MINISTRY OF EDUCATION AND TRAINING HANOI PEDAGOGICAL UNIVERSITY N2 DEPARTMENT OF MATHEMATICS TRAN QUANG KIEN FREE RESOLUTIONS AND BETTI NUMBERS OF GRADED MODULE GRADUATION THESIS Major: Algebra HA NOI – 2019 MINISTRY OF EDUCATION AND TRAINING HANOI PEDAGOGICAL UNIVERSITY N2 DEPARTMENT OF MATHEMATICS TRAN QUANG KIEN FREE RESOLUTIONS AND BETTI NUMBERS OF GRADED MODULE GRADUATION THESIS Major: Algebra Supervisor: Dr DO TRONG HOANG HA NOI - 2019 HA NOI – 2019 Contents Introduction Fundamental concepts 1.1 Graded modules 1.2 Graded complexes 10 Free resolutions 12 2.1 Graded free resolutions 12 2.2 Contructions 14 Betti numbers 21 3.1 Hilbert’s Syzygy Theorem 21 3.2 Betti numbers 23 Conclusion 27 References 28 INTRODUCTION The study of free resolutions is a core and beautiful area in commutative algebra The idea to associate a free resolution to a finitely generated module was introduced in two famous papers by Hilbert in 1890, 1893 Free resolutions provide a method for describing the structure of modules Base on the basic knowledge about algebraic and desiring comprehensive improvement of mathematics, I would like to choose a topic “free resolutions and Betti numbers of graded module” for my graduation thesis The main goal of this thesis is to describe the structure of graded finitely generated modules I will focus on the algebraic invariants associated the free resolutions Moreover, I also use computer softwares (CoCoA) to calculate algebraic invariants and verify mathematical issues which arises in the free resolutions The CoCoA software can be downloaded free in the website (http://cocoa.dima.unige.it) Throughout this thesis, we always denote R = k[x1 , , xn ] polynomial ring with n variables x1 , , xn over field k According to my understanding, we organize the thesis based on three books [3, 4, 5] In chapter 1, we will present some basic concepts for further exploration of our topic such as graded modules, graded complexes, Hilbert functions, Hilbert series Chapter will provide a graded free resolutions and a contruct it for a graded finitely generated R-module M Chapter is reserved for Betti numbers and relevant invariants This thesis was completed under the guidance of Dr Do Trong Hoang I would like to express my gratitude to him I would also like to thank the teachers of Hanoi Pedagogical University for helping me to have the knowledge and create conditions for me to this thesis Due to limitations in time and knowledge, the thesis can not avoid errors I hope to receive feedback from teachers and friends Chapter Fundamental concepts This chapter aims to provide all relevant definitions and some context regarding current research topics For a comprehensive introduction to these concepts see [4] 1.1 Graded modules This section will recall some definitions and notations about the graded R-modules and Nakayama’s lemma Definition 1.1.1 A ring R is called graded (or more precisely, Zgraded) if there exists a family of subgroups (Ri )i∈Z of R such that R = ⊕i Ri (as abelian groups), and Ri Rj ⊆ Ri+j for all i, j Remark 1.1.2 If R = ⊕i Ri is a graded ring, then R0 is a subring of R, ∈ R0 and Ri is an R0 -module for all i Proof Since R0 · R0 = R0 , R0 is closed under multiplication and thus is a subring of R We can write n xn , where each xn ∈ Rn Then for all i, we have xn xi xi = 1.xi = n Moreover, we have xi = xi x0 for all i Therefore, xn x0 x0 = 1.x0 = n xn = = n Hence = x0 ∈ R0 Now we prove that Ri is an R0 -module for all i Indeed, we have R0 · Rn ⊂ Rn for all i, this statement is proved Now, let R = ⊕i∈Z Ri be a graded ring Definition 1.1.3 A R-module M is called graded if it satisfies two following conditions: 1) M = ⊕i∈Z Mi (as abelian groups), and 2) Ri Mj ⊆ Mi+j , for i, j ∈ Z Then, the Mi is called the homogeneous component of M , and each m ∈ M is called a homogeneous element of degree i, denoted degM (m) = i, if m belongs to Mi Example 1.1.4 Let K be a field, and let R = K[x1 , , xn ] be a polynomial ring over K For c = (c1 , , cn ) ∈ Nn , let xc = xc11 xcnn Then R is a N-graded R-module, where Ri = rm xm | rm ∈ K and c1 + · · · + cn = i This is called the standard grading on the polynomial ring K[x1 , , xn ] Notice that R0 = K and deg xi = for all i Let M be Z-graded finite generated R-module We asume M = Rf1 + + Rfs Hence there exists an onto map Rs −→ M ei −→ fi Hence, dimK (Mi ) ≤ dimK (M ) ≤ s To measure the size of the module M , we should first measure the sizes of its graded components Definition 1.1.5 Let M be a Z-graded finite generated R-module The map H(−) : Z −→ Z i −→ dimK (Mi ) is called the Hilbert function of M Futhermore, HilbM (t) = HM (i) = i∈Z dimK (Mi ) i∈Z is called Hilbert series of M For p ∈ Z, denote by M (−p) the graded R-module such that M (−p)i = Mi−p for all i We say that M (−p) is the module M shifted p degrees, and call p the shift Its Hilbert function is HilbM (−p) (t) = HilbM (t) Note that degM (−p) (x) = a ⇒ degM (x) = a + p Example 1.1.6 LetR = K[x, y] and I = (x3 , y ) Then R/I is graded ring in degree with basis {1}, in degree with basis {x, y}, in degree with basis {x2 , xy}, in degree with basis {x2 y} The Hilbert series of R/I is: HilbR/I (t) = + 2t + 2t2 + t3 By the above example, Hilbert series of R/I(−7) is: HilbR/I(−7) (t) = t7 HilbR/I (t) = t7 + 2t8 + 2t9 + t10 In order to compute the Hilbert series, in practice we can use CoCoA software as follows: > Use R ::= QQ[x, y]; > I:=Ideal(x3 , y ); > Hilbert(R/I); H(0) = H(1) = H(2) = H(3) = H(t) = for t ≥ > HilbertSeries(R/I); (1 + 2x + 2x2 + x3 ) Definition 1.1.7 Let N and T be graded R-modules and a homomorphism ϕ : N → T For each m ∈ N , if deg (ϕ(m)) = i + deg(m) then we say that ϕ has degree i Example 1.1.8 Consider the homomorphism x x2 f : R[x]/(x ) −−−−−→ R/[x] Suppose that 1, x is the basis of R[x]/(x2 ) Hence deg(1) = 0, deg(x) = We have f (1) = x ⇒ deg(f (1)) = 1, f (x) = x2 ⇒ deg(f (x)) = Thus, deg(f (1)) = deg(1) + and deg(f (x)) = deg(x) + Therefore, f is graded and has degree Consider the homomorphism x f : R[x](x2 ) −−−−→ R/[x] Hence deg(1) = 0, deg(x) = We have f (1) = ⇒ deg(f (1)) = 0, f (x) = x ⇒ deg(f (x)) = Thus, deg(f (1)) = deg(1) + and deg(f (x)) = deg(x) + Therefore, f is graded and has degree Theorem 1.1.9 The following properties are equivalent M is a finitely generated graded R-module M ∼ = W/T , where W is a finite direct sum of shifted free R-module, T is a graded submodule of W , and the isomorphism has degree Proof (2) ⇒ (1): We know that the quotient of finitely generated graded R-module is finitely generated graded R-module, as required Now, we prove (1) ⇒ (2): Since M is a finitely generated graded R-module, so M = m1 R + + mk R, deg(mi ) = Therefore, di+1 (Fi+1 ) ⊆ (x, y)Fi for all i ≥ 0, so the resolution in the example 2.1.3 is minimal 2.2 Contructions We construct a graded free resolution of a graded finitely generated R-module M d d d (F• ) : → Fi →i Fi−1 → → F2 →2 F1 →1 F0 → M → Let M0 = M , M =< m1 , , mr >, deg(mi ) = Set F0 = R(−a1 ) ⊕ ⊕R(−ar ) Let fi be a basis element of R(−ai ), j = 1, r, then deg(fi ) = By Theorem 1.1.9, the following homomorphism of degree is defined well as follows: d0 : F0 −→ M fj −→ mj We have Ker(d0 ) = M1 , and so M1 is a graded finitely generated Rmodule So M1 =< l1 , , ls >, deg(li ) = bi Set F1 = R(−b1 ) ⊕ ⊕ R(−bs ) and gj is a basis element of R(−cj ), deg(gj ) = cj Define d1 : F1 −→ M1 gj −→ lj Then d1 is a surjective homomorphism of degree Consider the embedding map: d1 : F1 −→ F0 gj −→ lj 14 Assume by induction, that Fi and di are defined Set Mi+1 = Ker(di ) Mi+1 is a graded finitely generated R-module So M1 = t1 , , , deg(ti ) = dI Set Fi+1 = R(−d1 ) ⊕ ⊕ R(−dp ) and ui is a basis element of R(−di ) Define di+1 : Fi+1 −→ Mi+1 uj −→ tj Then di+1 is a surjective homomorphism of degree Consider the embedding map: di+1 : Fi+1 −→ Fi uj −→ tj By construction we have Ker(di ) = Im(di+q ) Example 2.2.1 Let A = K[x, y] and B = (x3 , xy, y ) We will construct a graded free resolution of A/B over A Set F0 = A and let d0 : A → A/B Then Ker(d0 ) = x3 , xy, y Set F1 = A(−3) ⊕ A(−2) ⊕ A(−5) and f1 , f2 , f3 are the basis elements of A(−3) ⊕ A(−2) ⊕ A(−5) We defined d1 : A(−3) ⊕ A(−2) ⊕ A(−5) → A f1 → x f2 → xy f3 → y We obtain that x3 xy y G = A(−3) ⊕ A(−2) ⊕ A(−5) −−−−−−−−−→ A → A/B → 15 Then we get Ker(d1 ) = af1 + bf2 + cf3 ∈ G | ax3 + bxy + cy = , where a, b, c ∈ A We have ax + bxy + cy = ⇔ ax3 = −y(bx + cy ), cy = −x(ax2 + by) From there, y | a and x | c Suppose that a = y˜ a, c = x˜ c Then a ˜x2 + b + c˜y = It implies that b = −˜ ax2 + c˜y So, we can write it into discrete sections as follows a ˜ = a y + a” b = b y + b”x2 + bx2 y c˜ = c + c”x2 We get (a + b + c”)x2 y + (a” + b”)x2 + (b + c )y = Thus,    a + b + c” = a” + b” =    b + c = Therefore, a , a”, b , b”, ¯b, c , c” = (0, 1, 0, −1, 0, 0, 0) , (0, 0, −1, 0, 0, 1, 0) , (1, 0, 0, 0, −1, 0, 0) , (0, 0, 0, 0, −1, 0, 1) It implies that (a, b, c) = σ1 = y, −x2 , , σ2 = 0, −y , x , σ3 = y , x2 y , , σ4 = 0, −x2 y , x3 = σ1 , σ2 We conclude that Ker (d1 ) = yf1 − x2 f2 , −y f2 + xf3 We have deg yf1 − x2 f2 = deg −y f2 + xf3 = 16 Set F2 = A (−4) ⊕ A (−6) and g1 , g2 are the basis elements of A (−4) and A (−6) Consider  x3     −x2   −y       x d2 : A(−4) ⊕ A(−6) −−−−−−−−−−→ A(−3) ⊕ A(−2) ⊕ A(−5), where g1 −→ yf1 − x2 f2 , and g2 −→ y f2 + xf3 Then Ker(d2 ) = ug1 + vg2 ∈ F2 | uyf1 + −ux2 − vy f2 + vxf3 = Hence   uy =   −ux2 − vy = ⇒    vx = u=0 v = We obtain F3 = Therefore,  y    −x2       −y   x → A(−4) ⊕ A(−6) −−−−−−−−−→ A(−3) ⊕ A(−2) ⊕ A(−5) x3 xy y −−−−−−−−−→ A → A/B → Example 2.2.2 Let A = K[x, y] and B = (x3 , xy, y ) Suppose we are given (say by computer) the non-graded free resolution  y    −x2       −y   x3 xy y x → A −−−−−−−−−→ A −−−−−−−−−→ A 17 of the module A/B over A We will determine the grading Denote by f1 , f2 , f3 the basis of A3 with respect to which the matrix of d1 is given Since f1 → x3 and deg x3 = f2 → xy and deg (xy) = f3 → y and deg y = and since we want d1 to be homogeneous of degree 0, we set deg(f1 ) = 3, deg(f2 ) = 2, deg(f3 ) = Therefore, the free A-module generated by f1 is A(−3), the free Amodule generated by f2 is A(−2), and the free A-module generated by f3 is A(−5) Thus, A3 is identified with A(−3) ⊕ A(−2) ⊕ A(−5) Furthermore, denote by g1 , g2 the basis of A2 with respect to which the matrix of d2 is given Since g1 →yf1 − x2 f2 deg yf1 − x2 f2 = deg (yf1 ) = deg(y) + deg(f1 ) = g2 → − y f2 + xf3 deg −y f2 + xf3 = deg y f2 = deg(y ) + deg(f2 ) = and since we want d2 to be homogeneous of degree 0, we set deg(g1 ) = and deg(g2 ) = Hence the free A-module generated by g1 is A(−4) and the free A-module generated by g2 is A(−6) Thus, A2 is identified with A(−4) ⊕ A(−6) 18 Therefore, we obtain the graded free resolution   y   −x2      4 −y   x3 xy y x → A(−4)⊕A(−6) −−−−−−−−−→ A(−3)⊕A(−2)⊕A(−5) −−−−−−−−−→ A In order to compute the free resolution, in practice we can use Cocoa software > Use R ::= QQ[x,y]; > I:=Ideal(x3 , xy, y ); > Res(I); − − > R(−4) ⊕ R(−6) − − > R(−2) ⊕ R(−3) ⊕ R(−5) Theorem 2.2.3 The graded free resolution constructed above is minimal if and only if at each step we choose a minimal homogenous system of generators of the kernel of the differential Proof We use the notation introduced in Construction 2.2.1 and set Ker(d1 ) = U We will prove the construct resolution is minimal On the contrary that for some i ≥ −1, we have chosen a non-minimal homogeneous system l1 , l2 , , ls of generators of Ker(di ) Assume l1 = rj lj 2≤j≤s for some rj ∈ R Since di+1 : Fi+1 → Fi gj → lj 19 So l1 = rj lj Then 2≤j≤s rj di+1 (gj ) ⇒ di+1 (g1 − di+1 (g1 ) = 2≤j≤s 2≤j≤s Hence, g1 − 2≤j≤s rj gj rj gj ) = ∈ Ker(di+1 ) = Im(di+2 ) Since the resolu- tion is minimal, we have that Im(di+2 ⊆ (x1 , , xn )Fi+1 Hence, g1 − 2≤j≤s rj gj ∈ (x1 , , xn )Fi+1 , which is a contradiction Now, suppose that at each step we choose a minimal homogeneous system of generators of the kernel of the differential We want to show that the obtained resolution is minimal Assume the contrary There exists an i ≥ −1 such that Im(di+2 ) (x1 , , xn )Fi+1 Therefore, Ker(di+1 ) = Im(di+2 ) contains a homogeneous element that is not in (x1 , , xn )Fi+1 We can assume that g1 − 2≤j≤s rj gj ∈ Ker(di+1 ) for some rj ∈ R Hence di+1 (g1 ) = rj di+1 (gj ) 2≤j≤s Hence, l1 = rj lj This contradicts to the fact that we have chosen 2≤j≤s l1 , , ls to be a minimal homogeneous system of generators of Ker(di ) 20 Chapter Betti numbers In this chapter, I will give the definition of the Betti numbers Hilbert’s Syzygy Theorem will also be presented Two invariants, regularity and projective dimension can be associated with the resolutions that measure ”shape” and ”size” if it is interpreted geometrically 3.1 Hilbert’s Syzygy Theorem We know that the free resolution of finitely generated module can be finite or infinite However, Hilbert’s Syzygy Theorem 3.1.4, which says that the minimal free resolution of every finitely generated graded module over a polynomial ring is finite Definition 3.1.1 The length of a free resolution (F• ) of a finitely generated M is max {i | rank Fi = 0} We say that the free resolution (F• ) is finite if its length is finite Otherwise, (F• ) is infinite The projective dimension of M is pdR (M ) = max {i| rank (Fi ) = 0} 21 Thus, pdR (M ) is the length of the minimal free resolution (F• ) of M Example 3.1.2 Let A = K[x, y] and B = (x3 , xy, y ) We have the free resolution is  y    −x2       −y   x → A(−4) ⊕ A(−6) −−−−−−−−−→ x3 xy y A(−3) ⊕ A(−2) ⊕ A(−5) −−−−−−−−−→ A → A/B → So, pdR (A/B) = Example 3.1.3 Let M = R = k[x]/(x2 ) The following sequence ·x ·x ·x ·x · · · −→ M −→ M −→ M −→ · · · is minimal free resolution Following the above example, length of the minimal free resolution can be infinity The idea to associate a resolution to a finitely generated R-module M was introduced in Hilbert’s famous papers [1, 2] Theorem 3.1.4 (Hilbert’s Syzygy Theorem) The minimal graded free resolution of a finitely generated graded R-module is finite and its length is at most n Proof We know that the Koszul complex K• (x1 , , xn ) is a graded free resolution of R/m = k of length n, where m = (x1 , , xn ) Thus pdR (k) = n Now we prove pdR (M ) ≤ n Assume that d d i (F• ) · · · −→ Fi −→ Fi−1 → · · · → F1 −→ F0 → M → 22 is the minimal graded free resolution of M Tensoring (F• ) with R/m = k yields the complex d ⊗1 d ⊗1 i · · · → Fi ⊗ k −→ Fi−1 ⊗ k → · · · → F1 ⊗ k −→ F0 ⊗ k → M ⊗ k → Using Im(di ) ⊆ mFi−1 for all i ≥ 1, one obtains that all the maps di ⊗ are zero Hence Ker(di+1 ⊗ 1) = Fi ⊗ R/m Im(di ⊗ 1) for i ≥ In particular, Fi ⊗ R/m ∼ = Fi /mFi = for i ≥ n + By Nakayama’s lemma 1.1.10, one obtains Fi = for all i ≥ n + 3.2 Betti numbers It is quite difficult to obtain a description of the differential in a graded free resolution In such cases, we can use the Betti numbers to obtain some information about the numerical invariants of the resolution Definition 3.2.1 The i-th Betti number of M over R is bR i (M ) = rank(Fi ) Remark 3.2.2 The Betti numbers not depend on the choice of the minimal graded fre resolution of M Proof By Theorem 2.2.3, we know that to construct a minimal graded free resolution of a graded finitely generated R-module, at each step of construction in section 2.2, we choose a minimal homogenous system of generators of the kernel of the differential Hence, there exists a unique minimal graded free resolution of a graded finitely generated R-module Therefore, The Betti numbers not depend on the choice of the minimal graded free resolution of M 23 We have that pdR (M ) is the length of the shortest graded free resolution of M So we have the following proposition: Proposition 3.2.3 The length of the minimal free resolution (F• ) of M is pdR (M ) = max i | bR i (M ) = Definition 3.2.4 The Poincar´e series of M over R is R PM (t) = i bR i (M )t i≥0 Example 3.2.5 Back to the example 2.1.3, we have the minimal graded free resolution of A/B is   y   −x2      4 −y   x → A(−4) ⊕ A(−6) −−−−−−−−−→ A(−3) ⊕ A(−2) ⊕ A(−5) x3 xy y −−−−−−−−−→ A We obtain pd(A/B) = 2, and moreover, b0 (A/B) = 1, b1 (A/B) = 3, b2 (A/B) = 2, and bi (A/B) = for all i ≥ Hence, the Poincare polynomial of A/B is PA/B (t) = + 3t + 2t2 In order to find Betti numbers of R/I, we can use CoCoA software as the following example: Example 3.2.6 Let R = k[x, y, z] and I = x2 , xy, xz, y > Use R == ZZ/32003[x, y, z]; 24 > I = Ideal(x2 , xy, xz, y ); > Res(I); > Describe it; We obtain the minimal graded free resolution of R/I is  z         x        −y       y    −x       z z −y −x y            0 −x 0 → R(−4) −−−−→ R4 (−3) −−−−−−−−−−−−−−−→ R4 (−2) → R Therefore, pd (R/I) = 3, b1 (R/I) = b2 (R/I) = 4, and b0 (R/I) = b3 (R/I) = So, R PR/I (t) = + 4t + 4t2 + t3 Definition 3.2.7 Let (F• ) be graded free resolution of a R-module M Since each free module Fi is a direct sum of modules of the form R(−p) We define the graded Betti numbers of M by bR i,p (M ) = number of summands in Fi of the form R(−p) Definition 3.2.8 The regularity is defined by reg(M ) = max{j − i bR i,j (M ) = 0}, and the graded Poincar´e series of M over R is R PM (t, z) = i p bR i,p (M )t z i≥0,p∈Z We typically write the Betti numbers in a matrix called the Betti table of M The entry in the i-th column and the p-th row is bi,i+p (M ) 25 ··· ··· i pd(M ) p bi,i+p (M ) reg(M ) Example 3.2.9 Back to the example 2.2.2, we have F0 = A, F1 = A(−3)⊕A(−2)⊕A(−5), and F2 = A(−4)⊕A(−6) Therefore, pdA (A/B) = A 2, b0 (A/B) = 1, b1 (A/B) = 3, b2 (A/B) = 2, and PA/B (t) = + 3t + 2t2 Moreover, b0,0 = 1, b1,2 = 1, b1,3 = 1, b1,5 = 1, b2,4 = 1, b2,6 = 1, and A reg(A/B) = 6, PA/B (t, z) = + tz + tz + tz + t2 z + t2 z We obtain the Betti table of A/B is 1 26 Conclusion In this thesis, we have systematically presented the following results: (1) I have recalled the definition and basic properties of some important concepts in homological algebra such as complexes, free resolutions (2) A graded free resolution of a graded finitely generated R-module is constructed (3) Hilbert’s Syzygy Theorem have also been presented and proved (4) The definition of Betti number and some information about the numerical invariants associated to the resolution (5) In order to compute the above invariants, CoCoA software could be easily used for studying It has already led to theoretical results 27 References [1] D Hilbert, Ueber die Theorie der algebraischen Formen, Math Ann 36 (4) (1890), 473–534 [2] D Hilbert, Ueber die vollen Invariantensysteme, Math Ann 42 (3) (1893), 313–373 [3] I Peeva, Graded syzygies, Algebra and Applications, 14 SpringerVerlag London, Ltd., London, 2011 [4] R Y Sharp, Steps in commutative algebra, London Mathematical Society Student Texts, Cambridge University Press, Cambridge, 1990 [5] R Villarreal , Monomial Algebras, Monographs and Textbooks in Pure and Applied Mathematics Vol 238, Marcel Dekker, New York, 2001 28 ... structure of modules Base on the basic knowledge about algebraic and desiring comprehensive improvement of mathematics, I would like to choose a topic ? ?free resolutions and Betti numbers of graded module? ??... a free resolution of M Definition 2.1.2 A free resolution (F• ) of a finitely generated graded module M is called graded if the modules Fi are graded and each di is a 12 graded homomorphism of. .. finite direct sum of shifted free R -module, T is a graded submodule of W , and the isomorphism has degree Proof (2) ⇒ (1): We know that the quotient of finitely generated graded R -module is finitely