Journal of Algebra 374 (2013) Contents lists available at SciVerse ScienceDirect. Journal of Algebra.

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1 Journal of Algebra 374 (03) 3 Contents lists available at SciVerse ScienceDirect Journal of Algebra Macaulay Lex rings Abed Abedelfatah Department of Mathematics, University of Haifa, Mount Carmel, Haifa 3905, Israel article info abstract Article history: Received 6 April 0 Availableonline5November0 Communicated by Bernd Ulrich Keywords: Hilbert function Lex ideals Piecewise lexsegment ideals Macaulay Lex ideals We present simple proofs of Macaulay s theorem and Clements Lindström s theorem. We generalize Shakin s theorem by proving that a stable ideal I of S is Macaulay Lex if and only if I is a piecewise lexsegment ideal. We also study Macaulay Lex idealsoftheform x e, xt n xen n, where e e n and t i < e i for all i, and generalize Clements Lindström s theorem. 0 Elsevier Inc. All rights reserved.. Introduction Let S = k[x,...,x n ] be a polynomial ring over a field k and I = d 0 I d be a graded ideal of S. The function Hilb I : N {0} N {0}, Hilb I (d) = dim k I d = I d is called the Hilbert function of I. In[5], Macaulay proved that if I is a graded ideal in S, then there exists a lex ideal L of S such that Hilb I = Hilb L, i.e., every Hilbert function of a graded ideal in S is attained by a lex ideal of S. LetM be a monomial ideal of S. WesaythatM and S/M are Macaulay Lex if every Hilbert function of a graded ideal in S/M is attained by a lex ideal of S/M. In [], Clements and Lindström proved that the ideal x e,...,xe n n is Macaulay Lex for e e n (where x i = 0). In the case e = = e n =, the result was obtained earlier by Katona [3] and Kruskal [4]. In[8], Shakin proved that a strongly stable ideal B of S is Macaulay Lex if and only if B is a piecewise lexsegment ideal. address: abedelfatah@gmail.com /$ see front matter 0 Elsevier Inc. All rights reserved.

2 A. Abedelfatah / Journal of Algebra 374 (03) 3 3 In Section, we present simple proofs of Macaulay s theorem and Clements Lindström s theorem. In Section 3, we generalize Shakin s theorem by proving that a stable ideal I of S is Macaulay Lex if and only if I is a piecewise lexsegment ideal. In Section 4, we study Macaulay Lex ideals of the form e x, xt n, where e e n and t i < e i for all i, and generalize Clements Lindström s theorem by proving the following theorem. Theorem.. Let I = x e, xt n, where e e n,t i < e i for all iandx = 0. I is Macaulay Lex if and only if I has one of the following forms: i (a) I = x e,...,xt xe n (b) I = x e x e, xe 3,...,xe xt n. Macaulay s and Clements Lindström s theorems First, we recall some definitions and notations from []. LetS = k[x,...,x n ] be a polynomial ring over a field k. S = d 0 S d is standard graded by deg(x i ) = forall i n. ForanysetA S, we denote by Mon(A) the set of all monomials of A. For any monomial x a = x a n,wedefinethe degree of x a to be deg(x a ) = n i= a i.wedefinethegraded lexicographic order on Mon(S) by setting x a = x a n < lex x b n = x b if either deg(x a )<deg(x b ) or deg(x a ) = deg(x b ) and a i < b i for the first index i such that a i b i. For any monomial u S, wesetm(u) = max{i: x i divides u}. Ad-vector space is a k-vector subspace of S d.ad-monomial space is a d-vector space spanned by monomials. A d-monomial space V d is called lexsegment if whenever V d x a < lex x b and x b S d we have x b V d. A d-monomial space V d is called strongly stable if x i u/x j V d for all u Mon(V d ) and all i < j such that x j divides u. A d-monomial space V d is called stable if x i u/x m(u) V d for all u Mon(V d ) and all i < m(u). A monomial ideal I = d 0 I d is called a lex ideal, or a (strongly) stable monomial ideal, if I d is lexsegment, or a (strongly) stable monomial space, for all d 0. The definitions above can be generalized to a quotient ring S/M, where M is a monomial ideal in S, see[7]. IfV d is a d-monomial space in S/M, wedenotebys V d the (d + )-monomial space generated by Example.. {x i z i n, z V d and x i z / M}. The ideal I = x 3, x x, x x 3, x x is lex in k[x, x, x 3 ]. The ideal I = x 3, x x, x x is strongly stable in k[x, x, x 3 ] but not a lex ideal (note that I x x < lex x x 3,butx x 3 / I). The ideal I = x, x x, x, x x 3 is stable in k[x, x, x 3 ] but not a strongly stable ideal (note that x x 3 I, butx x 3 = x (x x 3 /x )/ I). The ideal I = x, x x, x, x x 3 is lex in k[x, x, x 3 ]/ x x 3. The following lemma is helpful to check if the ring is Macaulay Lex. Lemma.. Let M be a monomial ideal in S. The following properties are equivalent. (a) S/M is Macaulay Lex. (b) If V d is a d-monomial space in (S/M) d and L d is the lexsegment space in (S/M) d,suchthat V d = L d, then S L d S V d.

3 4 A. Abedelfatah / Journal of Algebra 374 (03) 3 Proof. Assume that (a) holds. Let V d be a d-monomial space in (S/M) d and I S/M be the ideal generated by Mon(V d ). There exists a lex ideal L = j 0 L j, such that L j = I j,forall j 0. It follows that L d = I d = V d and S L d L d+ = I d+ = S V d. Assume that (b) holds. Let J = d 0 V d be a graded ideal in S/M. By Gröbner basis theory, we can assume that J is a monomial ideal. For each d 0, let L d be the lexsegment space in (S/M) d,such that L d = V d.since S L d S V d L d+ for all d 0, it follows that L = j 0 L j is an ideal. Clearly, it is a lex ideal and has the same Hilbert function as J. Definition.3. (See [6, Definition 3.].) Let M be any monomial ideal in S, R = S/M and i {,...,n}. Clearly, if V d is a d-monomial space of R, thenv d can be written uniquely in the form V d = 0 j d x d j W i j where W j is a j-monomial space in the ring S/(M + x i ). We say that V d is x i -compressed if each W j is lexsegment in S/(M + x i ). If K j is the lexsegment space in S/(M + x i ) such that K j = W j and L d = 0 j d xd j i K j, we say that L d is the x i -compression of V d. We say that a d-monomial space of R is compressed if it is x i -compressed for all i n. Example.4. The space V = (x 3, x x, x x 3, x x, x3 ) is compressed in k[x, x, x 3 ] 3. Note that V = ( ) ( ) x 3 x x x (x, x 3 ) x 3 () = ( x 3, ) ( ) x x 3 x x x (x ) x 3 () = ( x 3, x x, x x, ) ( x3 x3 x ). Let M be a monomial ideal. For any z Mon(S/M), wedenotebyt z the set T z = { x i z i m(z) and xi z / M }. Lemma.5. Let V d be a d-monomial space in R = S/M, where M is a monomial ideal in S. (a) If V d is a stable space in R d,then S V d = z Mon(V d ) T z. (b) If V d is a compressed space and n 3,thenV d is strongly stable. Proof. First we prove (a). It is sufficient to show that Mon(S V d ) = Mon ( z Mon(V d ) Let z Mon(V d ) such that x i z Mon(S V d ).Ifi m(z), thenx i z T z. Assume that i < m(z). Itis clear that w = x i z/x m(z) / M. SinceV d is stable, we have w = x i z/x m(z) Mon(V d ).So x i z = x m(z) w T w. We show that the union is disjoint. Let x i z T z and x j z T z such that z z and x i z = x j z. Assume that m(z )<m(z ).So j m(z )>m(z ).Sincex j x i z, it follows that i = j, and so z = z, T z ).

4 A. Abedelfatah / Journal of Algebra 374 (03) 3 5 a contradiction. Similarly, m(z ) m(z ). We conclude that m(z ) = m(z ).Buttheni = j and z = z, a contradiction. Now we prove (b). Let z V d and w = x i z/x j Mon(S/M), where i < j and x j divides w. There exists an index k n such that k i, j. Sincew > lex z, w and z have the same k-exponent and V d is compressed, it follows that w V d. A special case of the previous lemma, when M = x e,...,xe n n, was proved by Mermin and Peeva (Lemma 3.6 of [6]). We need the following lemma, which is very useful. Lemma.6. For every n, letf n be a class of monomial ideals in k[x,...,x n ]. Assume that the classes satisfying the following properties: (a) I is Macaulay Lex for all I F. (b) If I F n,n 3 and i {,...,n}, then there exists J F n such that k[x,...,x n ]/ ( I + x i ) = k[x,...,x n ]/ J. (c) For every I F n and a compressed space V d in (k[x,...,x n ]/I) d,wheren 3, ifz, w Mon((k[x,...,x n ]/I) d ) such that z > lex w V d and z / V d,then T z T w. Then, for all n, everyi F n is Macaulay Lex. Proof. We prove the lemma by induction on n. Letn >. Let I F n and V d be a d-monomial space in k[x,...,x n ]/I. By Lemma., it suffices to prove that S L d S V d, where L d is the lexsegment space in (k[x,...,x n ]/I) d such that L d = V d. We use an argument similar to that in the proof of [7, Lemma 3.], to assume that V d is compressed. Set V d (0) = V d.foreach i n, letv d (i) be the x i -compression of V d (i ). Obviously, V d (i) = V d (i ) for all i n. We prove that S V d (i) S V d (i ), forall i n. Assume that V d (i ) = 0 j d xd j i W j, where W j is a j-monomial space in k[x,...,x n ]/(I + x i ) and V d (i) = 0 j d xd j K i j.lets = k[x,...,ˆx i,...,x n ], where ˆx i means that x i is omitted. We have S V d (i) = x d+ i K 0 0 j d x d j i (S K j + K j+ ) S K d. By [6, Proposition.5], S K j and K j+ are lexsegments. So we have S V d (i) d = K 0 + { max S K j, K j+ } + S K d. j=0 By (b) and the inductive step we obtain that S K j S W j.then S V d (i) d W 0 + { max S W j, W j+ } + S W d j=0 d W 0 + S W j + W j+ + S W d j=0 = S V d (i ).

5 6 A. Abedelfatah / Journal of Algebra 374 (03) 3 In particular, we obtain that V d = V d (n) and S V d (n) S V d. So we may assume that V d is a compressed space. If L d = V d, then we are done. Otherwise, we denote by z the greatest monomial in L d such that z / V d and w the smallest monomial in V d.wealsodenotebym the vector space spanned by Mon(V d ) \{w} {z}. By (c) and Lemma.5, wehave S M = S V d T w + T z S V d. By repeating this argument, we obtain that S L d S V d. Lemma.7. Let M be a monomial ideal in S and V d a compressed space in (S/M) d.ifz, w Mon((S/M) d ) such that z = x a n > lex w = x b n V d and z / V d,thena i b i for all i. Proof. Assume, on the contrary, that a i = b i z V d, a contradiction. for some i. SinceV d is x i -compressed, we obtain that Lemma.8. Let I = x e, xt n, where e e n is a sequence of integers or, x i = 0, n 3 and t i < e i for all i. Assume that V d is a compressed space in (S/I) d and z, w Mon((S/I) d ) such that z = x a n > lex w = x b n V d and z / V d.then: (a) m(z) = n. (b) If t i = 0 for all i n,thena n > b n. Proof. We show that m(z) m(w). Assume, on the contrary, that m(z) <m(w). By Lemma.7, we have m(w) = n. Letz = x a xa s s, where a s 0 and s < n. We consider the following cases: Case. b = 0. So a 0, since V d is x -compressed. Since a < e e n, we obtain that u = x b xa xa s s x a b n is a monomial in S/I. Clearly,u > lex w.sincev d is x -compressed, we have u V d. Since V d is strongly stable, we obtain that z = x a b u/x a b n V d, a contradiction. Case. b 0 and b n < e n. Note that zx b n n /v / V d, for every monomial v k[x,...,x n ] of degree b n such that v divides z. By Lemma.7, a > b. So we can obtain a monomial u / V d of the form u = x b +c x c xc s s x b n n, where c > 0, 0 c j a j for j s and u S d.sincev d is x n -compressed, we have u V d, a contradiction. Case 3. b 0 and a i < t i for some i. The argument is similar to that in Case. Case 4. b 0, b n e n and a i t i for all i. Ifa < b,...,a s < b s,then b + +b s + b n > a + +a s + e n > a + +a s, a contradiction to our assumption that deg(z) = deg(w) = d. So there exists j s, such that a j b j.sincea i t i for all i, it follows that a j e j < e n. We obtain that u = zx a j b j n /x a j b j j is a monomial in S/I. Clearly,u > lex w.sincev d is x j -compressed, we have u V d.sincev d is a strongly stable space, we obtain that z = x a j b j j u/x a j b j n V d, a contradiction. This implies that m(z) m(w). SinceV d is x n -compressed, we conclude that m(z) = n. Now, we prove (b). By Lemma.7, a n b n. Assume, on the contrary, a n < b n. If b < t, then wx b n a n /x b n a n n V d is a monomial in S/I. SinceV d is x n -compressed, it follows that z V d, a contradiction. So, we may assume that b t and then b n < e n.

6 A. Abedelfatah / Journal of Algebra 374 (03) 3 7 If w = x b xb n n,thenz = wv/x b n a n n for some monomial v k[x,...,x n ] of degree b n a n.since V d is stable, we have z V d, a contradiction. So, we may assume that b k 0, for some k n. As before, we get a monomial u / V d of the form u = x b +c x c xc n n xb n n, where c > 0, 0 c j a j for j n and u S d.sincev d is x n -compressed, we have u V d, a contradiction. Theorem.9 (Macaulay). For every graded ideal J in S there exists a lex ideal L with the same Hilbert function. Proof. We apply Lemma.6 to {F n } n, where F n ={(0)}. We prove that the zero ideal (0) is Macaulay Lex in S = k[x, x ]. Let V d be a d-monomial space in S and L d be the lexsegment in S d such that V d = L d. By Lemma.5, S L d = L d +. It is clear that S V d V d +, so S V d S L d. Then the class F satisfies (a). It is clear that the classes F n, where n 3, satisfy (b). It remains to check the property (c). Let V d be a compressed space in (S/I) d, where S = k[x,...,x n ], n 3 and I F n. Assume that z, w Mon((S/I) d ) such that z > lex w V d and z / V d. We need to show that T z =n m(z) + T w =n m(w) +, i.e., m(z) m(w). This follows from Lemma.8. Theorem.0 (Clements Lindström). Let e e n be a sequence of integers or. Assume that I = x e,...,xe n n,wherex = 0. Then for every graded ideal J in S/I there exists a lex ideal L in S/I withthe i same Hilbert function. Proof. We apply Lemma.6 to {F n } n, where F n ={ x e,...,xe n n e e n }. First, we prove that the ideal I = x e, xe in S = k[x, x ] is Macaulay Lex. Let V d = (w,...,w s ) be a d-monomial space in S and L d = (m,...,m s ) be the lexsegment in S d such that V d = L d, where w i > lex w i+ and m i > lex m i+ for all i. Ifd < e, then d < e and S V d V d + = L d + = S L d. So we may assume that d e. By Lemma.5, every monomial in S L d is of the form x m i, for some i. Wedefinethefunction f : Mon(S L d ) Mon(S V d ) by { x w i, if m i = w i, f (x m i ) = x w i, if m i w i. We show that x w i is a monomial in S V d if m i w i and x m i Mon(S L d ).Otherwise,wehave x w i x e, xe and so x w i x e.thenx z x e for every monomial z > lex w i.sox m i x e, a contradiction. Clearly, f is injective. Then S L d S V d. This proves the case n =. The property (b) follows from the following isomorphisms k[x,...,x n ] x e,...,xe n n + x i = k[x,...,x n ] x e,...,xe i i, xe i+ i+,...,xe n n + x i = k[x,...,x i, x i+, x n ] x e,...,xe i i, xe i+ i+,...,xe n = x e,...,xe i k[x,...,x n ] i, xe i+ i n,...,x e n n. It remains to check the property (c). Let V d be a compressed space in (S/I) d, where S = k[x,...,x n ], n 3 and I F n. Assume that z, w Mon((S/I) d ) such that z > lex w V d and z / V d. We need to show that T z T w. By Lemma.8, m(z) = n. SoT z = or T z ={x n z}. Assume that T z. So,x n z is a monomial in S/I. By Lemma.8, x n w is also a monomial in S/I. This show that T z T w.

7 8 A. Abedelfatah / Journal of Algebra 374 (03) 3 3. Piecewise lexsegment ideals Definition 3.. Let I be a monomial ideal in S. Denote by G(I) the minimal generators of I. We say that I is piecewise lexsegment if whenever z > lex w G(I), deg(z) = deg(w) and m(z) m(w), wehave z I. It is clear that every piecewise lexsegment ideal is strongly stable. The proof of the next lemma is similar to that in Proposition 3.6 of [8]. Lemma 3.. Let I be a stable ideal in S and x n w Mon(I). Then either w Iorx n w G(I). In this section, we prove that if I is a stable ideal, then I is Macaulay Lex if and only if I is a piecewise lexsegment ideal. In [8], Shakin proved this statement for a strongly stable ideal. Lemma 3.3. Let I be a piecewise lexsegment ideal in S. Assume that V d is a compressed space in (S/I) d,n 3 and z, w Mon((S/I) d ) such that z = x a n > lex w = x b n V d and z / V d.then T z T w. Proof. We show that m(z) m(w). Assume, on the contrary, m(z) <m(w). Soa n = 0. Since V d is x n -compressed, we have b n 0, so m(w) = n. We consider the following cases: Case. b = 0. So a 0, since V d is x -compressed. Also a 0, since V d is x -compressed. Let v = zx n /x. If v I, then by Lemma 3., v G(I). Since m(z) m(v) and z > lex v, we have z I, a contradiction. Clearly, v / V d. After finitely many steps in this way, we obtain that u = x b xa xa s s x a b n / V d is a monomial in S/I. Clearly,u > lex w.sincev d is x -compressed, we have u V d, a contradiction. Case. b 0. A similar argument to that in Case shows that zx n /x i S/I, foralli < n such that x i divides z. We obtain that zx b n n /v / V d, for every monomial v k[x,...,x n ] of degree b n such that v divides z. So we can obtain a monomial u / V d of the form u = x b +c x c xc s s x b n n, where c > 0, 0 c j a j for j s and u S d.sincev d is x n -compressed, we have u V d, a contradiction. So m(z) m(w), and then m(z) = n. We obtain that T z = or T z ={x n z}. Assume that T z and x n w I. By Lemma 3., wehavex n w G(I). SinceI is a piecewise lexsegment ideal, it follows that x n z I, a contradiction. Theorem 3.4 (Shakin). If I is a piecewise lexsegment ideal, then I is Macaulay Lex. Proof. For all n, let F n to be the class of all piecewise lexsegment ideals in S = k[x,...,x n ].We apply Lemma.6 to {F n } n. Since every piecewise lexsegment ideal in k[x, x ] is lex, it follows that every I F is Macaulay Lex, by Theorem.9. Thisproves(a)ofLemma.6. Let I be a piecewise lexsegment ideal in S = k[x,...,x n ], i n and A i ={z G(I) x i z}. Then I + x i = J + x i, where J is the monomial ideal in S/x i generated by A i.clearly, J is a piecewise lexsegment ideal in S/ x i.thens/(i + x i ) = k[x,...,x n ]/ J for some J F n.thisproves(b)of Lemma.6. The property (c) follows from Lemma 3.3. Proposition 3.5. Assume that I is a Macaulay Lex and stable ideal. Then I is piecewise lexsegment ideal. Proof. Assume, on the contrary, that I is not a piecewise lexsegment. So there exists a minimal generator w I d of I and a smallest monomial z S d, such that w < lex z and m(z) m(w), butz / I. Let w = x b xb l, where b l l > 0. If m(z)<l, thenzx m(z)+ /x m(z) I and so z I, a contradiction. So, we may assume that z = x a xa l, where a l l > 0. Since w is a minimal generator, we have z/x l, w/x l / I.

8 A. Abedelfatah / Journal of Algebra 374 (03) 3 9 Let k be the minimal integer such that a k > b k,clearlyk < l. LetA be the subspace generated by {v Mon((S/I) d ) v > lex z/x l }.DenotebyL and V the subspaces generated by Mon(A) {z/x l } and Mon(A) {w/x l }, respectively. By the proof of Lemma.5 and the minimality of z, wehave Mon(S A) = { v Mon ( (S/I) d ) v >lex x t z/x l } where t = m(z/x l ).Sincew I, we obtain that x i w/x l I for all i l. So Mon(S V ) = Mon(S A) {x j w/x l j > l and x j w/x l / I}. We show that the map f :{x j w/x l j > l and x j w/x l / I} T z/xl \{z} defined by f (x j w/x l ) = x j z/x l is well defined. If x j z/x l = u I for some j > l, then, since I is stable, we get z = x l u/x j I, a contradiction. Clearly, f is injective. So we have S L = S A + T z/xl \{z} + {z} > S A + T z/xl \{z} S A + {x j w/x l j > l and x j w/x l / I} = S V. Since I is Macaulay Lex, we have S L S V, a contradiction. Combining Theorem 3.4 and Proposition 3.5, we obtain the following theorem. Theorem 3.6. Let I be a stable ideal in S. Then I is Macaulay Lex if and only if I is piecewise lexsegment. 4. A generalization of Clements Lindström s theorem In this section, we study Macaulay Lex ideals of the form I = x e, xt n where e e n and t i < e i for all i. We begin with the following lemma. Lemma 4.. Let I = x e,...,xt n, where e e n and t i < e i for all i. If I is Macaulay Lex, then I has one of the following forms: (a) I = x e,...,xt xe n (b) I = x e x e, xe 3,...,xe xt n Proof. Let s = max{i e i < }. Ifs =, then I = x e. So we may assume that s >. Assume that 0 t < e, and t k 0 for some k < s. Let z = x t s + i= x (e i ) s, w = x t x e s s and A ={v Mon(S/I) v lex z}. Clearly, z, w Mon(S/I). DenotebyL and V the spaces generated by A and A \{z} {w}, respectively. Note that wx i /x s I, forany i < s, and wx /x s > lex z. We conclude that L and V are stable spaces. Clearly, x s w / T w and T z ={x s z,...,x n z}. So

9 30 A. Abedelfatah / Journal of Algebra 374 (03) 3 S V = S L T z + T w < S L, a contradiction. We conclude that t k = 0, for all k < s. ThenI = x e,...,xt xe n For a monomial z and a d-monomial space V d,wedenotebyz > lex V d the property that z > lex w, for all w V d.westarttoprovetheconverseoflemma4.. Lemma 4.. The ideal I = x e xe of S = k[x, x ],where e e and t < e, is Macaulay Lex. Proof. Let V d = (w,...,w s ) be a d-monomial space in S/I and L d = (m,...,m s ) be the lexsegment in (S/I) d such that V d = L d, where w i > lex w i+ and m i > lex m i+ for all i. We consider the following cases of d. Case. d < e. Since e e, we obtain that S V d V d + = L d + = S L d. Case. e d < e + t. The function f : Mon(S L d ) Mon(S V d ) defined by f (x m j ) = x w j is well defined and injective. So S V d S L d. Case 3. d = e + t. Define the function f : Mon(S L d ) Mon(S V d ) by x w j, if m j = w j, f (x m j ) = x w j, x w j, if w j lex x t m j w j, if w j < lex x t m j w j. If w j lex x t and m j w j,thenx w j Mon(S/I). Otherwise, we obtain that x m j I, a contradiction. Similarly, if w j < lex x t and m j w j,thenx w j Mon(S/I). This show that f is well defined. Note that f is injective. So S V d S L d. Case 4. d e + t. Definethefunction f : Mon(S L d ) Mon(S V d ) by x w j, if m j = w j, f (x m j ) = x w j, if w j > lex x t d and m j w j, x w j, if w j < lex x t d and m j w j. As before f is well defined and injective. So S V d S L d. Lemma 4.3. The ideal I = x e x e, xe x t 3,...,xe x t xt n n, where e e n and t i < e i for all i, is Macaulay Lex. Proof. For every n, let F n be the class of all ideals in k[x,...,x n ] of the form e x x e, xe x t 3,...,xe x t xt n n.

10 A. Abedelfatah / Journal of Algebra 374 (03) 3 3 We apply Lemma.6 to {F n } n. The property (a) follows from Lemma 4.. Wechecktheproperty(b). Let I F n, where n 3, and i {,...,n}. Ifi = thens/i + x = k[x,...,x n ]. In this case, we take J = (0) F n (e = ). If i >, then we take J = x e x e,...,xe x t xt i i xe i i. We prove the property (c). Let z = x a n and w = x b n such as in Lemma.6. By Lemma.8 and Lemma.7, we obtain that m(z) = n and a > b, respectively. It follows that x n w / I. So T z T w. Lemma 4.4. The ideal I = x e, xt,...,xt xe n n, where e e n and t < e, is Macaulay Lex. Proof. For all n, let F n be the class of all ideals in k[x,...,x n ] of the form x e xe,...,xt xe n We apply Lemma.6 to {F n } n. We prove the property (c). Let z = x a n and w = x b n such as in Lemma.6. By Lemma.8, m(z) = n. SoT z = or T z ={x n z}. Assume that T z. So,x n z is a monomial in S/I. By Lemma.8, a n > b n.ifx n w I, thenb t and b n = e n. So a > t, and a n > e n. It follows that z I, a contradiction. So x n w is also a monomial in S/I. This show that T z T w. By combining the whole lemmas in this section, we have the following result. Theorem 4.5. Let I = x e, xt n, where e e n and t i < e i for all i. I is Macaulay Lex if and only if I has one of the following forms: (a) I = x e,...,xt xe n (b) I = x e x e, xe 3,...,xe xt n Acknowledgment I would like to thank the referee for his attention, comments and suggestions. References [] G.F. Clements, B. Lindström, A generalization of a combinatorial theorem of Macaulay, J. Combin. Theory 7 (3) (969) [] J. Herzog, T. Hibi, Monomial Ideals, Springer, 0. [3] G.O.H. Katona, A theorem of finite sets, in: Theory of Graphs, 968, pp [4] J.B. Kruskal, The number of simplices in a complex, in: Mathematical Optimization Techniques, 963, p. 5. [5] F. Macaulay, Some properties of enumeration in the theory of modular systems, Proc. Lond. Math. Soc. () (97) 53. [6] J. Mermin, I. Peeva, Lexifying ideals, Math. Res. Lett. 3 (/3) (006) 409. [7] J. Mermin, I. Peeva, Hilbert functions and lex ideals, J. Algebra 33 () (007) [8] D.A. Shakin, Piecewise lexsegment ideals, Sb. Math. 94 (003) 70.