N Example: The ring of polynomials S = k[x 1,, x n ] is graded by degree S = i N S i, S i = {homogeneous polynomials of degree i} Generally, for R = i R i, the elements of R i are called homogeneous of degree i Any f R can be uniquely expressed as a sum of homogeneous components f = i f i, where f i R i
Graded modules Let R = i R i be a graded ring A module M over R is (Z)-graded if it can be, as an additive abelian group, decomposed into a direct sum 6+ M = i Z M i such that R i M j M i+j for i, j Z
Graded modules Let R = i R i be a graded ring A module M over R is (Z)-graded if it can be, as an additive abelian group, decomposed into a direct sum 6+ M = i Z M i such that R i M j M i+j for i, j Z For M a graded module, a Z, we denote M(a) the module M with grading shifted by a, so M(a) d := M a+d
Graded modules Let R = i R i be a graded ring A module M over R is (Z)-graded if it can be, as an additive abelian group, decomposed into a direct sum 6 M = i Z M i such that R i M j M i+j for i, j Z For M a graded module, a Z, we denote M(a) the module M with grading shifted by a, so M(a) d := M a+d A (graded) R-module M is free if it is isomorphic to the direct sum of copies of R, ie m m M = R = R m or M = R( a i ) i=1 i=1
Free resolutions A free resolution of an R-module M is an exact complex (ie im ϕ i+1 = ker ϕ i ) F: ϕ m+1 F m ϕ m ϕ m 1 ϕ 2 ϕ 1 Fm 1 F1 F0 7+ Equivalently, we could say that the complex of free R-modules such that coker ϕ 1 F 0 / im ϕ 1 = M ϕ m+1 F m ϕ m ϕ m 1 ϕ 2 ϕ 1 ϕ 0 Fm 1 F1 F0 M 0 should be exact The image im ϕ i is called the i-th syzygy module of M
Free resolutions A free resolution of an R-module M is an exact complex (ie im ϕ i+1 = ker ϕ i ) F: ϕ m+1 F m ϕ m ϕ m 1 ϕ 2 ϕ 1 Fm 1 F1 F0 7+ Equivalently, we could say that the complex of free R-modules such that coker ϕ 1 F 0 / im ϕ 1 = M ϕ m+1 F m ϕ m ϕ m 1 ϕ 2 ϕ 1 ϕ 0 Fm 1 F1 F0 M 0 should be exact The image im ϕ i is called the i-th syzygy module of M A graded free resolution is a resolution of a graded module, where the maps ϕ i are homogeneous of degree zero
Free resolutions A free resolution of an R-module M is an exact complex (ie im ϕ i+1 = ker ϕ i ) F: ϕ m+1 F m ϕ m ϕ m 1 ϕ 2 ϕ 1 Fm 1 F1 F0 7 Equivalently, we could say that the complex of free R-modules such that coker ϕ 1 F 0 / im ϕ 1 = M ϕ m+1 F m ϕ m ϕ m 1 ϕ 2 ϕ 1 ϕ 0 Fm 1 F1 F0 M 0 should be exact The image im ϕ i is called the i-th syzygy module of M A graded free resolution is a resolution of a graded module, where the maps ϕ i are homogeneous of degree zero If F m = 0 and F m 1 0, then we say that the resolution is finite of length m
Free resolutions ϕ m+1 F ϕ m ϕ m 1 ϕ 2 ϕ 1 ϕ 0 m Fm 1 F1 F0 M 0 Construction of a free resolution for a graded module M over R Let {f i } be generators of M Denote a i := deg f i 8+ Put F 0 := i R( a i) and denote e i the generators Define ϕ 0 : F 0 M mapping e i f i The same can be done with the module ker ϕ 0 Find its basis, construct corresponding free module F 1 and a map ϕ 1 : F 1 ker ϕ 0 F 0 Repeat to construct the free resolution
Free resolutions ϕ m+1 F ϕ m ϕ m 1 ϕ 2 ϕ 1 ϕ 0 m Fm 1 F1 F0 M 0 Construction of a free resolution for a graded module M over R Let {f i } be generators of M Denote a i := deg f i Put F 0 := i R( a i) and denote e i the generators Define ϕ 0 : F 0 M mapping e i f i The same can be done with the module ker ϕ 0 Find its basis, construct corresponding free module F 1 and a map ϕ 1 : F 1 ker ϕ 0 F 0 Repeat to construct the free resolution Theorem (Hilbert syzygy theorem): Consider S := k[x 1,, x n ] Every finitely generated S-module has a finite graded free resolution of length less or equal to n by finitely generated free modules 8
Hilbert function Consider ring S := k[x 1,, x n ] and finitely generated S-module M Function H M : Z N, H M (s) := dim k M s is called the Hilbert function of M The submodule i=s M i is finitely generated, so all the dimensions dim k M s are finite 9+
Hilbert function Consider ring S := k[x 1,, x n ] and finitely generated S-module M Function H M : Z N, H M (s) := dim k M s is called the Hilbert function of M The submodule i=s M i is finitely generated, so all the dimensions dim k M s are finite Theorem(Hilbert) Suppose M has a finite graded free resolution ϕ m ϕ m 1 ϕ 2 ϕ 1 0 F n Fm 1 F1 F0 9 such that each F i is finitely generated, so F i =: j S( a i,j), then m H M (d) = ( 1) ( ) n + d i ai,j 1 n 1 i=0 j
Hilbert function Theorem(Hilbert) Suppose M has a finite graded free resolution ϕ m ϕ m 1 ϕ 2 ϕ 1 ϕ 0 0 F m Fm 1 F1 F0 M 0 such that each F i is finitely generated, so F i =: j S( a i,j), then m H M (d) dim k M d = ( 1) ( ) n + d i ai,j 1 n 1 i=0 j 10+
Hilbert function Theorem(Hilbert) Suppose M has a finite graded free resolution ϕ m ϕ m 1 ϕ 2 ϕ 1 ϕ 0 0 F m Fm 1 F1 F0 M 0 such that each F i is finitely generated, so F i =: j S( a i,j), then m H M (d) dim k M d = ( 1) ( ) n + d i ai,j 1 n 1 i=0 j 10 For d a i,j, we have ( ) n + d ai,j 1 n 1 = (n + d a i,j 1)(n + d a i,j 2) (d a i,j + 1), (n 1)! so the Hilbert function coincides with a polynomial for large d This polynomial is called the Hilbert polynomial
Hilbert function Theorem(Hilbert) Suppose M has a finite graded free resolution ϕ m ϕ m 1 ϕ 2 ϕ 1 ϕ 0 0 F m Fm 1 F1 F0 M 0 such that each F i is finitely generated, so F i =: j S( a i,j), then m H M (d) dim k M d = ( 1) ( ) n + d i ai,j 1 n 1 i=0 j Proof: Considering d-th grade of the complex, we get H M (d) = i ( 1) i H Fi (d) 11+ It holds that H S (d) = ( n d H Fi (d) = j H S( ai,j )(d) = j ) ( = n+d 1 ) n 1, so H S (d a i,j ) = j ( ) n + d ai,j 1 n 1
Minimal free resolutions Idea: Use the construction for a minimal set of generators of M Denote m := x 1,, x n the homogeneous maximal ideal of S := k[x 1,, x n ] A complex of graded S-modules δ F i i Fi 1 is called minimal if im δ i mf i for every i 11
Minimal free resolutions We can write k = S/m Indeed, a general element of S/m is f + m = f 0 + m, where f 0 is the 0-th coefficient of f So, k can be treated as an S-module: f α = f 0 α for f S and α k It holds that k S M M/mM via α x αx + mm Indeed, elements of k S M are classes of pairs (α, x) k M wrt the relation f (α x) := f 0 α x = α fx The right exact functor k S maps finitely generated module M onto finite-dimensional vector space M/mM 12
Minimal free resolutions Lemma (Nakayama): If M is finitely generated graded S-module and m 1,, m t M generate M/mM, then m 1,, m t generate M Proof: Denote N := Sm i M m 1,, m t generate M/mM, so for every x M there are f 1,, f t S such that ie M = N + mm x + mm = f 1 m 1 + + f t m t + mm, Denote M := M/N We have that M/m M = (M/N)/m(M/N) = M/(N + mm) = 0, so M = m M If we prove M = 0, then M = N If M = m M 0, then, since M is finitely generated, it must contain an element of least grade But this element cannot be inside m M 13
Minimal free resolutions Corollary: A graded free resolution δ F: F i i Fi 1 is minimal if and only if, for each i, the map δ i maps the basis of F i to a minimal set of generators of the image Proof: For every i, we can consider the following short exact sequence δ i+1 F i+1 Fi δ i im δi 0 Applying the right-exact functor k S, we get the following exact sequence F i+1 /mf i+1 δi+1 Fi /mf i δi im δi /m im δ i 0 14
Minimal free resolutions Corollary: A graded free resolution δ F: F i i Fi 1 is minimal if and only if, for each i, the map δ i maps the basis of F i to a minimal set of generators of the image Proof: We have F i+1 /mf i+1 δi+1 Fi /mf i δi im δi /m im δ i 0 By definition, the resolution F is minimal if and only if the map δ i+1 is zero This is equivalent to the fact that the map δ i maps isomorphically F i /mf i to im δ i /m im δ i, ie maps basis of F i /mf i to basis of im /m im δ Using Nakayama Lemma, this is equivalent to the fact that δ i maps the basis of F i to a minimal generating set of im δ i 15
Minimal free resolutions Theorem: The minimal graded free resolution of a finitely generated S-module M is unique up to isomorphism Any free resolution of M contains the minimal free resolution as a direct summand In particular the number of generators of each F i with grade j is uniquely determined by M Those numbers are called Betti numbers and denoted β i,j F i = S( j) β i,j j Proposition: We have β i,j = dim k Tor S i (k, M) j since Tor S i (k, M) is the homology of the complex 16 δi+1 F i+1 /mf i+1 Fi /mf i δi
Betti numbers We usually write the into a table called Betti diagram of the form 1 2 s i β 0,i β 1,i+1 β s,i+s i + 1 β 1,i β 1,i+2 β s,i+1+s j β j,i β j,j+1 β s,j+s Proposition: Let d N If β i,j = 0 for all j < d, then β i+1,j+1 = 0 for all j < d 17+
Betti numbers We usually write the into a table called Betti diagram of the form 1 2 s i β 0,i β 1,i+1 β s,i+s i + 1 β 1,i β 1,i+2 β s,i+1+s j β j,i β j,j+1 β s,j+s Proposition: Let d N If β i,j = 0 for all j < d, then β i+1,j+1 = 0 for all j < d Substituting into the formula for Hilbert function, we get H M (d) = ( ) n + d j 1 B j, where B j = ( 1) i β i,j n 1 j i 0 17