Citation Osaka Journal of Mathematics. 44(4)

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1 TitleCohen-Macaulay local rings of embed Author(s) Wang, Hsin-Ju Citation Osaka Journal of Mathematics. 44(4) Issue Date Text Version publisher URL DOI Rights Osaka University

2 Wang, H.-J. Osaka J. Math. 44 (2007), COHEN-MACAULAY LOCAL RINGS OF EMBEDDING DIMENSION e + d k HSIN-JU WANG (Received July, 2006, revised November 3, 2006) Abstract In this paper, we prove the following. Let (R, m) be a d-dimensional Cohen- Macaulay local ring with multiplicity e and embedding dimension Ú = e + d k, where k 3 and e k. If (m 3 Jm 2 ) = and m 3 Jm, where J is a minimal reduction of m, then 3 s + k, where s is the degree of the h-polynomial of R and is the Cohen-Macaulay type of R.. Introduction Let (R, m) be a d-dimensional Noetherian local ring of multiplicity e. The Hilbert function of R is by definition the Hilbert function of the associated graded ring of R: G := Å n0 m n m n+, i.e., H R (n) = dim Rm m n m n+. The Hilbert series of R is the power eries P R (z) = H R (n)z n. n0 It is known that there is a polynomial h(z) ¾ Z[z] such that P R (z) = h(z)( z) d and h() = e. This polynomial h(z) = h 0 + h z + + h s z s is called the h-polynomial of R. Let (R, m) be a d-dimensional Cohen-Macaulay local ring with embedding dimension Ú = e + d k, where k 3. Let J be a minimal reduction of m. Let be the Cohen-Macaulay type of R, h = Ú d and Ú i = (m i+ Jm i ) for every i; then there are at least two possible Hilbert series of RJ: P RJ (z) = + hz + z z k and P RJ (z) = + hz + (k )z 2. In the first case, R is stretched (cf. definition below) and we have m k Jm; in the second case, following [3], we say that R is short and we have m 3 Jm and Ú = k Mathematics Subject Classification. 3D40, 3H0.

3 88 H.-J. WANG Let (R, m) be a d-dimensional local Cohen-Macaulay ring of multiplicity e and embedding dimension Ú. If d = 0, then R is called stretched if e Ú is the least integer i such that m i+ = 0. If d 0, then R is stretched if there is a minimal reduction J of m such that RJ is stretched (cf. [6]), or equivalently, (m 2 + J)J is principal. Regular local rings are not stretched since fields are not stretched. However, for any d-dimensional local Cohen-Macaulay ring (R, m) having infinite residue field, if Ú = e + d with e or Ú = e + d 2 with e 2, then R is stretched. Moreover, if Ú = e + d 3 and R is Gorenstein, then R is stretched. These stretched rings have been studied in [6], [7] and [8]. In [4], Rossi and Valla extended the notion stretched. There they defined, for each m-primary ideal I, I is stretched if there is a minimal reduction J of I such that I 2 J = I J and (I 2 (J I + I 3 )) =. In [6], Sally studied the structure of stretched local Gorenstein rings, and use it to show in [8] that if (R, m) is a d-dimensional Gorenstein local ring with embedding dimension Ú = e + d 3, then the associated graded ring of R is Cohen-Macaulay. This result has been generalized by Rossi and Valla in [3] as follows. Theorem. ([3, Theorem 2.6]). If (R, m) is a d-dimensional Cohen-Macaulay local ring of multiplicity e = h + 3 and (m 3 Jm 2 ) =, then s + 2, where s is the degree of the h-polynomial of R. In [4], Rossi and Valla generalized Theorem. to stretched m-primary ideals. In this note, we are able to generalize Theorem. in a different manner in Section 4 as follows. In which, we do not assume R is stretched. In stead, we assume that R is short and Ú 2 =. Theorem.2. Let (R, m) be a d-dimensional Cohen-Macaulay local ring of multiplicity e = h + k, where k 3 and e k. If (m 3 Jm 2 ) = and m 3 Jm, where J is a minimal reduction of m, then 3 s + k, where s is the degree of the h-polynomial of R. In the final section, we provide several examples to answer some questions raised by Rossi and Valla in [3]. 2. One dimensional local Cohen-Macaulay ring We state several facts of one dimensional local Cohen-Macaulay rings. These results can be derived easily from [] and [5]. Lemma 2.. Let (R, m) be a one dimensional local Cohen-Macaulay ring; then (m n m n+ ) = e (m n+ Jm n ), where J is any minimal reduction of m. Lemma 2.2. Let (R, m) be a one dimensional Cohen-Macaulay local ring with embedding dimension 2. Then G(R) is Gorenstein.

4 COHEN-MACAULAY LOCAL RINGS 89 Corollary 2.3. Let (R, m) be a d-dimensional Cohen-Macaulay local ring with embedding dimension d +. Then G(R) is Gorenstein. 3. Cohen-Macaulay local rings of embedding dimension e + d k Let (R, m) be a d-dimensional Cohen-Macaulay local ring with embedding dimension Ú = e + d k, where k 3 and e k. Let be the Cohen-Macaulay type of R, h = Ú d and Ú i = (m i+ Jm i ) for every i. Let J be a minimal reduction of m; then one of the possible Hilbert series of RJ is + hz + (k )z 2. In this case, m 3 Jm and Ú = k. If k = 3, it is shown in [3, Theorem 2.6] that if Ú 2 = then s + 2, where s is the degree of the h-polynomial of R. We are able to generalize this result in this section. Theorem 3.. Let (R, m) be a d-dimensional Cohen-Macaulay local ring of multiplicity e = h + k, where k 3 and e k. If (m 3 Jm 2 ) = and m 3 Jm, where J is a minimal reduction of m, then 3 s + k, where s is the degree of the h-polynomial of R. REMARK 3.2. (i) Notice that the assumption Ú 2 = ensures that the depth of G is at leat d (cf. [3]). Therefore to show Theorem 3., we need only to consider the case when d =. (ii) If d =, then s is the least integer for which (m s m s+ ) = e. (iii) Notice that (m 2 Jm) = k. Moreover, if m 2 = Jm + (u,, u k ), then u,, u k is part of a generating set of the socle of R. By Remark 3.2, we may assume from now on that d = and Ú 2 =. Lemma 3.3. Let r be the reduction number of m with respect to J. If r 3, then Theorem 3. holds. Proof. If r 3, then m 4 = Jm 3, so that (m 3 m 4 ) = e, it follows that s 3 + k by the choice of s. By Lemma 3.3, we may assume in the sequel that r 4. Lemma 3.4. The following hold for R: (i) If m 3 = Jm 2 +(ab) for some b ¾ math f racm 2 and a ¾ m, then m i+ = Jm i +(a i b) for every i 2. (ii) If ym 2 Jm 2 for some y ¾ m, then y 3 ¾ Jm 2. In particular, there is an element y ¾ m such that m i+ = Jm i + (y i+ ) for every i 2. Proof. (i) If m i+ = Jm i + (a i b) for some i 2, then m i+2 = Jm i+ + a i bm Jm i+ + a i m 3 = Jm i+ + (a i b) m i+2.

5 820 H.-J. WANG (ii) Suppose that ym 2 Jm 2. Then there are u, Ú ¾ m such that uúy ¾ Jm 2 and m 3 = Jm 2 + (yuú). Therefore, m 4 = Jm 3 + (y 2 uú). It follows that y 2 u ¾ Jm 2 and m 3 = Jm 2 + (y 2 u). Thus, m 4 = Jm 3 + (y 3 u) and then y 3 ¾ Jm 2. Now, choose y ¾ m such that ym 2 Jm 2, then m i+ = Jm i + (y i+ ) for every i 2. Lemma 3.5. Let J = (x) be a minimal reduction of m. If there is an element y ¾ m such that m i+ = Jm i + (y i+ ) for every i 2, then y l x t is a generator of the module (J t m l +m l+t+ )(J t+ m l +m l+t+ ) whenever 2 l r, where r is the reduction number of m with respect to J. Proof. If not, y l x t ¾ J t+ m l + m l+t+, so that y r x t ¾ x t+ m r, it follows that y r ¾ Jm r, a contradiction. Therefore, the conclusion holds. Theorem 3.6. Let (R, m) be a one dimensional Cohen-Macaulay local ring of multiplicity e = h + k, where k 3 and e k. Assume that (m 3 Jm 2 ) = (m 4 Jm 3 ) = and m 3 Jm, where J = (x) is a minimal reduction of m. Then there is a basis x, y,, y, z,, z e k of m, elements u t+,, u k contained in m and elements c i j i =, È, k, j =,, j i contained in the ideal k (y 2,, y k, z,, z e k) with i= j i(k i) = e k such that J = (x) and the following hold: (i) m i+ = Jm i + (y i+ ) for every i 2. (ii) m 2 = Jm+(y 2, y y 2,, y y t, y t+ u t+,, y k u k ), where t = ((y m+ Jm)Jm). (iii) y 2, y y 2,, y y t, y t+ u t+,, y k u k, y k,, y is a generating set of the socle of R. (iv) y y i ¾ Jm for i t + and y z i ¾ Jm for every i. (v) y i m 3 Jm 3 for every Ë i 2 and z i m 3 Jm 3 for every i. È (vi) z,, z e k = i, j,k z(l) i j, ((ci j m+ Jm)Jm) = k i and m 2 k i = Jm+ l= z(l) i j c i j for every i =,, k and j =,, j i. (vii) c i j z (l) i ¼ j ¾ Jm if i ¼ i ¼ or i = i ¼ but j j ¼. (viii) y 3 ¾ J(z,, z e k) + Jm 2. Proof. By Lemma 3.4, there is an element y ¾ m such that (i) hold. Let t = ((y m + Jm)Jm); then there are y 2,, y k, u t+,, u k ¾ m such that m 2 = Jm + (y 2, y y 2,, y y t, y t+ u t+,, y k u k ) and y m + Jm = (y 2, y y 2,, y y t ) + Jm. We may assume that y 2y i ¾ Jm 2 for 2 i t by replacing y i by y i + y if necessary, and assume that y y j ¾ Jm for t + j k by replacing y j by y j + y + + t y t if necessary. It follows that y i m 3 = (y i y 3) + Jm3 = Jm 3 for every i k. Since the Cohen-Macaulay type of R is and y 2, y y 2,, y y t, y t+ u t+,, y k u k is part of a generating set of the socle of R, we may choose y k,, y, z,, z e k ¾ m such that y k,, y, z,, z e k is part of a generating set of m and y 2, y y 2,, y y t, y t+ u t+,, y k u k, y k,, y is a generating set of the socle of R. If z i y ¾ Jm for some i, then we may replace z i by

6 COHEN-MACAULAY LOCAL RINGS 82 z i + «y + + «t y t if necessary and assume that z i y ¾ Jm for every i. Therefore z i m 3 Jm 3 + z i y m 2 Jm 3. Hence, the basis x, y,, y, z,, z e k of m satisfies (i) to (v) so far. Claim. For any integer i =,, k, there is an integer j i, a basis x, y,, y, z,, z e k of m and elements c i j j =,, j i contained in the ideal (y 2,, y k, z,, z e k) such that not only (i) to (v) but also the following hold: (a) ((c i j m + Jm)Jm) = k i, m 2 = Jm + z () i j c (k i) i j,, z i j c i j. (b) c i j z (l) i j ¾ Jm for every l if j ¼ j ¼ and c i j z ¾ Jm for every generator of the ideal generated (l) by S i, where S i = z,, z e k z i ¼ j i ¼ i, j j i, l k i. Note that (vi) and (vii) follows from the Claim. Proof of the Claim. We proceed by induction on i. Let z be any generator of the ideal (z,, z e k). Since y z, y i z ¾ Jm for every i k, there is an element c ¾ (y 2,, y k, z,, z e k) such that cz ¾ Jm. If for any generating set z ¼,, z¼ e k of the ideal (z,, z e k) there is no element c ¾ (y 2,, y k, z,, z e k) such that m 2 = (cz ¼,, cz¼ k ) + Jm, then the Claim holds for i =. If not, we may assume that m 2 = (c z,, c z k ) + Jm for some c ¾ (y 2,, y k, z,, z e k). Set z (l) = z l. Let z be any generator of the È ideal (z k,, z e k). If c z ¾ Jm, then k there are elements «i such that c z i= c z È (i) ¾ Jm, so that we may replace k z by i= z(i) if necessary and assume that c z ¾ Jm. If for any generating set z ¼ k,, z¼ e k of the ideal (z k,, z e k) there is no element c ¾ (y 2,, y k, z,, z e k) such that m 2 = (cz ¼ k,, cz¼ 2k 2 ) + Jm, then again the Claim holds for i =. If not, we may use the same trick to find c 2, c 3, so that the Claim holds for i =. Suppose now È we have shown that the Claim holds for any integer i ¼ i for some i i. Let m = i ¼ = j i ¼(k i ¼ ) and S i = z m+,, z e k. If for any generating set z ¼ m+,, z¼ e k of the ideal generated by S i there is no element c ¾ (y 2,, y k, z,, z e k) such that m 2 = (cz ¼ m+,, cz¼ m+k i ) + Jm, then the Claim holds for i +. If not, we may assume that for some c i+, ¾ (y 2,, y k, z,, z e k), m 2 = (c i+, z m+,, c i+, z m+k i )+ Jm. Set z (l) i+, = z m+l. As before, we may assume that c i+, z ¾ Jm for every generator z of the ideal (z m+k i,, z e k). If for any generating set z ¼ m+k i,, z¼ e k of the ideal (z m+k i,, z e k) there is no element c ¾ (y 2,, y k, z,, z e k) such that m 2 = (cz ¼ m+k i,, cz¼ m+2k 2i 2 )+ Jm, then again the Claim holds for i +. If not, we may use the same trick to find c i+,2, c i+,3, so that the Claim hods for i +. The Claim is now fulfilled. To finish the proof, assume that È y 3 ¾ J(z,, z e k) + Jm 2. Then there are Æ i ¾ R not all in m such that y 3 e k i= x ¾ Jm 2. Let t be the smallest integer

7 822 H.-J. WANG È for which Æ t is a unit; then y 3 e k x ¾ Jm 2. Let z = c i j if z t = z (l) i j for È some l; then z e k x zy È 3 ¾ Jm 3, so that z e k ¾ m3 Jm È as z ¾ (y 2,, y k, z,, z e k). However, z e k ¾ Jm by the Claim, a contradiction. Therefore (viii) holds. Now, we are ready for: Proof of Theorem 3.. From the above, we may assume that d =, 2 and r 4, where r is the reduction number of some minimal reduction J of m. By Theorem 3.6, there is a basis x, y,, y, z,, z e k of m, elements u t+,, u k contained in m and elements c i j i =, È, k, j =,, j i contained in the k ideal (y 2,, y k, z,, z e k) with i= j i(k i) = e k such that J = (x) and the following hold: (i) m i+ = Jm i + (y i+ ) for every i 2. (ii) m 2 = Jm+(y 2, y y 2,, y y t, y t+ u t+,, y k u k ), where t = ((y m+ Jm)Jm). (iii) y 2, y y 2,, y y t, y t+ u t+,, y k u k, y k,, y is a generating set of the socle of R. (iv) y y i ¾ Jm for i t + and y z i ¾ Jm for every i. (v) y i m 3 Jm 3 for every Ë i 2 and z i m 3 Jm 3 for every i. È (vi) z,, z e k = i, j,k z(l) i j, ((ci j m+ Jm)Jm) = k i and m 2 k i = Jm+ l= z(l) i j c i j for every i =,, k and j =,, j i. (vii) c i j z (l) i ¼ j ¾ Jm if i ¼ i ¼ or i = i ¼ but j j ¼. (viii) y 3 ¾ J(z,, z e k) + Jm 2. If h, then s = h + k + k by [2] and we are done. Therefore, we may assume that h. To show that s + k, it is enough to show that (m +k m +k ) = e by Remark 3.2 (ii). Moreover, by Lemma 3.5, y +k, y +k 2 x,, y 2x +k 3 are generators of the module m +k (J +k 2 m+m +k ), therefore to show that (m +k m +k ) = e it is enough to show that y x +k 2, x +k, z x +k 2,, z e kx +k 2 is a linearly independent set in (x +k 2 m + m +k )m +k. Suppose not, there are «,, Æ i in R not all in m such that e k «y x +k 2 + x +k + i= x +k 2 ¾ m +k. Then e k «y r x +k 2 + y r x +k + i= x +k 2 y r ¾ m +r+k,

8 COHEN-MACAULAY LOCAL RINGS 823 so that «y r x +k 2 ¾ x +k m r as y z i ¾ Jm, it follows that «¾ m by the choice of r. Therefore x +k + È e k i= x +k 2 ¾ m +k. If Æ i ¾ m for every i, then x +k ¾ m +k, which is impossible. So, there is an integer i such that Æ i is a unit. By replacing z i by z i + Æ i x, we may assume that ¾ m. Hence È e k i= x +k 2 ¾ m +k. Let t be the smallest integer for which Æ t is a unit; then È e k x +k 2 ¾ m +k. Let «+k be the integer such that È e k If È e k È e k x +k 2 ¾ J +k 3 m 3, then È e k x ¾ Jm 2 by (viii), it follows that È e k x +k 2 ¾ J +k «m «J +k+ «m «. x ¾ m 3 = (y 3) + Jm2, so that ¾ m 2, a contradiction. Therefore, «4. Since m «= (y «) + Jm «and (m «Jm «) =, there is a unit such that () e k x «2 y «¾ Jm «. È Let z = c i j if z t = z (l) i j ; then z e k ¾ Jm by (vi) and (vii). Moreover, z k x «2 y «z ¾ Jm«. È Furthermore, y 3z ¾ Jm3 e k by (v), we have z x 3 «¾ m «. Therefore, there is an element 2 of R such that k (2) z x «3 2 y «¾ Jm «. From () and (2), we see that there is an element 3 of R such that (z 3 x) k x «4 ¾ m «. Let «4 + k 4 be the non-negative integer such that (z 3 x) È e k Since z is a unit 4 of R such that (3) (z 3 x) k x ¾ m +3 Ò Jm +2. È ¾ Jm, (z e k 3 x) ¾ Jm 2, exists. Moreover, there k x 4 y +3 ¾ Jm +2.

9 824 H.-J. WANG On the other hand, from (), we have y r «+ x «2 y r+ ¾ Jm r, or equivalently, y r «+ x «3 ¾ m r. Since m r = (y r ) + Jmr, there is an element 5 of R such that (4) y r «+ However, from (), we have (5) y r «Thus, from (4) and (5), we obtain that (6) y r «(y 6 x) x «3 5 y r ¾ Jm r. x «2 y r ¾ Jm r k x «4 ¾ m r, for some element 6 of R. Now, if we can show that (7) for some element and (z 3 x)ỹr 3 4 y +3 y r 3 y r 3 x ¾ m r y r 3 ¾ m r 3 Ò Jm r 4, then from (3) and (7), we see that (z 3 x)ỹr 3 x 4 y +3 y r 3 ¾ Jm r È e k x ¾ zm r + Jm r = Jm r by (v), therefore ¾ Jm r, which contradicts to the choice of r. Hence, we conclude that y x +k 2, x +k, z x +k 2,, z e kx +k 2 is a linearly independent set in (x +k 2 m + m +k )m +k.

10 COHEN-MACAULAY LOCAL RINGS 825 Finally, by (6), we may prove (7) by reverse induction. Suppose we have shown that for some Æ, Æ «4, y r Æ 3 x Æ ¾ m r for some element ỹr Æ 3 ¾ m r Æ 3 Ò Jm r Æ 4. Then there is an element 6 ¾ R such that (8) y ỹ r Æ 3 From (5) and (8), we see that for some element y r Æ 2 x Æ 6 y r ¾ Jm r. x Æ ¾ Jm r y r Æ 2 ¾ m r Æ 2 Ò Jm r Æ 3, it follows that y r Æ 2 x Æ ¾ m r. We end this section by providing the following example. EXAMPLE 3.7. Let K be a field and R = K [[x, y, z,, z k ]]I, where I is the ideal of R generated by the set z 3 xy, y 2, yz,, yz k, z z 2,, z z k z i z j 2 i j k. The it is easy to see the following hold: (i) R is a -dimensional Cohen-Macaulay local ring with maximal ideal m = (x, y, z,, z k )I. (ii) x is a regular element of R and x R is a minimal reduction of m. (iii) Ú = k +, h = k and e = 2k. (iv) m 3 xm, z 3 is a basis of m3 xm 2 and z 2, z z 2,, z z k is a basis of (m 2 xm). È (v) H R (z) = + (k + )z + (2k )z 2 ½ + i=3 2kzi = ( + kz + (k 2)z 2 + z 3 )( z) and H Rx R (z) = + kz + (k )z 2. (vi) s = r = 3. (vii) depth G = 0.

11 826 H.-J. WANG 4. Examples In [3], Rossi and Valla raised the following questions: QUESTION. Let (R, m) be a d-dimensional Cohen-Macaulay local ring with embedding dimension Ú = e + d 3. If h, then is depth G d? QUESTION 2. If (R, m) is a d-dimensional Cohen-Macaulay local stretched domain with multiplicity e = h + 3 and = 2, then is G Cohen-Macaulay? We give counterexamples to these questions as follows. EXAMPLE 4.. Let K be a field and R = K [[x, y, z, u, Ú]](u 3 xz, Ú 3 yz, u 4, Ú 4, uú, z 2, zu, zú); then (R, m) is a 2-dimensional Cohen-Macaulay local ring and x, y is a regular sequence of m, where m = (x, y, z, u, Ú)R. Moreover, h = 3, e = 6 and = 3 as u 2, Ú 2, z generates the socle of R. However, z ¾ (m 3 : (x, y)) and z ¾ m 2, therefore the depth of G is 0. EXAMPLE 4.2. Let K be a field and R = K [[t 5, t 6, t 4 ]]; then (R, m) is a onedimensional Cohen-Maculay local domain, where m = (t 5, t 6, t 4 )R. Let x = t 5, y = t 6 and z = t 4 ; then h = 2, e = 5 = h + 3 and = 2 as z, y 3 generates the socle of R. Moreover, and P Rx R (z) = + 2z + z 2 + z 3 P R (z) = + 2z + z2 + z 4. z Hence R is stretched and G is not Cohen-Macaulay. In fact, zx ¾ (m 4 : x) and zx ¾ m 3. ACKNOWLEDGMENT. I am grateful to the referee for a number of valuable suggestions that improved the paper a lot. References [] W. Heinzer, M.-K. Kim and B. Ulrich: The Gorenstein and complete intersection properties of associated graded rings, J. Pure Appl. Algebra 20 (2005), [2] J. Herzog and R. Waldi: A note on the Hilbert function of a one-dimensional Cohen-Macaulay ring, Manuscripta Math. 6 (975), [3] M.E. Rossi and G. Valla: Cohen-Macaulay local rings of embedding dimension e + d 3, Proc. London Math. Soc. (3) 80 (2000),

12 COHEN-MACAULAY LOCAL RINGS 827 [4] M.E. Rossi and G. Valla: Stretched m-primary ideals, Beiträge Algebra Geom. 42 (200), [5] J.D. Sally: Super-regular sequences, Pacific J. Math. 84 (979), [6] J.D. Sally: Stretched Gorenstein rings, J. London Math. Soc. (2) 20 (979), 9 26 [7] J.D. Sally: Tangent cones at Gorenstein singularities, Compositio Math. 40 (980), [8] J.D. Sally: Good embedding dimensions for Gorenstein singularities, Math. Ann. 249 (980), Department of Mathematics National Chung Cheng University Chiayi 62 Taiwan hjwang@math.ccu.edu.tw