Hilbert functions and Betti numbers of reverse lexicographic ideals in the exterior algebra

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1 Turk J Math 36 (2012), c TÜBİTAK oi: /mat Hilbert functions an Betti numbers of reverse lexicographic ieals in the exterior algebra Marilena Crupi, Carmela Ferró Abstract Let K be a fiel, V a K -vector space with basis e 1,...,e n an let E be the exterior algebra of V. We stuy the class of reverse lexicographic ieals in E. We analyze the behaviour of their Hilbert functions an Betti numbers. Key Wors: Grae ring, exterior algebra, monomial ieal, minimal resolution 1. Introuction Let K be a fiel, V a K -vector space with basis e 1,...,e n an let E be the exterior algebra of V. In [1, Theorem 4.4] Aramova, Herzog an Hibi prove that in the exterior algebra the lexicographic ieals give the maximal Betti numbers among all grae ieals with a given Hilbert function. The result hols in any characteristic. Such a result is the analogue of a result prove for grae ieals in a polynomial ring by Bigatti [3] an Hulett [6] in characteristic zero, an by Parue [7] in any characteristic. Afterwars, Deery [5] has prove an analogue of Bigatti, Hulett an Parue s result about minimal Betti numbers. He showe that revlex segment ieals give the lowest Betti numbers among all stable ieals with the same Hilbert function in a polynomial ring. Our aim is to prove a similar result for grae ieals in the exterior algebra. We introuce the reverse lexicographic ieals in the exterior algebra E, stuy their Hilbert functions an prove that the reverse lexicographic ieals have minimal Betti numbers for given Hilbert functions. In this context the stable ieals (see [1] an [2]) play an important role. In [1], Aramova, Herzog an Hibi gave an explicit minimal free resolution for this class of monomial ieals an obtaine a formula for their Betti numbers (Theorem 2.2). This formula will be a funamental tool in the article. The paper is organize as follows. Section 2 contains preliminary notions an results. In Section 3 we introuce the reverse lexicographic segment an analyze its shaow (Definition 3.5). As in the polynomial case the shaow of a reverse lexicographic segment is not necessarily a reverse lexicographic segment. Therefore we etermine the conitions uner which this property hols (Theorem 3.6 an Corollary 3.8). We characterize sets of monomials of E that generates a reverse lexicographic ieal AMS Mathematics Subject Classification: 13A02, 15A75, 18G

2 Section 4 is evote to the stuy of the behaviour of the Hilbert functions of reverse lexicographic ieals (Propositions 4.2 an 4.3). In Section 5 we stuy the grae Betti numbers of reverse lexicographic ieals in the exterior algebra E. The main statement is the following: Theorem 1.1 Let I E be a strongly stable ieal an J E a reverse lexicographic ieal. Suppose that H E/J () =H E/I (), for all. Thenβ i,j (E/I) β i,j (E/J), for all i an j. The key result is Theorem 5.3, which is obtaine by a certain ecomposition of a subset of monomials of the same egree in the exterior algebra. 2. Preliminaries an notation Let K be a fiel. We enote by E = K e 1,..., e n the exterior algebra of a K -vector space V with basis e 1,...,e n. For any subset σ = {i 1,...,i } of {1,...,n} with 1 i 1 <i 2 <... < i n we write e σ = e i1... e i an call e σ a monomial of egree. The set of monomials in E forms a K -basis of E of carinality 2 n. In orer to simplify the notation we put fg = f g for any two elements f an g in E. An element f E is calle homogeneous of egree j if f E j, where E j = j V. An ieal I is calle grae if I is generate by homogeneous elements. If I is grae, then I = j 0 I j,wherei j is the K -vector space of all homogeneous elements f I of egree j. Let e σ = e i1 e i2 e i be a monomial of egree. We efine supp(e σ )={i 1,i 2,...,i } = {i : e i ivies e σ }, an we write m(e σ )=max{i : i supp(e σ )}. If I E is a monomial ieal, we enote by G(I) the unique minimal set of monomial generators of I. The function H E/I (j) = im K (E/I) j,j = 0, 1,... is calle the Hilbert function of E/I an the polynomial H E/I = j 0 H E/I(j)t i is calle the Hilbert series of E/I. The possible Hilbert functions of grae algebras of the form E/I are escribe by the Kruskal-Katona Theorem [1, Theorem 4.1], which is the precise analogue to Macaulay s theorem (see, e.g., [4]). For later use we recall the next efinitions, which are quote from [1], [2]. Definition 2.1 Let I E be a monomial ieal. I is calle stable if for each monomial e σ I an each j<m(e σ ) one has e j e σ\{m(eσ)} I. I is calle strongly stable if for each monomial e σ I an each j σ one has that e i e σ\{j} I for all i<j. Now let I be a grae ieal of E. Consier the minimal grae free resolution of E/I over E F :... F F1 F0 E/I

3 Each of the moules F i is of the form j E( j) βi,j (E/I) an the maps i are escribe by matrices with homogeneous coefficients in E. Moreover, the resolution of E/I is always infinite, unless I = (0). Inee ker i 0foralli since the kernel of i contains the submoule (e 1 e 2 e n )F i. Viewing K as a left E - moule via the canonical epimorphism, we have that β i,j (E/I)=im K Tor E i (E/I,K) j. The numbers β i,j (E/I) are calle the grae Betti numbers. In [1], Aramova, Herzog an Hibi stuie the minimal grae free resolution of E/I when I is a stable ieal of E an prove a formula for computing the grae Betti numbers of E/I. More precisely [1, Corollary 3.3]: Theorem 2.2 Let I E be a stable ieal. Then β i,j+i (E/I) = u G(I) j+1 m(u)+i 2, for all i 1. m(u) 1 3. Revlex ieals in the exterior algebra In this section we introuce the reverse lexicographic ieals in the exterior algebra E = K e 1,..., e n an stuy some of their properties. Let M enote the set of all monomials of egree 1inE. We write > revlex for the reverse lexicographic orer (revlex for short) on the finite set M, i.e., if u = e i1 e i2 e i an v = e j1 e j2 e j are monomials belonging to M with 1 i 1 <i 2 <... < i n an 1 j 1 <j 2 <...<j n, then u> revlex v if i = j,i 1 = j 1,...,i s+1 = j s+1 an i s <j s for some 1 s. From now on, in orer to simplify the notation, we will write > instea of > revlex. Definition 3.1 A nonempty set M M is calle a reverse lexicographic segment of egree (revlex segment of egree, for short) if for all v M an all u M such that u>v, we have that u M. Example 3.2 Let E = K e 1,e 2,e 3,e 4,e 5. The subset of M 2 W = {e 1 e 2,e 1 e 3,e 2 e 3,e 1 e 4 } is a revlex segment of egree 2, whereas the subset of M 4 W = {e 1 e 2 e 3 e 4,e 1 e 2 e 4 e 5,e 1 e 3 e 4 e 5 } is not a revlex segment, because e 1 e 2 e 3 e 5 >e 1 e 2 e 4 e 5 an e 1 e 2 e 3 e 5 / W. Definition 3.3 Let I = j 0 I j be a monomial ieal of E. We say that I is a reverse lexicographic ieal of E if, for every j, I j is spanne by a revlex segment (as K -vector space). 368

4 Example 3.4 Let E = K e 1,e 2,e 3,e 4. The ieal I =(e 1 e 2,e 2 e 3 e 4 ) E is not a revlex ieal since I 3 = e 1 e 2 e 3,e 1 e 2 e 4,e 2 e 3 e 4 is not spanne as K -vector space by a revlex segment. In fact, e 1 e 3 e 4 >e 2 e 3 e 4 an e 1 e 3 e 4 / I 3. The ieal I =(e 1 e 2,e 1 e 3 e 4 ) E is a revlex ieal. In fact I 2 = e 1 e 2 is spanne as K -vector space by a revlex segment of egree 2; I 3 = e 1 e 2 e 3,e 1 e 2 e 4,e 1 e 3 e 4 is spanne as K -vector space by a revlex segment of egree 3; an I 4 = e 1 e 2 e 3 e 4 is spanne as K -vector space by a revlex segment of egree 4. It is clear that reverse lexicographic strongly stable stable. From now on, for the sake of simplicity, given a monomial ieal I = j 0 I j of E we will say that I j is a reverse lexicographic segment of egree j if I j is spanne as K -vector space by a reverse lexicographic segment of egree j. Definition 3.5 Let M be a subset of monomials of E.Sete i = {e 1,...,e i }. We efine the set e i M = {ue j : u M, j/ supp(u), j =1,...,i}. Note that e i M = if, for every monomial u M an for every j =1,...,i, one has j supp(u). If M is a set of monomial of egree <n, e n M is calle the shaow of M an is enote by Sha(M): Sha(M) ={ue j : u M, j / supp(u), j =1,...,n}. Note that, if M is a revlex segment of egree, then Sha(M) nees not be a revlex segment of egree +1. For example, if E = K e 1,e 2,e 3,e 4,e 5 an M = {e 1 e 2,e 1 e 3 }, then Sha(M) ={e 1 e 2 e 3,e 1 e 2 e 4,e 1 e 3 e 4,e 1 e 2 e 5,e 1 e 3 e 5 } is not a revlex segment of egree 3. Infact e 2 e 3 e 4 >e 1 e 2 e 5 but e 2 e 3 e 4 / Sha(M). Theorem 3.6 Let M be a revlex segment of egree of E an let i be an integer such that n>i>+1. Suppose e i+1 M. The following conitions are equivalent: (1) e i+1 M is a revlex segment of egree +1. (2) e i e i ( 1) e i 2 e i 1 M. 369

5 Proof. Set M = e i+1 M. (1) (2). Suppose that e i e i ( 1) e i 2 e i 1 / M. Let u be the smallest monomial that belongs to M. Then u>e i e i ( 1) e i 2 e i 1 an m(u) i 1. Let w be the greatest monomial of egree such that u>w. It follows that w e i e i ( 1) e i 2 e i 1 an m(w) i 1. Therefore i/ supp(w) ansowe i 0. Clearly, we i >ue i+1. Hence, since ue i+1 M an since M is a revlex segment by assumption, we have that we i M. Therefore there exist v u an j {1,...,i+1}, j/ supp(v), such that we i = ve j. (1) Since v u, it follows that m(v) i 1 an consequently i/ supp(v). Hence, from (1) we have i = j an w = v.thenw = v u, which contraicts the choice of w. Thus e i e i ( 1) e i 2 e i 1 M. (2) (1). Let u be the smallest monomial that belongs to M. We will show that every monomial w of egree +1 such that w>u is an element of M. We have that u = e i1 e i e i+1, with e i1 e i M. Let w = e j1 e j+1 with w > u. Then m(w) =j +1 m(u) =i +1. (Case 1). Suppose m(w) =j +1 = i +1. We have w = e j1 e j e i+1 >u= e i1 e i e i+1 e j1 e j >e i1 e i. Since M is a revlex segment, it follows that e j1 e j M. Hence w = e j1 e j e j+1 M an M is a revlex segment of egree +1. (Case 2). Suppose m(w) =j +1 <i+1. In this case e j1 e j e i e i ( 1) e i 2 e i 1.Infactj <j +1 i. Therefore, since M is a revlex segment, e j1 e j M an w M. Remark 3.7 In Theorem 3.6 we may assume i>+ 1, since otherwise the problem is trivial. As consequences of Theorem 3.6 we obtain the following corollaries. Corollary 3.8 Let M be a revlex segment of egree of E such that <n 2. The following conitions are equivalent: (1) Sha(M) is a revlex segment of egree +1. (2) e n (+1) e n 3 e n 2 M. Corollary 3.9 Let M = {e σ1,...,e σt } be a set of monomials of E an let 1 =min{eg(e σi ):i =1,...,t} an 2 =max{eg(e σi ):i =1,...,t}, with 2 <n 2. ThenI =(M) is a revlex ieal if an only if (1) I j is a revlex segment for 1 j 2 ; (2) e n (2+1) e n 3 e n 2 M. 370

6 4. Hilbert functions of revlex ieals In this section we put our attention on the behaviour of the Hilbert functions of revlex ieals in the exterior algebra E = K e 1,e 2,...,e n. For a finite subset S of E,weenotebyM(S) the set of all monomials in S an we enote by S its carinality. Proposition 4.1 Let M be a revlex segment of egree <n 2 of E.ThenSha(M) is a revlex segment if n 2 an only if M. Proof. From Corollary 3.8, Sha(M) is a revlex segment if an only if the monomial e n (+1) e n 3 e n 2 belongs to M.Setu = e n (+1) e n 3 e n 2. It follows that u M if an only if all the monomials e i1 e i2 e i with 1 i 1 <i 2 <...<i n 2areinM. Hence Sha(M) is a revlex segment if an only if M n 2. Proposition 4.2 Let I E be a grae ieal. Let be a positive integer such that <n 2. Suppose (1) I is a revlex segment, (2) H E/I () n n 2. Then H E/I ( +1) H E/I (). n n 2 Proof. Since H E/I () = im K E im K I = im K I, we have im K I by the assumption. It follows that Sha(I ) is a revlex segment (Proposition 4.1) an e n (+1) e n 3 e n 2 M(I ). Let u = e i1 e i, 1 i 1 <... < i n, be the smallest monomial in M(I ). Then u e n (+1) e n 3 e n 2 an therefore m(u) {n 2,n 1,n}. (Case 1). Suppose m(u) {n 2,n 1}. Thenue n is the smallest monomial in M(I +1 ) an all larger monomials are in M(I +1 ). Set A = {w M +1 : w<ue n } an A = {v M : v<u}. Note that A = H E/I (). If w = e τ <ue n,then n supp(w) an consequently e τ\{n} <u. Therefore there is an injection from A to A i.e. A A. It follows that H E/I ( +1) A H E/I (). (Case 2). Suppose m(u) =n. ThenI +1 = M +1, H E/I ( + 1) = 0 an the claim is trivially true. For a grae ieal I = j 0 I j such that I s 0. in E,theinitial egree of I, enote by ineg(i), is the minimum s 371

7 Proposition 4.3 Let I E be a revlex ieal with initial egree <n 2. Then H E/I ( +1) H E/I (), for all. Proof. If >, then the claim follows from Proposition 4.1 an Proposition 4.2. So suppose =. If H E/I () ( n n 2 ),thenhe/i ( +1) H E/I () by Proposition 4.2. So assume H E/I () > ( n ( ) n 2 ). It follows that u = en (+1) e n 3 e n 2 / M(I ). Let v = e i1 e i be the smallest monomial in G(I) ofegree, then v>u implies m(v) m(u) an therefore m(v) / {n 1,n}. Consier w ve n.sincei +1 is a revlex segment then w M(I +1 ). Set A = {u M +1 : u <ve n }. We have that H E/I ( +1) A an if u A,thenm(u )=n. Set u = v e n. From u = v e n <ve n it follows that v <v an v M \ M(I ). This shows that there is an injection from A to A = M \ M(I ). Hence H E/I ( +1) A A = H E/I (). 5. Grae Betti numbers In this section we stuy the grae Betti numbers of revlex ieals in the exterior algebra E = K e 1,..., e n. Let I E be a monomial ieal an 1 i n. We efine the following sets: G(I; i) ={u G(I) : m(u) =i}, m i (I) = G(I; i), m i (I) = j i m j (I). With every subset of monomials of E we can associate a ecomposition as follows. Let M be a set of monomials of egree of E.Wehave M = M 0 M 1 e n, where M 0 is the set of all monomials u M such that m(u) n 1anM 1 is the set of all monomials w E of egree 1 with m(w) n 1 such that we n M. Such a ecomposition will be calle the e n ecomposition of M, an will be enote by {M 0, M 1 }. Example 5.1 Let E = K e 1,e 2,e 3,e 4,e 5. Let M = {e 1 e 2 e 3,e 1 e 2 e 4,e 1 e 3 e 4,e 2 e 3 e 4,e 1 e 2 e 5, e 1 e 3 e 5,e 2 e 3 e 5 }. Then the e 5 -ecomposition of M is M = M 0 M 1 e 5 where M 0 = {e 1 e 2 e 3,e 1 e 2 e 4,e 1 e 3 e 4,e 2 e 3 e 4 } an M 1 = {e 1 e 2,e 1 e 3,e 2 e 3 }. Remark 5.2 Let M be a set of monomials of egree of E an {M 0, M 1 } be the e n -ecomposition of M. We can observe that if i<n,thenm i (M) =m i (M 0 ). In particular, m n 1 (M) = M 0. Moreover, m n (M) = M. We also have m (M) {0, 1}, since the only monomial u of egree such that m(u) = is u = e 1 e. Moreover m i (M) =0,fori<. 372

8 Theorem 5.3 Let J E be a revlex ieal generate in egree an let I E be a strongly stable ieal generate in the same egree, such that im K J im K I.Then m i (J) m i (I), for 1 i n. Proof. The proof will procee by inuction on n. Suppose n>3 an assume that the assertion is true for n 1. We have m n (J) =im K J im K I = m n (I). So suppose i = n 1 anlet{m 0, M 1 } be the e n ecomposition of G(J) an{n 0, N 1 } the e n ecomposition of G(I). If M 1 =, thenwehave M 0 =im K J im K I N 0 an so m n 1 (J) = M 0 N 0 = m n 1 (I). If M 1, sincej is a revlex ieal generate in egree, then M 0 contains all monomials w of egree such that m(w) n 1. Hence M 0 N 0 an so m n 1 (J) = M 0 N 0 = m n 1 (I). Suppose i<n 1. From the n 1casewehave M 0 N 0. Since M 0 is a revlex segment an N 0 is strongly stable, by the inuction hypothesis m i (M 0 ) m i (N 0 ), for 1 i n 1, an therefore m i (J) =m i (M 0 ) m i (N 0 )=m i (I). Example 5.4 Let E = K e 1,e 2,e 3,e 4,e 5. Let J =(e 1 e 2 e 3,e 1 e 2 e 4,e 1 e 3 e 4,e 2 e 3 e 4,e 1 e 2 e 5,e 1 e 3 e 5,e 2 e 3 e 5 ) be a revlex ieal of egree 3, an I =(e 1 e 2 e 3,e 1 e 2 e 4,e 1 e 3 e 4,e 1 e 2 e 5,e 1 e 3 e 5 ) a strongly stable ieal generate in egree 3. We have m i (J) =m i (I) =0,fori =1, 2; m 3 (J) =1=m 3 (I); m 4 (J) =4 m 4 (I) =3 an m 5 (J) =7 m 5 (I) =5. Lemma 5.5 Let I E be a strongly stable ieal generate in egree an let I +1 be the ieal generate by the elements of I +1.Then m i (I +1 )=m i 1 (I), for all i. Proof. See [1, Theorem 4.3]. Hence we can state the following theorem. Theorem 5.6 Let I E be a strongly stable ieal an J E a revlex ieal. Suppose that H E/J () =H E/I (), for all. Then β i,j (E/I) β i,j (E/J), for all i an j. 373

9 Proof. The proof is similar to that of [1, Theorem 4.4]. Set G(I) = {u G(I) :egu = }. From Theorem 2.2, we have β i,j+i (E/I) = u G(I) j+1 m(u)+i 2, for all i 1. m(u) 1 Since G(I) j+1 = G(I j+1 ) G(I j ){e 1,...,e n } we may write the above sum as a ifference β i,j+i (E/I) = C D, with C = = = u G(I j+1 ) n u G(I j+1 :t) m(u)+i 2 m(u) 1 = n m t (I j+1 ) n (m t (I j+1 ) m t 1 (I j+1 )) n + i 2 = m n (I j+1 ) + n 1 [ ] t + i 2 (t +1)+i 2 (m t (I j+1 )) t n 1 + n 1 n + i 2 = m n (I j+1 ) (m t (I j+1 )) n 1 t an D = u G(I j ){e 1,...,e n} m(u)+i 2 m(u) 1 n 1 = (m t 1 (I j+1 )), t=2 from Lemma 5.5. The number of generators of I j+1 an J j+1 are equal for all j,thenm n (I j+1 )= m n (J j+1 ), an it follows from Theorem 5.3 that m i (J j+1 ) m i (I j+1 ) for all i. Therefore if we compare the above expressions we have: 374

10 n 1 n + i 2 β i,j+i (E/I) =m n (I j+1 ) (m t (I j+1 )) + n 1 t n 1 (m t 1 (I j+1 )) t=2 n 1 n + i 2 m n (J j+1 ) (m t (J j+1 )) + n 1 t n 1 (m t 1 (J j+1 )) = β i,j+i (E/J). t=2 The require inequality follows. Acknowlegements We thank the referee for his careful reaing an useful comments. References [1] Aramova, A., Herzog, J., Hibi, T.: Gotzman Theorems for Exterior algebra an combinatorics. J. of Algebra 191, (1997). [2] Aramova, A., Herzog, J., Hibi, T.: Squarefree lexsegment ieals. Math.Z. 228, (1998). [3] Bigatti, A. : Upper bouns for the Betti numbers of a given Hilbert function. Comm. in Algebra 21, (1993). [4] Bruns, W., Herzog, J.: Cohen-Macaulay rings 39, Revise Eition, Cambrige [5] Deery, T.: Revlex segment ieals an minimal Betti numbers. Queen s Papers 129, (1996). [6] Hulett, H.: Maximum Betti numbers for a given Hilbert function. Comm. in Algebra 21, (1993). [7] Parue, K.: Deformation classes of grae moules an maximal Betti numbers Ill. J. of Math. 40, N. 4, (1996). Marilena CRUPI, Carmela FERRÓ University of Messina, Department of Mathematics Viale Ferinano Stagno Alcontres, Messina-ITALY s: mcrupi@unime.it, cferro@unime.it Receive: