Some algebraic invariants of mixed product ideals
aa r X i v : . [ m a t h . A C ] N ov SOME ALGEBRAIC INVARIANTS RELATED TO MIXEDPRODUCT IDEALS
CRISTODOR IONESCU AND GIANCARLO RINALDO
Abstract.
We compute some algebraic invariants (e.g. depth, Casteln-uovo - Mumford regularity) for a special class of monomial ideals, namelythe ideals of mixed products. As a consequence, we characterize theCohen-Macaulay ideals of mixed products. Introduction and preliminaries
The class of ideals of mixed products is a special class of square-freemonomial ideals. They were first introduced by Restuccia and Villarreal (see[6] and [7]), who studied the normality of such ideals. They gave a completeclassification of normal mixed products ideals, as well as applications ingraph theory.Let S = K [ x , y ] be a polynomial ring over a field K in two disjoint setsof variables x = ( x , . . . , x n ), y = ( y , . . . , y m ). The ideals of mixed products are the ideals I q J r + I s J t , q, r, s, t ∈ N , q + r = s + t, where I l (resp. J p ) is the ideal of S generated by all the square-free mono-mials of degree l (resp. p ) in the variables x (resp. y ). We set I = J = S .By symmetry, essentially there are 2 cases: i ) L = I k J r + I s J t , ≤ k < s,ii ) L = I k J r , k ≥ r ≥ . Our aim is to investigate some algebraic invariants of this type of ideals, suchas depth and Castelnuovo-Mumford regularity. One case is already known.Namely in [3], Herzog and Hibi proved the case m = 0 (see Proposition 2.2).In the general case, we use also some techniques used by Herzog, Restucciaand the second author in the paper [4]. Note that our results are true ina slightly more general situation than the one considered in [6], namely wedon’t use the condition q + r = s + t. All over the paper we shall use the following notation.
Notation 1.1.
Let K be a field and K [ x , y ] = K [ x , . . . , x n , y , . . . , y m ] thepolynomial ring in n + m variables over K. By I k we shall mean the monomialideal generated by all the square-free monomials of degree k in the variables The first author was supported by the CNCSIS grant ID-PCE no. 51/2007. x , . . . , x n and by J l we shall mean the monomial ideal generated by allthe square-free monomials of degree l in the variables y , . . . , y m . For amonomial ideal I , we shall denote by G ( I ) the minimal monomial system ofgenerators of I. We recall the following
Definition 1.2.
Let I = ( x α , . . . , x α q ) ∈ K [ x ] = K [ x , . . . , x n ] be a square-free monomial ideal, with α i = ( α i , . . . , α i n ) ∈ { , } n . The Alexander dual of I is the ideal I ∗ = q \ i =1 m i , where m i = ( x j : α i j = 1) . Remark 1.3.
It is easy to see that I k = I ∗ n − k +1 , k = 1 , . . . , n. It follows that I k = ∩ ( x i , . . . , x i n − k +1 ) , where the intersection is taken over all subsets with n − k + 1 elements { i , . . . , i n − k +1 } ⊆ { , . . . , n } . Hencedim( S/I k ) = k − . Regularity of ideals of mixed products
In this section we want to study the Castelnuovo-Mumford regularity ofthe mixed product ideal I q J r + I s J t .We remind that, if F is the minimal graded free resolution of a givengraded finitely generated S − module M F : 0 → b g ⊕ i =1 S ( − d g i ) φ g → · · · → b k ⊕ i =1 S ( − d k i ) φ k → · · · → b ⊕ i =1 S ( − d i ) φ → M → , then the Castelnuovo-Mumford regularity of M is given byreg( M ) := max( d k i − k | k > , i = 1 , . . . , b k ) . If J ⊆ K [ z , . . . , z s ] is a graded ideal, if all the generators of J have the samedegree d and also reg J = d , then J has a d-linear free resolution .We recall the following (see for example [5, Th. 5.56]). Theorem 2.1 (Eagon-Reiner Theorem) . Let I be a square-free monomialideal. Then S/I is Cohen-Macaulay if and only if I ∗ has a linear free reso-lution. Proposition 2.2.
Suppose that m = 0 , so that S = K [ x , . . . , x n ] . Then: a) S/I k is Cohen-Macaulay for every k ; b) reg( I k ) = k . c) pd( S/I k ) = n − k + 1 . OME ALGEBRAIC INVARIANTS RELATED TO MIXED PRODUCT IDEALS 3
Proof : a) See [3, Ex. 2.2].b) By Remark 1.3 we have dim
S/I k = k −
1. Therefore, by [5, Theorem5.59] and a) we getreg( I k ) = pd( S n /I ∗ k ) = pd( S n /I n − k +1 ) = k. c) pd( S/I k ) = reg( S/I ∗ k ) = reg( S/I n − k +1 ) = n − k + 1 . Lemma 2.3.
Let I and J be graded ideals in S such that Tor S ( S/I, S/J ) =0 . Assume that reg( I ) = q and reg( J ) = r. Then reg( I + J ) = q + r − . Proof : Let F = ( F i ) and G = ( G j ) be the graded minimal free resolutionsof S/I, resp.
S/J and consider the tensor product of complexes F ⊗ G , i.e.( F ⊗ G ) k = ⊕ i + j = k F i ⊗ G j . Then, Tor S ( S/I, S/J ) = 0 implies that F ⊗ G is the minimal free resolutionof S/ ( I + J ). Let P := X i,j β i,i + j ( S/I ) x i y i + j and Q := X i,j β i,i + j ( S/J ) x i y i + j be the graded Poincar´e series of S/I and resp.
S/J.
Then
P Q is the gradedPoincar´e series of S/ ( I + J ) . Sincereg(
S/I ) = deg y ( P ) , reg( S/J ) = deg y ( Q )and reg( S/ ( I + J )) = deg y ( P Q ) = deg y ( P ) + deg y ( Q ) , we get thatreg( I + J ) = reg( S/ ( I + J )) + 1 = reg( S/I ) + reg(
S/J ) + 1 == reg( I ) − J ) − q + r − . Corollary 2.4. reg( I q + J r ) = q + r − . Lemma 2.5. reg( I q J r ) = q + r .Proof : We have the exact sequence0 → I q J r → I q ⊕ J r → I q + J r → . Therefore, by Corollary 2.4 and of [2, Cor. 20.19] we have q + r ≤ reg( I q J r ) ≤ max(reg( I q ⊕ J r ) , reg( I q + J r ) + 1) , and the assertion follows. CRISTODOR IONESCU AND GIANCARLO RINALDO
Remark 2.6.
We observe that, when we consider the mixed product ideal I q J r + I s J t , we may fix q < s and t < r , since otherwise one of the two summandscontains the other. Remark 2.7.
Let I be a monomial ideal and G ( I ) = { u , . . . , u q } be theset of minimal generators of I. Consider the exact sequence0 → Z ( I ) → R q φ → I → , where Z ( I ) is the first syzygy of I. We recall that if we put σ ij = u i ( u i , u j ) e u j − u i ( u i , u j ) e u i , ≤ i, j ≤ q, where φ ( e u i ) = u i , φ ( e u j ) = u j , then { σ ij } ≤ i,j ≤ q is a system of generators of Z ( I ) [2, Lemma 15.1]. For simplicity, we denote this system of generatorsof Z ( I ) by SG ( I ) . Theorem 2.8.
Suppose that q < s and t < r . Then reg( I q J r + I s J t ) = r + s − . Proof : We show first that reg( I q J r + I s J t ) ≤ r + s −
1. In fact we havethe exact sequence0 → I q J r ∩ I s J t → I q J r ⊕ I s J t → I q J r + I s J t → . From the assumption we get I q J r ∩ I s J t = I q ∩ J r ∩ I s ∩ J t = I s J r . Applying [2, Cor. 20.19] we obtainreg( I q J r + I s J t ) ≤ max(reg( I s J r ) − , reg( I q J r ⊕ I s J t )) == max( s + r − , q + r, s + t ) = r + s − . To complete the proof it is sufficient to show that there exists an element f ∈ Z ( I q J r + I s J t ), that has degree r + s − Z ( I q J r + I s J t ).Since G ( I q J r + I s J t ) = G ( I q J r ) ∪ G ( I s J t ) , we consider an element f ∈ Z ( I q J r + I s J t ), f = v ( u, v ) e u − u ( u, v ) e v ,u = x i · · · x i q y j · · · y j r ∈ G ( I q J r ) ,v = x i · · · x i s y j · · · y j t ∈ G ( I s J t ) , such that i < i < . . . < i q < i q +1 < . . . < i s and j < j < . . . < j t 1. We observe alsothat ( u, v ) has the minimal degree between the greatest common divisorsof pairs of monomials respectively in G ( I q J r ) and G ( I s J t ), and it cannotexist a syzygy of bigger degree that is a generator of Z ( I q J r + I s J t ) , sincereg( I q J r + I s J t ) ≤ r + s − . Let g be a generator of Z ( I q J r + I s J t ). Theneither g is in SG ( I q J r ) ∪ SG ( I r J s ), or g has degree r + s − f . If there exists g of degree r + s − g of degree r + s − , f has to be generated by elementsin Z ( I q J r ) ⊂ Z ( I q J r + I s J t ) and elements in Z ( I s J t ) ⊂ Z ( I q J r + I s J t ).Since Z ( I q J r ) ∩ Z ( I s J t ) = ∅ we obtain that x i q +1 · · · x i s e u ∈ Z ( I q J r ) and y j t +1 · · · y j r e v ∈ Z ( I s J t ) , that is false.3. Depth and height of ideals of mixed products In this section we want to study the Krull dimension and the depth ofthe ring S/ ( I q J r + I s J t ). As a consequence we obtain conditions for theCohen-Macaulayness of this ring. We start with an easy result. Proposition 3.1. (1) If q = 0 , then dim( S/J r ) = n + r − . (2) If r = 0 , then dim( S/I q ) = m + q − . (3) If q > and r > , then dim( S/I q J r ) = n + m − min( n − q + 1 , m − r + 1) . Proof : 1) and 2) follow immediately from Remark 1.3.3) We have to show thatheight( I q J r ) = min( n − q + 1 , m − r + 1) . We observe that I q , resp. J r , is an unmixed ideal, whose associated primeideals have height n − q + 1, resp. m − r + 1. The prime decomposition ofthe radical ideal I q J r is I q J r = I q ∩ J r = ( t \ i =1 P i ) ∩ ( s \ j =1 Q j )where Ass( S/I q ) = { P , . . . , P t } and Ass( S/J r ) = { Q , . . . , Q s } . The asser-tion follows. Theorem 3.2. Let ≤ q < s , ≤ t < r . Then (1) dim S/ ( I s + J r ) = r + s − ; (2) dim S/ ( I q J r + I s ) = n + m − min( n − q + 1 , m + n − ( r + s ) + 2) ; (3) dim S/ ( I q J r + I s J t ) = n + m − min( n − q +1 , m − t +1 , m + n − ( r + s )+2) .Proof : 1) We have S/ ( I s + J r ) ∼ = K [ x , . . . , x n ] /I s ⊗ K K [ y , . . . , y m ] /J r and the assertion follows by Remark 1.3 and by [7, Ex. 2.1.14]. CRISTODOR IONESCU AND GIANCARLO RINALDO 2) Since I q ⊃ I s , we have I q J r + I s = I q ∩ ( J r + I s ) . The assertion follows from the fact that each face ideal of J r + I s is thedisjoint union of a face ideal of I s with a face ideal of J r .3) We remark that I q J r + I s J t = I q ∩ ( J r + I s ) ∩ J t I q ⊃ I s , J t ⊃ J r . Now we continue as in case 2). Lemma 3.3. depth( S/I q J r ) ≥ q + r − .Proof : We consider the exact sequence0 → ( I q + J r ) /J r → S/J r → S/ ( I q + J r ) → . By [1, Prop. 1.2.9] we havedepth( I q + J r ) /J r ≥ min(depth( S/J r ) , depth( S/ ( I q + J r ) + 1)) . By Remark 1.3 and Lemma 2.2 we obtain that depth( S/J r ) = n + r − 1. Wealso observe that S/ ( I q + J r ) ∼ = K [ x , . . . , x n ] /I q ⊗ K K [ y , . . . , y m ] /J r . Then from [7, Th. 2.2.21] we getdepth S/ ( I q + J r ) = depth( K [ x , . . . , x n ] /I q ) + depth( K [ y , . . . , y m ] /J r ) == q − r − q + r − . Thereforedepth(( I q + J r ) /J r ) ≥ min( n + r − , q + r − 1) = q + r − . If we consider the exact sequence0 → I q /I q J r → S/I q J r → S/I q → S/I q J r ) ≥ min(depth( I q /I q J r ) , depth( S/I q )) , and since I q /I q J r = I q /I q ∩ J r ∼ = ( I q + J r ) /J r , we get depth( S/I q J r ) ≥ min( q + r − , m + q − 1) = q + r − . Theorem 3.4. (1) If q = 0 , then depth( S/J r ) = n + r − . (2) If r = 0 , then depth( S/I q ) = m + q − . (3) Suppose that q > and r > . Then depth( S/I q J r ) = q + r − . OME ALGEBRAIC INVARIANTS RELATED TO MIXED PRODUCT IDEALS 7 Proof : 1) and 2) follow immediately from the proof of 2.2.3) It remains to show that depth( S/I q J r ) ≤ q + r − 1. We prove this byinduction on r and q . Let r = 1, q = 1 . We claim thatTor m + n − ( K, S/I J ) = 0which implies the inequality. We shall construct an element[ z ] ∈ H := H m + n − ( K ( S ; x , y ) ⊗ S/I J ) , where K ( S ; x , y ) is the Koszul complex of S with respect to the sequence x , . . . , x n , y , . . . , y m .We fix the lexicographic term ordering such that x > · · · > x n > y > · · · > y m and let z := m X k =1 ( − k +1 y k ( n ^ i =1 e i ) ∧ ( ^ j = k f j ) . We want to show that z is a cycle. ∂ ( z ) = m X k =1 ( − k +1 y k ∂ [( n ^ i =1 e i ) ∧ ( ^ j = k f j )]= m X k =1 ( − k +1 y k [( − n ( n ^ i =1 e i ) ∂ ( ^ j = k f j )] + m X k =1 ( − k +1 y k [ ∂ ( n ^ i =1 e i ) ^ j = k f j ] . It is easy to see that the second sum vanishes, since x i y k ∈ I J , for all i = 1 , . . . , n, k = 1 , . . . , m. Therefore ∂ ( z ) = ( − n ( m X k =1 ( − k +1 y k n ^ i =1 e i ∂ ( ^ j = k f j )) , hence we have to show that(3.1) m X k =1 ( − k +1 y k ∂ ( ^ j = k f j ) = 0 . A summand of (3.1) is of the form( − k +1 y k ( − π ( l )+1 y l ( ^ j = k, l f j ) , where π ( l ) is the position of f l in ( V j = k f j ). We observe that if k < l then π ( l ) = l − 1, while if k > l then π ( l ) = l . We also observe that in (3.1) thereexists only one other summand( − k ′ + π ( l ′ ) y k ′ y l ′ ( ^ j = k ′ , l ′ f j ) , with l ′ = k and k ′ = l . It is easy to show that( − l + π ( k ) y l y k + ( − k + π ( l ) y k y l = 0 CRISTODOR IONESCU AND GIANCARLO RINALDO Indeed, either k < l , that is π ( k ) = k and π ( l ) = l − 1, or k > l , that is π ( k ) = k − π ( l ) = l .Now we want to show that z is not a boundary. We observe that, if wewrite z = ( e ∧ . . . ∧ e n ) ∧ ( m X k =1 ( − k +1 y k ( ^ j = k f j )) , (1) The coefficients of z are polynomials of total degree 1;(2) The multidegree of the coefficients of z with respect to x ismultideg x ( z ) = (0 , . . . , ∈ Z n . Suppose now that b is a boundary such that ∂ ( b ) = z . Then b = γe ∧ . . . ∧ e n ∧ f ∧ . . . ∧ f m , γ ∈ S and ∂ ( b ) = γ n X k =1 ( − k +1 x k ( ^ i = k e i ) ∧ ( m ^ j =1 f j )++ γ m X l =1 ( − n + l +1 y l ( n ^ i =1 e i ) ∧ ( ^ j = l f j ) . Suppose that γ = 0. From the fact that the multidegree of z with respectto x is (0 , . . . , γx i = 0 for all i = 1 , . . . , n . It follows that γx i ∈ I J and the total degree of γ is at least 1. Therefore γy l has totaldegree at least 2, for all l = 1 , . . . , m . But the coefficients of z have totaldegree 1 , therefore γ = 0.Now suppose that 1 < i ≤ m and that depth( S/I J i − ) = i − 1. We wantto show that depth( S/I J i ) = i. Let S l = K [ x , . . . , x n , y , . . . , y l ], with l = 1 , . . . , m and let L i := I J i ⊆ S l .L + j := I J j ⊆ S l +1 . We have the exact sequence0 → L + i /L i S l +1 → S l +1 /L i S l +1 → S l +1 /L + i → . We observe that L + i = L i S l +1 + y l +1 L i − S l +1 and L + i /L i S l +1 = ( L i S l +1 + y l +1 L i − S l +1 ) /L i ∼ = ∼ = y l +1 L i − S l +1 /L i S l +1 ∩ y l +1 L i − S l +1 . Since L i ⊂ L i − , y l +1 ∩ L i − S l +1 = y l +1 L i − S l +1 and y l +1 is not a zero-divisor on S l +1 , we obtain L + i /L i S l +1 ∼ = L i − S l +1 /L i S l +1 . Therefore S l +1 /L + i ∼ = S l +1 /L i − S l +1 . OME ALGEBRAIC INVARIANTS RELATED TO MIXED PRODUCT IDEALS 9 By the induction hypothesis we havedepth( S l /L i − ) = i − y l +1 is a regular element of S l +1 /L i − S l +1 we obtain the assertion.If we consider the ideal I i J r , with the same argument, by induction on i ≥ S/I q J r ) = q + r − Corollary 3.5. S/ ( I q J r ) is Cohen-Macaulay if and only if one of the fol-lowing conditions hold: (1) q = 0;(2) r = 0;(3) q > , r > , m = r and n = q. Proof : The cases q = 0 or r = 0 being clear, we can suppose that q > r > . By 3.1 we havedim( S/I q J r ) = m + n − min( n − q + 1 , m − r + 1)and by 3.4 we get depth( S/I q J r ) = q + r − . It follows that if S/ ( I q J r ) is Cohen-Macaulay then m + n − min( n − q + 1 , m − r + 1) = q + r − . Suppose that n − q + 1 ≤ m − r + 1 . Then from above we get m = r and thenclearly n = q . The case m − r + 1 ≤ n − q + 1 leads to the same condition.The converse is obvious. Lemma 3.6. Let ≤ q < s and ≤ t < r . Then (1) depth S/ ( I q J r + I s J t ) ≥ min( q + r − , s + t − S/ ( I q J r + I s ) ≥ q + r − . Proof : 1) We have the exact sequence0 → S/I q J r ∩ I s J t → S/I q J r ⊕ S/I s J t → S/ ( I q J r + I s J t ) → I q J r ∩ I s J t = I s J r (see the proof of Theorem 2.8). By [1, Prop. 1.2.9]we obtaindepth( S/ ( I q J r + I s J t )) ≥ min(depth( S/I s J r ) − , depth(( S/I q J r ) ⊕ ( S/I s J t ))) == min( s + r − , min( q + r − , s + t − q + r − , s + t − . 2) In this case we have0 → S/I s J r → S/I q J r ⊕ S/I s → S/ ( I q J r + I s ) → S/ ( I q J r + I s )) ≥ min(depth( S/I s J r ) − , depth(( S/I q J r ) ⊕ ( S/I s ))) == min( s + r − , min( q + r − , s + m − q + r − . Theorem 3.7. Let ≤ q < s , ≤ t < r . Then: (1) depth S/ ( I s + J r ) = s + r − ; (2) depth S/ ( I q J r + I s ) = q + r − S/ ( I q J r + I s J t ) = min( q + r, s + t ) − . Proof : 1) This was already proved in the proof of Lemma 3.3.2) Since by Lemma 3.6 we have that depth S/ ( I q J r + I s ) ≥ q + r − , weonly have to prove thatTor m + n − ( q + r − ( K, S/ ( I q J r + I s )) = 0 . We consider the exact sequence(3.2) 0 → S/I s J r → S/I q J r ⊕ S/I s → S/ ( I q J r + I s ) → . Let k = n + m − ( q + r − 1) and consider the long exact sequence of Torinduced by the sequence (3.2). We get the exact sequence · · · → Tor k ( K, S/I s J r ) → Tor k ( K, S/I q J r ⊕ S/I s ) φ → φ → Tor k ( K, S/ ( I q J r + I s )) → · · · By hypothesis we have s > q and by Theorem 3.4 we getpd S/I s J r = m + n − ( s + r − < k. Therefore we have that Tor k ( K, S/I s J r ) = 0, that is φ is injective. We havealso thatTor k ( K, S/I q J r ⊕ S/I s J t ) ∼ = Tor k ( K, S/I q J r ) ⊕ Tor k ( K, S/I s J t )and by Theorem 3.4 we obtain Tor k ( K, S/I q J r ) = 0. Therefore by theinjectivity of φ we obtain the assertion.3) Let us suppose that q + r ≤ s + t . Since by Lemma 3.6 we have thatdepth S/ ( I q J r + I s J t ) ≥ q + r − S/ ( I q J r + I s J t ) ≤ q + r − 1, that is equivalent to sayTor m + n − ( q + r − ( K, S/ ( I q J r + I s J t )) = 0 . We consider the exact sequence(3.3) 0 → S/I s J r → S/I q J r ⊕ S/I s J t → S/ ( I q J r + I s J t ) → . Let k = m + n − ( q + r − 1) and consider the long exact sequence of Torinduced by sequence (3.3). We get · · · → Tor k ( K, S/I s J r ) → Tor k ( K, S/I q J r ⊕ S/I s J t ) ψ → ψ → Tor k ( K, S/ ( I q J r + I s J t )) → · · · By hypothesis we have s > q and by Theorem 3.4 we getpd S/I s J r = m + n − ( s + r − < k. Therefore we have that Tor k ( K, S/I s J r ) = 0, that is ψ is injective. We havealso thatTor k ( K, S/I q J r ⊕ S/I s J t ) ∼ = Tor k ( K, S/I q J r ) ⊕ Tor k ( K, S/I s J t ) OME ALGEBRAIC INVARIANTS RELATED TO MIXED PRODUCT IDEALS 11 and by Theorem 3.4 we obtain Tor k ( K, S/I q J r ) = 0. Therefore by theinjectivity of ψ we obtain the assertion. Corollary 3.8. Let ≤ q < s , ≤ t < r . Then: (1) S/I q , S/J r and S/ ( I q + J r ) are always Cohen-Macaulay; (2) S/ ( I q J r ) is Cohen-Maculay if and only if m = r and n = q ; (3) S/ ( I q J r + I s ) is Cohen-Macaulay if and only if s = q + 1 and r = m ; (4) S/ ( I q J r + I s J t ) is Cohen-Macaulay if and only if r = m, s = n,t = m − , q = n − .Proof : (1) and (2) see Corollary 3.5.(3) Suppose that R := S/ ( I q J r + I s ) is Cohen-Macaulay. Then from 3.7 and3.2 we get n + m − min( n − q + 1 , n + m − ( r + s ) + 2) = q + r − . We have 2 cases:a) m + q ≥ r + s − 1. Then dim( R ) = m + q − R ) = q + r − . Since R is Cohen-Macaulay we get m = r . From this it follows that s ≤ q + 1and consequently s = q + 1 . b) m + q < r + s − 1. Then dim( R ) = r + s − 2. Since R is Cohen-Macaulay it follows at once that s = q + 1. Further we get m < r , which isa contradiction.The converse is obvious.(4) Let R := S/ ( I q J r + I s J t ) . and suppose that R is Cohen-Macaulay. Thenfrom 3.7 and 3.2 we have thatmin( q + r, s + t ) − n + m − min( n − q + 1 , m − t + 1 , n + m − ( r + s ) + 2) . We have to consider 2 cases:a) q + r ≤ s + t. We shall consider 3 subcases, as follows.a1) n − q + 1 ≤ m − t + 1 and n − q + 1 ≤ n + m − r − s + 2 . Since R is Cohen-Macaulay we obtain that m = r. From n − q ≤ n + m − r − s + 1it follows that q ≥ s − 1, that is q = s − . Now from s − r ≤ s + t we get r − ≤ t and consequently t = r − m − . Further we obtain immediately n = q + 1 . a2) n + m − r − s + 2 ≤ m − t + 1 and n + m − r − s + 2 ≤ n − q + 1 . Thismeans that m + q ≤ r + s − ,n + t ≤ r + s − . Since R is Cohen-Macaulay we get immediately that q = s − r ≥ m , that is r = m. From q + r ≤ s + t we have that t ≥ r − 1, that is t = m − . Now it follows at once that s = n. a3) m − t + 1 ≤ n − q + 1 and m − t + 1 ≤ n + m − r − s + 2 . Since R is Cohen-Macaulay we get q + r = n + t. Then r = n − q + t ≥ m − t + t = m, that is r = m. From m − t ≤ m + n − r − s + 1we get q + r = t + n ≥ r + s − . It follows at once that q ≥ s − 1, that is q = s − . Now, from q + r = n + t ≤ s + t we obtain s = n and then t = q + r − n = s − r − s = r − . b) s + t ≤ q + r. One has to consider the same 3 subcases as in case a).The proof is exactly the same.The converse is clear. References [1] W. Bruns, J. Herzog, Cohen-Macaulay rings , Cambridge Univ. Press, Cambridge,1997.[2] D. Eisenbud, Commutative Algebra with a View Toward Algebraic Geometry , SpringerVerlag, New-York, 1994.[3] J. Herzog, T. Hibi, Cohen-Macaulay polymatroidal ideals , Eur. J. Comb., (2006),513-517.[4] J. Herzog, G. Rinaldo, G. Restuccia, On the depth and regularity of the symmetricalgbera , Beitr. Algebra Geom., (2006), 29-51.[5] E. Miller, B. Sturmfels, Combinatorial Commutative Algebra , Springer-Verlag, Berlin,2005.[6] G. Restuccia, R. Villarreal, On the normality of monomial ideals of mixed products ,Commun. Algebra, (2001), 3571-3580.[7] R. Villarreal, Monomial algebras , Marcel Dekker, New-York, 2001. Cristodor Ionescu, Institute of Mathematics Simion Stoilow of the Roma-nian Academy, P.O. Box 1-764, 014700 Bucharest. Romania E-mail address : [email protected] Giancarlo Rinaldo, Dipartimento di Matematica, Universit`a di Messina ,Contrada Papardo, salita Sperone, 31, 98166 Messina. Italy E-mail address ::