Integral closure of powers of the graded maximal ideal in a monomial ring

Transcription

1 ISSN: Integral closure of powers of the graded maximal ideal in a monomial ring Vincenzo Micale Research Reports in Mathematics Number 1, 003 Department of Mathematics Stockholm University

2 Electronic versions of this document are available at Date of publication: January 8, Mathematics Subject Classification: Primary 13B5, Secondary 13F0. Keywords: integral closure, monomial ideal. Postal address: Department of Mathematics Stockholm University S Stockholm Sweden Electronic addresses:

3 Integral closure of powers of the graded maximal ideal in a monomial ring Vincenzo Micale Abstract In this paper we study the integral closure of ideals of monomial subrings R of k[x 1, x,..., x d ] spanned by a finite set of distinct monomials of the polynomial ring. We generalize a well known result for monomial ideals in the polynomial ring to rings R as above proving that the integral closure of a monomial ideal is monomial and we give a geometric description of the integral closure of an ideal of R. Then we focus our attention to the study of the integral closure of powers of the graded maximal ideal of R in two particular cases. MSC: 13B5; 13F0 1 Introduction Let H = {h 1, h,..., h m } be a finite set of distinct monomials in k[x 1, x,..., x d ] and let R = k[h] = k[h 1, h,..., h m ] k[x 1, x,..., x d ] be the monomial subring spanned by H. Furthermore we suppose that the complement to N d of the set of exponents of all monomials in R is finite. In this paper we study the integral closure of ideals in R. In Section we give the concepts of multidegree of a monomial and of integral closure and normality of an ideal. In Section 3 we generalize a well known result for monomial ideals in the polynomial ring to rings R as above proving that the integral closure of a monomial ideal is monomial and we give a geometric description of the integral closure of an ideal of R. In Section 4 we focus our attention to the study of the integral closure of powers of the graded maximal ideal of R in two particular cases. Preliminaries Let R be a ring as in the Introduction. We can associate to every monomial ux a1 1 xa d d in R, with u k \ {0}, the power product x a1 1 xa d d. Let m be a monomial in R and let x a1 1 xa d d be the associated power product. We call vmicale@dipmat.unict.it 1

4 (a 1, a,..., a d ) N d the multidegree of m and we denote it by mdeg(m). If m 1 and m are monomials in R, then m 1 and m have the same multidegree if there exists u k \ {0} such that m 1 = um. Let I be an ideal of R. We denote by I the integral closure {z R z n + r 1 z n r n = 0, for some r i I i }. By definition I I and, in general, it may happen that I I. An ideal I in R is called normal if I j = I j for every j 1. 3 Integral closure of monomial ideals Now we generalize a well known result for monomial ideals in the polynomial ring to rings R as in the Introduction; we have used some ideas from []. Theorem 3.1. Let I be a monomial ideal in R. Then I is a monomial ideal. Proof. Since it is well known that I is an ideal in R, we only need to prove that I is generated by monomials. Let z I, z = m m s, where the m i s are monomials with different multidegrees. We want to prove that m i I for every i = 1,..., s. By induction, it suffices to verify that m i I for some i since I is an ideal in R. To this aim we prove that there exists N > 0 such that m N i I N for some i = 1,... s (hence m i I since it is root of Z N m N i = 0). From the definition of integral closure, we have (m m s ) n + l 1 (m m s ) n l n = 0, (l i I i ). Let us consider the multidegree of m n 1. There must exist another term in the equation above which has the same multidegree and it must be of the form b i1 m j1,1 1 m j1,s s, where b i1 I i1 and s k=1 j 1,k = n i 1 (we note that if s = 1, then z = m 1 I). Since the set of elements of the same multidegree as m n 1 is a 1-dimensional vector space over R and since b i1 m j1,1 1 m j1,s s and m n 1 have the same multi- s = m n 1 for some u k \ {0}. Using the same degree, we get ub i1 m j1,1 1 m j1,s argument as above, we get that for every v = 1,..., s, m n v = c iv m jv,1 1 ms jv,s with c iv I iv, s k=1 j v,k = n i v. Since m n v is a monomial, we get (after cancelling out common terms) that m n1,v v = c iv k v mj v,k k for some n 1,v. We note that 0 j v,k < n 1,v for every v and k. Indeed if, for example, j,1 = n 1,1, then m n1,1 1 = c i1 m n1,1 with c i1 k, whence mdeg(m n1,1 1 ) = mdeg(m n1,1 ), that is mdeg(m 1 ) = mdeg(m ). A contradiction. Hence we have a system of s equalities m n1,1 1 = c i1 k 1 mj 1,k k m n1, = c i k mj,k k m n1,s s = c is k s mj s,k k

5 We use an induction on s to prove that there exists N > 0 such that m N i I N for some i = 1,... s. If s = 1, then m n1,1 1 = c n1,1 I n1,1. Suppose now such N exists for systems as above with s 1 equalities. Consider the system above. For every v =,..., s, we first raise, mv n1,v to n 1,1 and m n1,1 1 to j v,1, then we substitute m n1,1jv,1 1 in m n1,vn1,1 v with (c i1 k 1 mj 1,k k ) jv,1 and finally we cancel out common terms (it is easy to check that, by j v,k < n 1,v for every v and k, we never cancel out m n1,vn1,1 v ). Hence, for every v =,..., s, we get m n,v v = d iv m kv, kv,v m v m kv,s s for some n,v and with d iv I n,v (kv,+ +kv,v 1+kv,v+1+ +kv,s) kv,v (where m v means that we delete mv kv,v in the product). Using induction we get the proof. 3.1 A geometric description Our next aim is to have a geometric description of the integral closure of ideals in R. Let a i N d, i = 1,..., r and let conv(a 1,..., a r ) = { r λ i a i i=1 r λ i = 1, λ i Q 0 } be the convex hull (over the rational numbers) of a 1,..., a r. For α = (α 1,..., α d ) N d we set x α = x α1 1 xα d d. Proposition 3.. Let I be a monomial ideal in R generated by x a1,..., x ar. Then the exponents a such that a monomial x a in R belongs to the integral closure I are the integer points in the convex hull of the union of the set b + N d, where b is an exponent of an element in I. Proof. Let x a R such that a is in the convex hull of the union of the set b + N d, where b is an exponent of an element in I. Hence a = r i=1 λ ia i with x ai I, r i=1 λ i = 1 and λ i Q 0. Let m an integer such that mλ i N for every i = 1,..., r. Then, by r i=1 mλ i = m, we get (x a ) m = (x r i=1 λiai ) m = (x a1 ) mλ1 (x ar ) mλr I m. So x a I. Vice versa if x a I, then, by the proof of Theorem 3.1, we get x am I m, that is x am = x b1 x bm with x bi I. By a = m i=1 1 m b i, we get that a is in the convex hull of the union of the set b + N d, where b is an exponent of an element in I. 4 Integral closure of powers of the graded maximal ideal m in special cases We recall that we are interested in monomial subrings R = k[h 1, h,..., h m ] of k[x 1, x,..., x d ] spanned by a finite set H of monomials and such that the complement to N d of the set of exponents of all power products in R is finite. R is a graded ring with graded maximal ideal m = (h 1, h,..., h m ). i=1 3

6 In this section we focus our attention to the study of the integral closure of powers of the graded maximal ideal m of R in two particular cases. We remark that, by definition of integral closure of an ideal, m = m. 4.1 The first case In this first case we restrict to rings R = k + (x δ1,..., x δt )k[x], subalgebras of k[x 1, x ] = k[x] such that the complement to N of the set of exponents of all power products in R is finite. Let m denote the graded maximal ideal of R. By m = (x δ1,..., x δt )k[x], we get m r = ((x δ1,..., x δt )k[x]) r = (x δ1,..., x δt ) r k[x]. Let a 1 = min{α 1 0 x α1 1 m}, a = min{α 0 x α 1 x α m, for some α < a 1 } As the complement to N of the set of exponents of all power products in R is finite, such a i exists for i = 1,. By definition of a, we get x γ 1 xa m for some positive integer γ < a 1. Proposition 4.1. Suppose that a i, for i = 1,. If there exists γ as above such that a 1 γ, then m j \ m j for every j. Proof. Let j and let us consider m j = x ja1 1 1 x a 1. By definition of a 1 and by j, we have x ja1 1 1 x a 1 R. Since a 1 γ, x ja1 1 x a = x(j 1)a1 1 (x a1 1 x a ) mj. Finally by a, x ja1 1 x a m j. Since (ja 1 1, a 1) = λ 1 (ja 1, a )+λ (ja 1, a ) with λ 1 = λ = 1/, we get, by Proposition 3., that m j m j. But m j / m j as, by definition of a, x α 1 xa 1 m j only if α ja 1. Remark 4.. We can change the role of x 1 with that one of x in Proposition 4.1. Indeed let b = min{β 0 x β m}, b 1 = min{β 1 0 x β1 1 xβ m, for some β < b }. As above such b i exists for i = 0, 1. By definition of b 1, we get x b1 1 xγ m for some γ < b. Using the same argument as for a i, we get that if there exists γ as above such that b γ, then x b1 1 1 x jb 1 m j \ m j for every j. Example 4.3. Let R = k + (x 9, x 1 x7, x3 1 x6, x5 1 x3, x9 1 )k[x 1, x ] and let us consider m = (x 9, x 1 x7, x3 1 x6, x5 1 x3, x9 1 )k[x 1, x ] the graded maximal ideal of R as in Figure 1. Then a 1 = 9, a = 3, b 1 =, b = 9 and {x j9 1 1 x, x 1x j9 1 } m j \ m j for every j. Remark 4.4. We can generalize Proposition 4.1 to the d-dimensional case. Let R = k + (x δ1,..., x δt )k[x] be a subalgebra of k[x 1, x,..., x d ] = k[x] and let a 1 = min{α 1 0 x α1 1 m}. 4

7 Suppose a 1 and that there exists i with i d such that a i := min{α i 0 x α 1 xαi i m, for some α < a 1 }. By definition of a i there exists a positive integer γ i < a 1 such that x γi 1 xai i m. Suppose a 1 γ i. Since (for every j ) (ja 1 1, 0,..., 0, a i 1, 0,..., 0) = 1 (ja 1, 0,..., 0, a i, 0,..., 0)+ 1 (ja 1, 0,..., 0, a i, 0..., 0) and by definition of a i, we get that for every j the element x ja1 1 1 x ai 1 i m j \ m j. It is straightforward to change the role of x 1 with that one of each x i, i {,..., d}. Example 4.5. Let R = k + (x 8 1, x3, x4 1 x5 3, x9 3 )k[x 1, x, x 3 ] and let us consider m = (x 8 1, x 3, x 4 1x 5 3, x 9 3)k[x 1, x, x 3 ] the graded maximal ideal of R. Then a 1 = 8, a = 3 and a 3 = 5 and, by Remark 4.4, x 8j 1 1 x, x8j 1 1 x 4 3 m j \ m j for every j. x 9 x 8 x 7 x 6 x 1 x7 x 3 1 x6 x 5 x 4 x 3 x 5 1 x3 x x 1 x 1 x 1 x 3 1 x 4 1 x 5 1 x 6 1 x 7 1 x 8 1 x 9 1 FIGURE 1 Let us come back to the -dimensional case, that is R subalgebras of k[x 1, x ]. Our next aim is to show that if a power t (with t ) of the graded maximal ideal m is integrally closed, then every other power l, with l t, of m is integrally 5

8 closed (cf. Theorem 4.11). As corollaries to this we give a characterization and a sufficient condition for m to be normal (cf. Corollaries 4.13 and 4.15). To this aim we need a little amount of work. From now on we always denote any power of m by J. We note that if x ai 1 xbi and xaj 1 xbj are different minimal generators of J as a k[x]-module, then a i a j and b i b j and, furthermore, a i < a j implies b i > b j. We say that (a i, b i ) (a j, b j ) in N if a i < a j. Let x ai 1 xbi and xaj 1 xbj be two generators of J as a k[x]-module with (a i, b i ) (a j, b j ) and with the property that if x α 1 xβ is any other element of J, then (b j b i )α + (a i a j )β + a i (b i b j ) + b i (a j a i ) 0 that is, (α, β) is not under the straight line in R connecting (a i, b i ) and (a j, b j ). We call the pair (a i, b i )(a j, b j ) of elements of N as above special pair of generators of J as a k[x]-module, (spg(j)). Example 4.6. Let R = k + (x 7, x 1x 5, x 3 1x 4, x 5 1x, x 7 1)k[x 1, x ] and consider m = (x 7, x 1 x5, x3 1 x4, x5 1 x, x 7 1 )k[x 1, x ] the graded maximal ideal of R as in Figure. It is easy to check that the only spg(m) are (0, 7)(5, 1) and (5, 1)(7, 0). x 7 x 6 x 5 x 4 x 1 x5 x 3 1 x4 x 3 x x 1 x 1 x 1 x 3 1 x 5 1 x x 4 1 x 5 1 x 6 1 x 7 1 FIGURE Lemma 4.7. Let J be generated by x a1 1 xb1, xa 1 xb,..., xar 1 xbr as a k[x]-module with (a 1, b 1 ) (a, b ) (a r, b r ). Then it is possible to choose (a i1, b i1 ) (a i, b i ) (a is, b is ) among the elements of N as above such that (a i1, b i1 )(a i, b i ), (a i, b i )(a i3, b i3 ),..., (a is 1, b is 1 )(a is, b is ) are spg(j) with (a i1, b i1 ) = (a 1, b 1 ) and (a is, b is ) = (a r, b r ) Proof. Since the complement to N of the set of exponents of all power products in R is finite, then a 1 = 0 = b r. If (a 1, b 1 )(a i, b i ) is not a spg(j) for 6

9 every i =,..., r 1, then (by definition of spg(j)) (a 1, b 1 )(a r, b r ) is a spg(j) and we get the proof. Hence suppose there exists i 1 < r such that (a 1, b 1 )(a i1, b i1 ) is a spg(j). As above, if (a i1, b i1 )(a k, b k ) is not a spg(j) for every k = i 1 + 1,..., r 1, then (a i1, b i1 )(a r, b r ) is a spg(j) and we get the proof. If not, using the same argument as above we get, after a finite number of steps (as the number of generators of J as a k[x]-module is finite), the proof. Lemma 4.8. Let (a i, b i )(a j, b j ) be a spg(m), then (la i, lb i )(la j, lb j ) is a spg(m l ). Proof. Since (la i, lb i ) (la j, lb j ) whenever (a i, b i ) (a j, b j ), to get the proof we need that if x α 1 xβ ml, then (α, β) is not under the straight line in R connecting (la i, lb i ) and (la j, lb j ) and that x lai 1 xlbi and x laj 1 xlbj are generators for m l as a k[x]-module. Let x α 1 x β ml, hence (α, β) = l k=1 (α k, β k ) and x α k 1 xβ k m. Since (a i, b i )(a j, b j ) is a spg(m), (b j b i )α k + (a i a j )β k + a i (b i b j ) + b i (a j a i ) 0 for every k = 1,..., l. Hence (lb j lb i ) l[(b j b i ) l α k + (la i la j ) k=1 l α k + (a i a j ) k=1 l β k + la i (lb i lb j ) + lb i (la j la i ) = k=1 l β k + l[a i (b i b j ) + b i (a j a i )]] = k=1 l[(b j b i )α 1 + (a i a j )β 1 + a i (b i b j ) + b i (a j a i ) + + (b j b i )α l + (a i a j )β l + a i (b i b j ) + b i (a j a i )] 0. Suppose x laj 1 xlbj is not a generator for m l as a k[x]-module, then there exists x a 1 xb ml such that either a = la j and b < lb j or a < la j and b = lb j. Since, in this case, (a, b) is under the straight line in R connecting (la i, lb i ) and (la j, lb j ), we get (lb j lb i )a + (la i la j )b + la i (lb i lb j ) + lb i (la j la i ) < 0 that is a contradiction to what we proved above, since if x α 1 x β ml, then (α, β) can not be under the straight line in R connecting (la i, lb i ) and (la j, lb j ). Corollary 4.9. If (a i1, b i1 ) (a i, b i ) (a is, b is ) are as in Lemma 4.7 with J = m, then (la i1, lb i1 )(la i, lb i ), (la i, lb i )(la i3, lb i3 ),..., (la is 1, lb is 1 )(la is, lb is ) are spg(m l ) (with (la i1, lb i1 ) = (la 1, lb 1 ) and (la is, lb is ) = (la r, lb r )). Remark Let (la i, lb i )(la j, lb j ) be a spg(m l ) and let r(x, Y ) = (b j b i )X + (a i a j )Y + l[a i (b i b j ) + b i (a j a i )] = 0 the straight line in R connecting (la i, lb i ) and (la j, lb j ). It is straightforward to prove that for every k such that 0 k < l, the integer point ((l k)a i, (l k)b i ) + (ka j, kb j ) = ((l k)a i + ka j, (l k)b i + kb j ) is in the straight line in R with equation r(x, Y ). 7

10 Let m = (x a1 1 xb1,..., xar 1 xbr )k[x 1, x ] and x γ m l \ m l (hence l ) with γ = (γ 1, γ ). By Proposition 3., γ is an integer point in the convex hull of the union of the set b + N d, where b is an exponent of an element in m l. Hence, by Corollary 4.9, there exists (la i, lb i )(la i+1, lb i+1 ) spg(m l ) such that (γ 1, γ ) is not under the straight line in R connecting (la i, lb i ) and (la i+1, lb i+1 ) and such that la i < γ 1 < la i+1. So γ is in the triangle in R with vertices (la i, lb i ), (la i+1, lb i+1 ), (la i+1, lb i ) (we note that γ cannot be out of the triangle since x γ / m l and x lai 1 xlbi, x lai+1 1 x lbi+1 are generators for m l as a k[x]-module by the proof of Lemma 4.8). Finally, since γ is in the triangle in R with vertices (la i, lb i ), (la i+1, lb i+1 ), (la i+1, lb i ) and since (cf. Remark 4.10), for every k with 1 k < l, ((l k)a i, (l k)b i ) + (ka j, kb j ) is in the straight line connecting (la i, lb i ) and (la i+1, lb i+1 ), we get or γ is in the triangle in R with vertices (la i, lb i ), ((l k)a i, (l k)b i ) + (ka j, kb j ), ((l k)a i + ka j, lb i ) γ is in the triangle in R with vertices ((l k)a i, (l k)b i ) + (ka j, kb j ), (la j, lb j ), (la j, (l k)b i + kb j ). Theorem Let m = (x a1 1 xb1,..., xar 1 xbr )k[x 1, x ] and suppose there exists t such that m t = m t. Then m l = m l for every l t. Proof. It is enough to prove that if m t = m t with t, then m t+1 = m t+1. Suppose x γ m t+1 \m t+1 and let l = t+1 (hence l 3). By what is written above, for a fixed k with 1 k < l we have either or γ is in the triangle in R with vertices (la i, lb i ), ((l k)a i, (l k)b i ) + (ka j, kb j ), ((l k)a i + ka j, lb i ) γ is in the triangle in R with vertices ((l k)a i, (l k)b i ) + (ka j, kb j ), (la j, lb j ), (la j, (l k)b i + kb j ). Let us consider the first case. Hence (γ 1, γ ) = λ 1 (la i, lb i )+λ ((l k)a i +ka j, (l k)b i +kb j )+λ 3 ((l k)a i +ka j, lb i ), λ 1 + λ + λ 3 = 1, λ i Q 0. Let δ = (δ 1, δ ) = (γ 1, γ ) (a i, b i ). Since x γ / m l, necessary x δ / m l 1. But δ = (δ 1, δ ) = λ 1 ((l 1)a i, (l 1)b i ) + λ ((l k 1)a i + ka j, (l k 1)b i + kb j )+ hence λ 3 ((l k 1)a i + ka j, (l 1)b i ), λ 1 + λ + λ 3 = 1, λ i Q 0, 8

11 δ is in the triangle in R with vertices ((l 1)a i, (l 1)b i ), ((l k 1)a i, (l k 1)b i ) + (ka j, kb j ), ((l k 1)a i + ka j, (l 1)b i ), that is δ is an integer point in the convex hull of the union of the set b + N d, where b is an exponent of an element in m l 1. By x δ R and Proposition 3., we get x δ m l 1 = m l 1. Absurd. Similarly we get the proof for the other case. The statement of Theorem 4.11 is, in general, not true for other kind of rings R as in the Introduction (cf. Remark 4.33). Remark 4.1. We note that it is not true in general that if m t = m t, then m l = m l for some l < t. Indeed, let n 1 and R = k+(x n 1, xn 1 1 x, x 1 x n 1, x n )k[x 1, x ]. In Remark 4.0 we show that m k = m k if and only if k n. As corollary to Theorem 4.11 and by m = m, we get a criterion for m to be normal. Corollary The graded maximal ideal m is normal if and only if m = m. Example Let R = k + (x 8, x 1x 6, x 1 x5, x4 1 x4, x8 1 x, x9 1 x, x 11 1 )k[x 1, x ] as in Figure 4. By Proposition 3., m = (x 16, x 1 x 14, x 1x 1, x 3 1x 11, x 4 1x 10, x 6 1x 9, x 8 1x 8, x 10 1 x 7, x 11 1 x 6, x 13 1 x 5, x 15 1 x 4, x 17 1 x 3, x 18 1 x, x0 1 x, x 1 )k[x 1, x ] = m and, by Corollary 4.13, we get that m is normal. x 16 x 15 x 14 x 13 x 1 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x x 1 x 1 x 1 x 3 1 x4 1 x5 1 x6 1 x7 1 x8 1 x9 1 x10 1 x11 1 x1 1 x13 1 x14 1 x15 1 x16 1 x17 1 x18 1 x19 1 x0 1 x1 1 x 1 FIGURE 4 9

12 We note that in the Example 4.14, x 7 1 x3 / R, while (7, 3) is an integer point in the convex hull of the union of the set b + N d, where b is an exponent of an element in m. By the proof of Theorem 4.11, we get the following corollary. Corollary If for every integer point a in the convex hull of the union of the set b + N d, where b is an exponent of an element in m, we get x a m, then m is normal. Example Let be k+(x 7, x 1x 6, x 3 1x 5, x 7 1x 4, x 11 1 x 3, x 14 1 x, x 18 1 x, x 1 )k[x 1, x ]. Since for every integer point a in the convex hull of the union of the set b + N d, where b is an exponent of an element in m = (x 7, x 1 x6, x3 1 x5, x7 1 x4, x11 1 x3, x14 1 )k[x 1, x ], we get x a m, then, by Corollary 4.15, m is normal. x 18 1 x, x A class of examples 1 x, Let n = (x 1, x ) be the graded maximal ideal of the polynomial ring k[x 1, x ] = k[x]. We look for rings R = k + (x δ1,..., x δt )k[x] with graded maximal ideal m = (x δ1,..., x δt )k[x] such that m = n n (n 1). Indeed by Proposition 3. and by Corollary 4.13, we have that m is normal. Let a 1 = min{α 1 0 x α1 1 m} and b = min{β 0 x β m} as at the beginning of Subsection 3.1. By m = n n = (x 1, x ) n, we necessarily have that a 1 = n = b and that, if x r = x r1 1 xr is a generator for m as a k[x]-module, then r 1 + r n. Furthermore if r 1 + r > n, then this generator is uninteresting in our discussion as if (x δ1,..., x δt ) k[x] = (x 1, x ) n then a := ({x δ1,..., x δt } \ {x r }) k[x] is equal to (x 1, x ) n. Indeed if there exists x c1 1 xc m \ a with x c1 1 xc generator for m as a k[x]-module, then x c1 1 xc = (xr1 1 xr )(xb1 1 xb ). This is absurd as n = c 1 + c = r 1 + r + b 1 + b > n. Hence we can assume r 1 + r = n Finally by Proposition 4.1 and Remark 4., x1 n 1 x, x 1 x n 1 R. Moreover, since we can suppose r 1 + r = n whenever x r = x r1 1 xr is a generator for m as a k[x]-module, we have that x1 n 1 x, x 1 x n 1 are generators of m as a k[x]-module. By what is written above we can translate the problem to a merely combinatorial problem just considering the powers of the x s in the generators of m as a k[x]-module. Indeed we look for a class of sets X with {0, 1, n 1, n} X {0, 1,..., n} such that X := X + X = {0, 1,..., n}. From now on, given two integers a and b with a b, we denote the set of integers between a and b (included) by [a, b]. Proposition Let X = {0, 1,..., h 1 1, h 1, h,..., h z = n h 1, n h 1 + 1,..., n} with h 1 1 and h i+1 h i h for every i [1, z 1] (*). Then X = [0, n]. Proof. We show that for every x [0, n], there exist x 1, x X such that x 1 + x = x. If x [0, h 1 ], then x 1 = x and x = 0. If x [h 1, h z ], then there exists i such that h i x h i+1. If x = h i or x = h i+1, then x 1 = x and x = 0. Suppose hence that h i < x < h i+1, that is 10

13 h i + 1 x h i+1 1. By (*), h i+1 1 h i + h 1, hence h i + 1 x h i + h 1. So x = h i + h with 0 h h 1 and we can assume x 1 = h i and x = h. If x [h z, n], then x 1 = x and x = 0. If x [n, n+h 1 ], then x = n+h with 0 h h 1. Hence x 1 = n and x = h. If x [n + h 1, n + h z ], then there exists i such that n + h i x n + h i+1. If x = n+h i, then x 1 = n and x = h i. Suppose hence that n+h i < x n+h i+1. So n + h i h i+1 < x h i+1 n. By (*), n h 1 1 n + h i h i+1 and this implies n h 1 1 < x h i+1 n. Hence x h i+1 X and we can assume x 1 = h i+1 and x = x h i+1. Finally if x [n + h z, n] = [n h 1, n], then x = n h 1 + h with 0 h h 1. Since n h 1 n h 1 + h n, then n h 1 h X. Hence x 1 = n h 1 + h and x = n. Corollary The graded maximal ideal m of R = k + (x n 1, xn 1 1 x,..., x n h1 1 x h1, xn h 1 x h,..., xn hz 1 x hz, xh1 1 1 x n h1+1,..., x n )k[x 1, x ] with h 1 1 and h i+1 h i h for every i [1, z 1] is normal. We now generalize Proposition Indeed we look for rings R with graded maximal ideal m such that m k = n kn, (n 1). By Proposition 3. and by Theorem 4.11, this implies m l = m l for every l k. As above we have that a 1 = n = b, that x1 n 1 x, x 1 x n 1 R are generators of m as a k[x]-module and that if x r1 1 xr is a generator for m as a k[x]-module, then we can assume r 1 + r = n. Hence we look for a class of sets X with {0, 1, n 1, n} X [0, n] such that kx = [0, kn]. Proposition Let X be a set with {0, 1, n 1, n} X [0, n] and such that px = [0, pn]. Then for every q p, qx = [0, qn]. Proof. It is enough to prove the proposition for q = p + 1. By [0, (p + 1)n] = {{0} + px} {{n} + px} (p + 1)X, we get the proof. Remark 4.0. We note that if X = {0, 1, n 1, n}, then kx = [0, kn] if and only if k n. Indeed for every i = 0,..., n, [i(n 1) + (n i ) 0, in + (n i ) 1] = [in i, (i + 1)n (i + )] (n )X. Hence [0, (n )n] = n i=0 [in i, (i + 1)n (i + )] (n )X. Moreover n / αx when α n 3. In particular, if k n, then kx = [0, kn] for every set X such that {0, 1, n 1, n} X [0, n]. Hence if m is the graded maximal ideal of R for which x n 1, x1 n 1 x, x 1 x n 1, x n are part of a set of generators for m as a k[x]- module, then m k = m k for every k n. As a particular case of Proposition 4.19 we get that the class of set X as in Proposition 4.17, satisfies kx = [0, kn] for every k. Now we find a class of sets X such that kx = [0, kn] but, in general, (k 1)X [0, (k 1)n]. To this aim we generalize the class X of Proposition

14 Proposition 4.1. Let X = {0, 1,..., h 1 1, h 1, h,..., h z = n h 1, n h 1 1,..., n} such that h 1 and h i+1 h i h 1 + k 1 for every i [1, z 1] (**). Then kx = [0, kn]. Proof. The case k = is Proposition We suppose hence k 3. Let Y := {0, 1,..., h 1, n h 1,..., n} X. So ky = {0,..., kh 1, n h 1,..., n+(k 1)h 1, (n h 1 ),..., n+(k )h 1,..., k(n h 1 ),..., kn} kx. To get the proof we need to cover all the holes in kx between in + (k i)h 1 and (i + 1)(n h 1 ) for every 0 i k 1. We start covering all the holes in kx between kh 1 and n h 1. We note that (k 1)Y = {0,..., (k 1)h 1, n h 1,..., n + (k )h 1, (n h 1 ),..., n + (k 3)h 1,..., (k 1)(n h 1 ),..., (k 1)n} (k 1)X. For every l [1, z 1] we have {h l }+[0, (k 1)h 1 ] = [h l, h l +(k 1)h 1 ] kx. By (**) and by h 1 and k 3, we get h l+1 h l + (k 1)h 1. Hence {h 1, h,..., h z } + [0, (k 1)h 1 ] = {h 1, h 1 + 1,..., h z + (k 1)h 1 } = [h 1, h z + (k 1)h 1 ] kx. By k 3 and h z = n h 1, we get h 1 kh 1 and n h 1 h z + (k 1)h 1, that is [kh 1, n h 1 ] [h 1, h z + (k 1)h 1 ] and we have covered all the holes in kx between kh 1 and n h 1. Making exactly the same sort of calculus as above you can check that {h 1, h,..., h z }+[i(n h 1 ), in+(k i 1)h 1 ] covers all the holes in kx between in + (k i)h 1 and (i + 1)(n h 1 ) for every 1 i k 1. Hence [0, kn] = kx. Corollary 4.. Let R = k + (x n 1, x1 n 1 x,..., x n h1 1 x h1, xn h 1 x h,..., x n hz 1 x hz, xh1 1 1 x n h1+1,..., x n )k[x 1, x ] with h 1, h i+1 h i h 1 +k 1 for every i [1, z 1]. Then for every l k, the l-th power of the graded maximal ideal m of R is integrally closed. Remark 4.3. We note that h 1 in Proposition 4.1 must be greater than 1. Indeed if X = {0, 1, 4, 7, 9, 10} (hence n = 10, k = 3 and h 1 = 1), then, 5 / 3X. Remark 4.4. If X is a set in the class as in Proposition 4.1, then we know kx = [0, kn]. Anyway in general (k 1)X [0, (k 1)n] (in particular the converse to Proposition 4.19 does not hold). Indeed let X = {0, 1,, 6, 10, 1, 13, 14} (hence n = 14, h 1 = and k = 3). By Proposition 4.1, 3X = [0, 3n] = [0, 4]. Anyway 5, 9, 17 / X. Let V = [0, n] and k and let us consider the collection k of subset of V such that F k if and only if 0, 1, n 1, n / F and k(v \ F ) = [0, kn]. Since F k whenever F G for some G k and since {i} k for every i V \{0, 1, n 1, n}, then k is a (finite) simplicial complex on V \{0, 1, n 1, n}. We first find a lower and an upper bound for dim := sup{dim(f ) F }, where dim(f ) := F 1. Theorem 4.5. dim n 1+ 16n+9. Furthermore, if n 4 then n 8n + 1 dim. 1

15 Proof. Let F be a face in with V \ F = m. As F is a face, (V \ F ) = [0, n] = n + 1. Since there are no more than (m 1)m sums of mutually different numbers from X along with no more than m sums of equal numbers, we get that n + 1 m(m+1), hence m 1+ 16n+9. So dim ((n + 1) 1+ 16n+9 ) 1 = n 1+ 16n+9. Let now n 4 and let us consider X = {0, 1,..., h 1 1, h 1, h,..., h z = n h 1, n h 1 + 1,..., n} with h 1 1, h i+1 h i = h for every i [1, z ] (note that z 1 as n 4) and h z h z 1 h By Proposition 4.17, X = [0, n]. Since in X we exclude (h 1 + 1) + n (h1+1) h 1+1 = (h 1 + 1) + n h 1+1 integers from [0, n], we get n (h 1 + 1) n h number of vertices in a maximal face of. Let us consider f(h 1 ) := n (h 1 + 1) n h + as a function of h The derivative f (h 1 ) = 0 if and only if h 1 = n 1. Since f( n 1) = n 8n+ and n h n 1+1 h n 1+1 h 1, we have dim 1+1 n 8n + 1. Remark 4.6. We note that since, for n 0, n 8n + 1 n 8n and n 1+ 16n+9 n 4n, then the class of sets X as in the second part of the proof of Theorem 4.5 is a almost extremal class of examples. Now we generalize part of Theorem 4.5 finding a lower bound for k. Proposition 4.7. Let n 4. Then dim k n ( n + k k) n+k 4 3 n +k Proof. Let X = {0, 1,..., h 1 1, h 1, h,..., h z = n h 1, n h 1 + 1,..., n} with h 1, h i+1 h i = h 1 +k 1 for every i [1, z ] and h z h z 1 h 1 +k 1. By Proposition 4.1, kx = [0, kn]. Since in X we exclude (h 1 +1)+ n (h1+1) h 1+k 1 integers from [0, n], we get n (h 1 + 1) n (h1+1) h 1+k 1 number of vertices in a maximal face of k. Let us consider f(h 1 ) := n (h 1 +1) n (h1+1) h 1+k 1 as a function of h 1. Using the same argument as in Theorem 4.5, we get the proof. We note that for n = 1,, 3 then k = { } for every k ; for n = 4, k = {, {}} for every k ; for n = 5, = {, {}, {3}} and k = {, {, 3}} for every k 3; for n = 6, = {, {}, {3}, {4}, {, 4}}, 3 = {, {}, {3}, {4}, {, 3}, {, 4}, {3, 4}}, k = {, {}, {3}, {4}, {, 3}, {, 4}, {3, 4}, {, 3, 4}} for every k 4. Our next aim is to prove that if k 3, then k is connected for every n and that is connected for every n 5, 6. We need to show that if n 7 then for every a, b V \ {0, 1, n 1, n} there exist γ 1,..., γ m V \ {0, 1, n 1, n} such that {a, γ 1 },{γ 1, γ },...,{γ m, b} are in k for every k. Lemma 4.8. If F i, then F l for every l i. Proof. This follows by definition of face of i and by Proposition

16 Lemma 4.9. Let n 7 and a, b V \ {0, 1, n 1, n} with a < b. Then {a, b} k, with k 3, and {a, b} if and only if the pair (a, b) is different from (, 3) and from (n 3, n ). Proof. Let (a, b) be different from (, 3) and (n 3, n ) (this is possible as n 7). We need to prove that {a, b} is a face in, that is (V \{a, b}) = [0, n]. Let x [0, n] and suppose first that x n. If x a, b, then x = x + 0 (V \ {a, b}). If x = a, then x = (x 1) + 1 (V \ {a, b}). Finally if x = b, then x = (b 1) + 1 (V \ {a, b}) if a b 1 and x = (b ) + (V \ {a, b}) if a = b 1. Suppose now n < x n 4. Since x n n 4, we get n > x n, n 1 > x n + 1 and n x n +. Hence we can write x in at least three different ways as a sum of two natural number less or equal to n. Precisely x = n + (x n), x = (n 1) + (x n + 1) and x = (n ) + (x n + ). This implies that in at least one of the sums above the two summands are both different from a and b. Thus x (V \ {a, b}). Suppose now n 3 x n. Since {n, n 1, n} V \ {a, b} or {n 3, n 1, n} V \ {a, b}, then x (V \ {a, b}). Hence {a, b}. To get the proof we need to show that {, 3} and {n 3, n } are not in. Indeed if X = [0, n]\{, 3} then [0, n]\x = {3} and if X = [0, n]\{n 3, n } then [0, n] \ X = {n 3}. Let now k 3. By the first part of the proof and by Lemma 4.8 we know that if a, b V \ {0, 1, n 1, n} with (a, b) different from (, 3) and from (n 3, n ), then {a, b} k. Let V \ {, 3} = {0, 1, 4, 5, 6,..., n}. By 3{0, 1, 4, 5, 6} = [0, 1], by 3[4, n] = [1, 3n] and by Lemma 4.8, we have {, 3} k. Finally let V \ {n 3, n } = {0, 1,..., n 4, n 1, n}. By 3[0, n 4] = [0, 3n 1], by 3{n 5, n 4, n 1, n} = [3n 1, 3n] and by Lemma 4.8, we have {n 3, n } k. Theorem If k 3, then k is connected for every n. is connected for every n 5, 6. Proof. By what is written before Lemma 4.8, we need to show that if n 7, then k is connected for every k. The case k 3 is immediate by Lemma 4.9. Let k =. Given a, b V \ {0, 1, n 1, n}. If (a, b) (, 3), (n 3, n ), then, by Lemma 4.9, {a, b}. Otherwise, since n 7, there exists γ V \ {0, 1, a, b, n 1, n }. By Lemma 4.9, {a, γ} and {γ, b} are in. 4. The second case Let us denote the set of all power products in the indeterminates x 1, x,..., x d of degree i in k[x 1, x,..., x d ], that is {x i 1, x i 1 1 x,..., x i d }, by F i. Let g 1, g,..., g n be positive natural numbers with g 1 < g < < g n and gcd(g 1, g,..., g n ) = 1. In this second case we study the integral closure of powers of the graded maximal ideal of R = k[f g1, F g,..., F gn ]. 14

17 We define the k-algebra homomorphism ψ : k[x 1, x,..., x d ] k[t] with ψ(f(x 1,..., x d )) = f(t,..., t). We also define the k-algebra homomorphism with φ(l(t)) = l(x 1 ). φ : k[t] k[x 1, x,..., x d ] Theorem Let R = k[f g1, F g,..., F gn ] and T = k[t g1, t g,..., t gn ]. Let m = (F g1, F g,..., F gn ) and M = (t g1, t g,..., t gn ) denote the graded maximal ideal of R and T respectively. Then for every natural number a, m a \ m a if and only if M a \ M a. Proof. It is straightforward to verify that, since gcd(g 1, g,..., g n ) = 1, the complements to N d and N respectively of the set of exponents of all power products in R and in T are finite. Suppose there exists z m a \ m a for some a. By Theorem 3.1, we can suppose that z is a monomial in k[x 1, x,..., x d ], that is z = x j1 i 1 x jp i p, 1 i 1 < < i p d. We note that ψ(z) = t j1 t jp / M a (since z / m a ). We will show that ψ(z) M a. By z m a, there exist c 1, c,..., c m with c i (m a ) i such that z m +c 1 z m c m = 0. Hence 0 = ψ(z m + c 1 z m c m ) = ψ(z m ) + ψ(c 1 )ψ(z m 1 ) + + ψ(c m ) = ψ(z) m + ψ(c 1 )ψ(z) m ψ(c m ). Since c i (m a ) i, we get ψ(c i ) (M a ) i. Thus ψ(z) M a. Suppose now M a \ M a for some a. By Theorem 3.1 we can suppose t b M a \ M a for some b. Let us consider x b 1. Clearly x b 1 / m a (if not ψ(x b 1) = t b M a ). We will show that x b 1 ma. By t b M a there exist c 1, c,..., c m with c i (M a ) i such that (t b ) m + c 1 (t b ) m c m = 0. Hence 0 = φ((t b ) m +c 1 (t b ) m 1 + +c m ) = φ((t b ) m )+φ(c 1 )φ((t b ) m 1 )+ +φ(c m ) = Since φ(c i ) (m a ) i, we get x a 1 ma. (x b 1) m + φ(c 1 )(x b 1) m φ(c m ). Corollary 4.3. Let (R, m) and (T, M) be rings as in Theorem Then m is normal if and only if M is normal. A study on normal graded maximal ideal for rings k[t g1, t g,..., t gn ] with gcd(g 1, g,..., g n ) = 1 can be found in [1]. 15

18 Remark Let R = k[f 6, F 7, F 11 ]. Then, using Theorem 4.31, it is easy to check that m = m, but x 1 m 3 \ m 3. Hence the statement of Theorem 4.11 is not true for this kind of rings. Corollary Let R be a ring as above with graded maximal ideal m = (F g1, F g,..., F gn ). (i) If m is normal, then g = g and g n < g 1. (ii) m is normal if and only if m a+1 = m a m for every a 0. (iii) Let m i denote the maximal ideal of k[f g1, F g,..., F gi ]; if m i is not normal for some i < n, then m n (= m) is not normal. Proof. (i) This follows by Corollary 4.3 and [1, Proposition 3.1]). (ii) Use Corollary 4.3 and [1, Theorem 3.5]. (iii) By Corollary 4.3 and [1, Theorem 3.14]. In the 3-generated case (n = 3) we can give a concrete characterization for normal graded maximal ideal m. Suppose g = g and g 3 < g 1 (if one of these two conditions is not satisfied, then m is not normal by (i) of Corollary 4.34) and let α be the unique integer such that (α 1)g 3 < αg 1 and αg 3 (α + 1)g 1. Corollary Let R, g and g 3 as above. Then m is normal if and only if αg 3 (α + 1)g. Proof. This follows by Corollary 4.3 and [1, Theorem 3.5]. Example Let k[f 10, F 11, F 17 ] = k[x 10 1, x9 1 x,..., x 10 d, x11 1, x10 1 x,..., x 11 d, x17 1, x16 1 x,..., x 17 d ]. Then α = and, by Corollary 4.35, we get m = (F 10, F 11, F 17 ) is not normal. Using the same argument as above it is easy to check that in k[f 10, F 11, F 16 ], m = (F 10, F 11, F 17 ) is normal. References [1] V. Micale, Normal maximal ideal in one-dimensional local rings, International Journal of Commutative rings, 1, (00). [] R. Villarreal, Monomial Algebras, Marcel Dekker,