Complete Intersections in Regular Local Rings ∗

J. K. Verma

Department of Mathematics Indian Institute of Technology Powai, Mumbai 400 076 E-mail: [email protected]

Let R be a 3-dimensional regular local ring. Let p be a dimension one prime of R. We are concerned with two problems: (1) when is p a complete intersection and (2) when is p a set theoretic complete intersection ? We prove a result of Cowsik and Nori to answer (1) and several results answering (2). Cowsik observed that if the symbolic Rees algebra of p is Noetherian then p is a set-theoretic complete intersection. Thus it becomes important to know when the symbolic Rees algebra is Noetherian. We will present Huneke’s elegant criterion for its Noetherian property.

1 Reductions of Ideals

D.G. Northcott and D. Rees [NR] introduced the concept of reduction of an ideal. An ideal J contained in an ideal I of a commutative ring R is called a reduction of I if JIn = In+1 for some n IN. This relationship is preserved under ring homorphisms and ring extensions. If I is a∈ zero dimensional ideal of a local ring then the reduction process simplifies I without changing its multiplicity.

Proposition 1.1. Let J I be m-primary ideals of a local ring (R, m). If J is a reduction of I then e(I) = e⊆(J).

Proof. Let JIn = In+1. Then for all m

`(R/In+m) `(R/J m) `(R/Im). ≥ ≥

Hence PI (n + m) PJ (m) PI (m) where P denotes the Hilbert polynomial. This ≥ ≥ shows that PI (m) and PJ (m) have equal degrees and leading coefficients. Definition 1.2. A reduction J of I is called a minimal reduction of I if no ideal properly contained in J is a reduction of I.

∗Notes of lectures given in the NBHM Instructional School on Complete Intersections at IISc, Ban- galore May 20 – June 8, 1996.

1 2

Lemma 1.3. If K is a reduction of J and J is a reduction of I then K is a reduction of I.

Proof. Let KJ m = J m+1 and JIn = In+1. Then KIm+n = KJ mIn = Im+n+1.

Lemma 1.4. Let (R, m) be a local ring and J I be ideals of R. Then J is a reduction of I iff J + Im is a reduction of I. ⊆

Proof. Let JIn = In+1. Then JIn + mIn+1 = In+1 , hence (J + mI)In = In+1. Conversely let (J + Im)In = In+1. By Nakayama’s lemma, JIn = In+1.

Definition 1.5. For an ideal I of a local ring (R, m), put F (I) = ∞ In/Inm. We say M n=0 F (I) is the fiber ring of I. The Krull dimension of F (I) is called the analytic spread of I. This will be denoted by `(I).

Theorem 1.6. Let I be an ideal of a local ring (R, m) with residue field k. For a I put ∈ a∗ = residue class of a in I/mI. Let a1, a2, . . . , as I. Then the following are equivalent: ∈

(i) (a1∗, a2∗, . . . , as∗) is a zero dimensional ideal of F (I).

(ii) J = (a1, . . . , as) is a reduction of I.

n 1 n n Proof. The nth homogeneous component of K := (a1∗, . . . , as∗) is (JI − + mI )/mI . n 1 n n Thus K is zero dimensional iff for all n large JI − + mI = I . By (1.4) the last equation holds iff J is a reduction of I.

Corollary 1.7. Every reduction J of I contains a minimal reduction of I. Let a1, a2,..., as be chosen from J such that the following hold:

(a) a1∗, . . . as∗ are k-linearly independent (b) dim F (I)/(a1∗, . . . , as∗) = 0 (c) s is minimal with respect to (b).

Then a1, a2, . . . , as is a minimal system of generators of a minimal reduction of I con- tained in J.

Proof. Put K = (a1, a2, . . . , as). Observe that K mI = mK. This is equivalent to Ker(K/mK I/mI) = 0. This is a consequence of∩ (a). By (1.6), (b) implies that K −→ is a reduction of I. Suppose that K 0 K is a reduction of I. Then K 0 + mI = K + mI ⊂ by (c). Hence K (K 0 + mI) K = K 0 + mI K = K 0 + mK. By Nakayama’s lemma ⊂ ∩ ∩ K = K 0 . It is clear that a1, . . . , as minimally generate K. In fact a1, . . . , as are part of a minimal basis of I.

Proposition 1.8.Let (R, m) be a local ring with infinite residue field k. Let a1, . . . , as ∈ I, an ideal of R. Then a1∗, . . . , as∗ form a homogeneous system of parameters of F (I) if and only if J = (a1, . . . , as) is a minimal reduction of I. 3

Proof. Since k is infinite, it is possible to choose a homogeneous system of parameters of F (I) from the degree one component of F (I). Hence every minimal reduction of I is minimally generated by dim F (I) = `(I) elements. If a1∗, . . . , as∗ form a homogeneous sys- tem of parameters of F (I) then s = dim F (I) and F (I)/(a1∗, . . . , as∗) is zero dimensional. Hence a1, . . . , as generate a minimal reduction of I. Conversely if J = (a1, . . . , as) is a minimal reduction of I then dim F (I)/(a1∗, . . . , as∗) = 0 and s is minimal with respect to this property. Hence a1∗, . . . , as∗ constitute a homogeneous system of parameters. Theorem 1.9. For an arbitrary ideal I of a local ring (R, m) the following inequalities are satisfied. altI := sup htp p is a minimal prime of I `(I) µ(I). { | } ≤ ≤ Proof. We may assume that R/m is infinite. Let J be a minimal reduction of I. Then the equation JIn = In+1 holds for some n which gives V (I) = V (J). Therefore by Krull’s altitude theorem altI = altJ µ(J) = `(I). Since dim F (I) dim I/Im, we get `(I) µ(I). ≤ ≤ ≤ 2 Reductions and integral dependence

Definition 2.1. Let R be a commutative ring, I an R-ideal. An element x R is called n n 1 ∈ integral over I if there exist elements a1, a2, . . . , an so that x + a1x − + + an = 0, i ··· where ai I for i = 1, 2, . . . , n. ∈ Proposition 2.2. The set of integral elements, I,¯ over I is an ideal of R.

Proof. Consider the Rees algebra R(I) = ∞ Intn of I, where t is an indeterminate. M n=0 n n 1 Let x be integral over I satisfying the equation x + a1x − + + an = 0, for some i n n 1 i ···n i n ai I , i = 1, 2, . . . , n. Then (xt) + (a1t)(xt) − + + (ait )(xt) − + + ant = 0. Hence∈ xt is integral over R(I). If x, y I¯ then xt,··· yt are integral over··· R(I). Thus xt + yt is integral over R(I). Let u R and∈ ut be integral over R(I). Then there exist n∈ n 1 b1, b2, . . . , bn R(I) such that (ut) + b1(ut) − + + bn = 0. Equating coefficient of tn we obtain∈un + b un 1 + + b = 0 where b···are defined by b = b tj where 1n − nn ij i X ij j ··· ¯ ¯ ¯ bij I for i = 1, 2, . . . , n. This shows that u I. In particular x + y I. If x I and c ∈R, it is easy to see that cx I.¯ Hence I¯ is∈ an ideal. ∈ ∈ ∈ ∈ Proposition 2.3. Let I be an ideal of a commutative ring R. Then x I¯ iff I is a reduction of (I, x). ∈

n n 1 i n Proof. Suppose x + a1x − + + an = 0 for some ai I , i = 1, 2, . . . , n. Then x n 1 n ···1 n ∈ ∈ I(I, x) − which yields I(I, x) − = (I, x) . Conversely suppose that I is a reduction of n 1 n n m n 1 (I, x) and I(I, x) − = (I, x) . Then x = i=1 aibi where ai I and bi (I, x) − . n 1 n 1 j j P ∈ ∈ Thus bi = − aijx − − for some aij I , j = 0, 1, . . . , n 1 and i = 1, 2, . . . , m. Hence Pj=0 n m n 1 n 1 j ∈ ¯ − x − aiaijx − − = 0. Thus x I. − Pi=1 Pj=0 ∈ 4

Proposition 2.4. Let I J be ideals of a commutative ring such that J is finitely generated. Then I is a reduction⊆ of J iff J I.¯ ⊆ ¯ Proof. Let J = (I, x1, x2, . . . , xm). Let J I. Then x1 is integral over I, hence I is a ⊆ reduction of (I, x1) by (2.3). Now apply induction on m to see that I is a reduction of n 1 n J. Conversely let I be a reduction of J. Then for an indeterminate t, (It)(Jt) − = (Jt) for some n. Therefore R[Jt] is a finite R[It]-module. Hence xt is integral over R[It] for any x J. Therefore x I.¯ ∈ ∈ 3 Analytic spread of monomial ideals and certain determinantal ideals

The Cowsik-Nori theorem on complete intersections requires one to calculate the analytic spread of an ideal I in a regular local ring (R, m). If R is a power series ring or a localisation of a polynomial ring at a prime ideal then the fiber ring of I can be presented as a quotient of polynomial ring. This presentation can be effectively calculated by using the package Macaulay. In this section we show how to calculate analytic spread of monomial ideals and determinantal ideals in a polynomail ring.

Theorem 3.1. Let k be a field and x1, . . . , xn be indeterminates, S = k[x1, . . . , xn] and R = Sm where m = (x1, . . . , xn). Let f1, f2, . . . , fr be homogeneous polynomials in S of degree d. Put I = (f1, , fr)R. Then ··· (a) k[f1, f2, . . . , fr] F (I), hence `(I) = trdegkk[f1, . . . , fr]. ai1 a'i2 ain (b) Let fi = X1 X2 ...Xn , i = 1, 2, . . . , r. Then `(I) = rank (aij). (c) (Cowsik-Nori) Let X = (Xij) be an r s matrix of indeterminates. Put S = × k[X], m = (X) and R = Sm. Let t < min(r, s). Then `(It(X)R) = rs.

Proof. (a) Let k[I] denote the subring k[f1, f2, . . . , fr] of k[X1,X2,...,Xn] where I = (f1, f2, . . . , fr) . Let Ig denote the k-vector space generated by φ(f1, . . . , fr) where φ is a homogeneous polynomial of degree g. Then k[I] = ∞ I . It is easy to see that M g g=0 g g dimk Ig = dimk I /mI for all g , hence k[I] F (I). Thus `(I) = trdeg k[I] = dim k[I]. ' k ai1 ai2 ain (b) Now let fi = Xi X2 ...Xn for i = 1, 2, . . . , r. We will show that rank (aij) = dim k[I]. Since A = (aij) is an integer matrix it can be transformed into a diagonal matrix with row and column operations. Hence it is enough to see that the transcendence degree of k(I) does not change under row and column operations. We list the operations and their corresponding effect on k(I): 5

Operation Effect on k(I) interchange row Ri with Rj fi and fj are interchanged 1 change Ri to Ri fi changes to − fi α change Rj to Rj + αRi fj is replaced by fi fj interchange column Ci to Cj Xi and Xj are interchanged 1 change Ci to Ci replace Xi by Xi − α change Cj to Cj + αCi replace Xj by XjXi The row operations do not change k(I). The last column operation changes k(I) α to β(k(I)) where β : k(X1,X2,...,Xn) k(X1,...,Xj 1,Xi Xj,...,Xn) maps Xj to α → − Xi Xj and fixes other Xi’s. Since β is an isomorphism , trdegkβ(k(I)) = trdegkk(I).

(c) Let 1, 2,..., ` be the t t minors of A. Then `(It(A)) = dim k[ 1, 2,..., `] = 4 4 4 × 4 4 4 trdegkk( 1,..., `). Once we show that k(Xij) is algebraic over k( 1, 2,..., `) the result will4 follow.4 We shall first prove the special case r = s = t 4+ 1.4Let d :=4 detA i+j and ij = ( 1) times the minor of A obtained by deleting the i th row and the j th 4 − column. Let B denote the adjoint of A, i.e. ( ji). Then BA = diag(d, d, . . . , d), which t 4 implies that d = detB F := k( 1,..., `) so that F (d) is algebraic over F. Next ∈ 4 4 we show that k(Xij) is algebraic over F (d). The equation BA = diag(d, . . . , d) gives, by Cramer’s rule, that Xij F (d) for all j and i. In the general case consider a submatrix ∈ A of (t + 1) (t + 1) size. The entries Xij of A are algebraic over the subfield generated over k by all× the t t minors of A. This implies, by varying A over all the (t+1) (t+1) × × submatrices of A, that k(Xij) is algebraic over the quotient field of F (I).

4 Stability of Ass(R/In) and depth R/In

One of the results which we are aiming at is an elegant characterization by Cowsik and Nori [CN] of complete intersection radical ideals in regular local rings : among radical ideals in regular local rings the only complete intersection ideals I are those for which R/In are Cohen-Macaulay for every n. This may be regarded as a partial converse to n Macaulay’s unmixed theorem. Burch’s inequality `(I) dim R minn(depthR/I ) plays a crucial role in its proof. Our approach to this inequality≤ is due− to Brodmann [B] who first establishes the stability of depth R/In and Ass(R/In). We begin by proving these results. We follow the treatment given in [Mc].

Lemma 4.1. Let R = ∞ R be a Noetherian homogeneous graded ring. Then the M n n=0 degrees of nonzero elements of 0 : R1 are bounded.

Proof. We need to prove that there is an ` such that (0 : R1) Rn = 0 for all ∩ n `. Let 0 : R1 = (a1, a2, . . . , as) where a1, a2, . . . , as are homogeneous elements. Put ≥ ` = 1 + max(deg ai). Let x = riai Rn where n `. Then we may assume that each X ∈ ≥ ri is homogeneous and hence deg ri 1. But then airi = 0 for each i, hence x = 0. ≥ 6

Corollary 4.2. Let I be an ideal in a Noetherian ring. Then there is an ` such that for all n `, (In+1 : I) I` = In. ≥ ∩ n+1 ` n Proof. Apply the lemma to grI (R). Let x (I : I) I and x I . Then ` < n. Let x Ik Ik+1 so that ` k < n. As xI∈ In+1 I∩k+2 andx ¯ 6∈ Ik/Ik+1 we have 0 =x ¯ ∈ (0 :\I/I2) (Ik/Ik+1≤) = 0, a contradiction.⊆ ⊆ ∈ 6 ∈ ∩ Lemma 4.3. Let R = ∞ R be a Noetherian graded ring. Let I be a homogeneous M n n=0 ideal and a be a homogeneous element. Let (I : a) S = φ for a multiplicatively closed ∩ subset S of R0. Then there is a homogeneous element b such that (I : ab) is a prime and (I : ab) S = φ. ∩ Proof. The set (I : ab0 ) (I : ab0 ) is prime for some homogeneous b0 R and (I : { | ∈ ab0) S = φ has a maximal element say I : ab. We show I : ab is prime. Let x, y be homogeneous∩ } elements of R such that x, y I : ab but abxy I. Hence x (I : aby) I : ab. Hence I : aby has an s S. Hence abys6∈ I which gives ∈y (I : abs).∈Again there\ is a t S (I : abs). Hence st∈ (I : ab) S which∈ is a contradiction.∈ ∈ ∩ ∈ ∩ Theorem 4.4. Let R = ∞ R be a Noetherian homogeneous graded ring. Then there M n n=0 is an m such that AssR (Rn) = AssR (Rm) for all n m. 0 0 ≥ Proof. Let P ∞ Ass (R ). Then P = (0 : c) for some homogeneous c R. Then [ R0 k R0 ∈ k=0 ∈ P = (0 : c)R R0. By (4.3) there is a homogeneous d R such that P ∗ = (0 : cd) is ∩ ∈ a prime ideal in R and P = P R . Since Ass(R) is a finite set, ∞ Ass (R ) is a ∗ 0 [ R0 k ∩ k=0 finite set. Let ` be as in lemma (4.1) and n `. If P AssR Rn,P = (0 : c)R for some ≥ ∈ 0 0 c Rn. Since n `, P = (0 : cR1)R . But cR1 Rn+1 , hence P AssR Rn+1. Since ∈ ≥ 0 ⊆ ∈ 0 ∞ AssR (R ) is finite, the result follows. [ 0 k k=0 Corollary 4.5. If I is an ideal in a Noetherian ring then the sequences AssIn/In+1 and Ass(R/In) stabilize.

Proof. The exact sequence 0 In/In+1 R/In+1 R/In 0, −→ −→ −→ −→ n+1 n n n+1 n 1 n n n+1 gives Ass(R/I ) Ass(R/I ) Ass(I /I ). But Ass(I − /I ) = Ass(I /I ) Ass(R/In) for large⊆ n. Thus Ass∪(R/In) = Ass(R/In+1) for all large n. ⊆ n Notation. The stable value of AssR/I is denoted by A∗(I). Theorem 4.6 (Brodmann). Let I be an ideal of a local ring (R, m). Then depth R/In stabilizes to a value := β(I) depth R. ≤ 7

n n Proof. Let h(I) := infn depth R/I . If h(I) = 0 then depth R/I = 0 for infinitely n many n. Hence m A∗(I) which implies depth R/I = 0 for n large. Now let h(I) > 0. ∈ n n Then m A∗(I). Let x m p p A∗(I) . Then depth R/(x, I ) = depth R/I 1 for n large.6∈ By induction∈ hypothesis\ ∪{ | ∈ on h(I)}, depth R/(x, In) stabilizes. Hence so does− depth R/In to a value atmost depth R.

Lemma 4.7. (B¨oger’sinequality) For any ideal I of a local ring (R, m),

(i) `( I+q ) `(I) for any minimal prime q and equality holds for some such q. q ≤ (ii) `(I) dim R. ≤

Proof. (i) Since µ(In + q/q) µ(In), dim F ((I + q)/q) dim F (I) for any minimal ≤ 1 ≤ prime q of R. Let N = (m, It, t− ) be the unique maximal homogeneous ideal of the 1 1 2 2 extended Rees ring = R[It, t− ]. Put J = + Rt− + m + mIt + mI t + . R ··· ··· Thus `(I) = htN/J := n. Let J P0 Pn = N with Pi primes in . Let ⊂ ⊂ · · · ⊂ R Q P0 be a minimal prime of . It is clear that q = Q R is a minimal prime of R ⊆ n n R ∩ 1 and Q = (q I )t . Clearly ht(N/Q)/(J + Q)/Q = n. Put 0 = R/q[(I + q/q)t, t− ] L ∩ 1 2 2 R 1 N 0 = + (R/q)t− + (m/q) + (I + q/q)t + (I + q/q)t + and J 0 = + R/qt− + ··· ··· ··· (m/q) + (Im + q/q)t + (mI2 + q/q)t2 + . Then htN 0 /J 0 = n = `(I). (ii) By (i) there is a minimal prime··· q such that `(I) = `((I + q)/q). Since ht m/q dim R, we may assume that R is a domain. We may also assume that R/m = k ≤ is infinite. Let L = (a1, a2, . . . , an) be a minimal reduction of I where n = `(I). Then F (L) is a polynomial ring. Hence mR[Lt] is a prime ideal in R[Lt] and mR[Lt] R = m. The altitude formula applied to the extension R R[Lt] gives the inequality ∩ ⊂

ht m + trdegR[Lt]/R ht mR[Lt] + trdegkF (L) 1 + `(I). ≥ ≥ Hence `(I) ht m = dim R. ≤ Theorem 4.8. Let I be an ideal of a local ring (R, m). Let β(I) = stable value of depth R/In. Then `(I) dim R β(I). ≤ − Proof. Apply induction on β(I). If β(I) = 0 then the result follows from the above lemma. If β(I) > 0. then there is an r such that m Ass(R/In) for all n r. 6∈ ≥ Pick x m P P A∗(I) . Then β(I) 1 = β(I, x/xR). The choice of x gives In : x =∈In for\ ∪{ all |n ∈r. Let us} now prove− that `(I, x/xR) = `(I), consider the n th graded component of ≥F (I, x/xR).

(In, x) In In = (mIn, x) ' In (mIn + xR) mIn + In xR ∩ ∩ Since In : x = In for n large, In xR Inx mIn. Thus F (I, xR/xR) and F (I) have same Hilbert polynomial, hence∩ equal⊆ dimension.⊆ By induction we get

`(I) = `(I, x/xR) dim R/xR β((I, x)/xR) = dim R β(I) ≤ − − 8

5 Cowsik-Nori-Dade Theorem

In this section we shall prove the above theorem which gives a very interesting char- acterization of complete intersections in regular local rings in terms of analytic spread. The result has played a decisive role in many subsequent studies related with Cohen- Macaulayness of various graded rings and symbolic powers of prime ideals. E. C. Dade [D] also proved this and many other results in his 1960 Princeton thesis. Cowsik and Nori [CN] rediscovered it.

Theorem 5.1. Let I be a radical ideal of height r in a Cohen-Macaulay local ring (R, m) with infinite residue field. Let RP be regular for all minimal primes P of I. Then I is a complete intersection if and only if htI = `(I).

Proof. Let J be a minimal reduction of I. Then µ(J) = r so J is generated by a regular sequence. Suppose J is properly contained in I. Then there is a prime Q ∈ AssR(I/J) Ass(R/J) Supp(I/J). Since R is Cohen-Macaulay, each associated prime ⊆ ∩ of J has height r. Since I is a radical ideal, IQ = QRQ. Regularity of RQ implies that IQ is generated by r elements. Hence IQ has no proper reduction. But Q Supp (I/J), ∈ hence IQ properly contains JQ which is a reduction. This contradiction implies that I = J and hence I is a complete intersection. The converse follows from Theorem 1.9.

Corollary (5.2).Under the assumptions of the theorem (5.1), R/In is Cohen-Macaulay for all n iff I is a complete intersection.

n n Proof. By Burch’s inequality `(I) dimR minn depth R/I . If R/I is C-M for all n, `(I) dim R dim R/In = dim R≤ dim R/I− = htI since R is Cohen-Macaulay. Hence `(I) = htI≤ which− implies that I is a− complete intersection. The converse is Macaulay’s unmixedness theorem. Let (R, m) be a local domain with R/m infinite. Let P be a prime ideal in R and P (n) denote the nth symbolic power of P, i.e., P (n) = r R rs P n for some s R P . It is natural to ask when P n = P (n) for all n. The{ next∈ result| gives∈ a criterion ∈ \ } in terms of the associated graded ring gr (R) := ∞ P n/P n+1. P M n=0

Corollary 5.3.Let P be a prime ideal of a Cohen-Macaulay local domain such that RP is regular and dim R/P = 1. Thus the following are equivalent: (a) P n = P (n) for all n. (b) grP (R) is a domain. (c) `(P ) = htP. (d) P is a complete intersection.

i i+1 j j+1 i i+1 Proof. (a) = (b). Let f ∗ P /P , g∗ P /P be nonzero. Then f/1 PP PP ⇒j j+1 ∈ i+j ∈ i+j+1 ∈ \ and g/1 PP PP . Then fg/1 PP PP since grRP (RP ) is a domain. Thus (i+∈j) \(i+j+1) ∈ \ fg P P , hence f ∗g∗ = 0. ∈ (b)\ = (a). Apply Induction6 on n. Let P k = P (k) k n 1. Let f P (n) ⇒ ∀ ≤ − ∈ 9

n 1 n (n) n and f ∗ = image of f in P − /P . As f P , there is an s R P such that sf P . ∈ n ∈1 \n ∈ Let s∗ = image of s in R/P. Thus s∗f ∗ = image of sf in P − /P = 0. Hence f ∗ = 0 as n grP (R) is a domain. Hence f P . (a) = (c). If P n =∈P (n) for all n, Ass(R/P n) = P for all n. Hence 1 depth R/P n ⇒ dim R/P n = 1. Hence R/P n is Cohen-Macaulay{ } for all n. Thus P is≤ a complete intersection.≤ Thus (c) holds by (1.9). (c) = (d) By theorem (5.1) (d) =⇒ (a) Since P is a complete intersection, P n is unmixed for all n by Macaulay’s unmixedness⇒ theorem. This implies that P n = P (n) for all n.

Example 5.4. Let X = (Xij) be an r s matrix of indeterminates. Put P = It(X) × where 2 t min(r, s). Hochster [H] showed that when t = min(r, s) then grP (R) is a domain.≤ Let≤ us show that this is not true when t < min(r, s). In this case `(P ) = rs m as seen in (3.1) . Let N = P grP (R). Then dim grP (R)/N = `(P ) = rs = dim grP (R). Thus N is a minimal prime, so grP (R) is not a domain.

6 Symbolic Rees Algebras and Set-theoretic Com- plete Intersections.

Let p be a prime ideal of a Noetherian ring R. The symbolic powers of p can be put together to form an R-algebra. Let t be an indeterminate. The graded ring ∞ p(n)tn M n=0 denoted by s(p) is called the symbolic Rees algebra of p. This was introduced by Rees in [R] in connectionR with Zariski’s generalization of Hilbert’s fourteenth problem. Rees gave a counterexample to Zariski’s conjecture by constructing a non-noetherian symbolic Rees algebra. He showed that if p is the prime ideal of a non-torsion point of an elliptic curve C in the projective plane then s(p) is non-Noetherian. Here the local ring (R, m) is the local ring at the origin of the coneR over C. On the other hand, Cowsik [C] observed a striking relationship between set- theoretic complete intersections and the noetherian property of s(p). He showed that if p is a nonmaximal prime ideal of a d-dimensional local ring (R,R m) such that R/m is infinite then noetherian property of s(p) implies that p is radical of an ideal generated R by d 1 elements chosen from p. Thus it became important to know when s(p) is Noetherian.− Cowsik conjectured [C] that if R is a three dimensional regular localR ring and p is a height two prime of R then s(p) is Noetherian. Several authors [E, GN, H1,R H2, HU, M, S1, S2] proved this conjecture in special cases. P. Roberts [Ro1, Ro2] constructed examples of prime ideals in polynomial and power series rings whose symbolic Rees algebras are not Noetherian thus settling Cowsik’s conjecture. Recently Goto, Nishida and Watanbe [GNW] constructed infinitely many examples of the defining primes of monomial space curves, whose symbolic Rees algebra is not Noetherian. The simplest of these examples is p(25, 72, 29) = (X11 YZ7,Y 3 X4Z4,Z11 X7Y 2) in k[[X,Y,Z]] where k is a field of characteristic zero.− Our− objective− in these notes is to discuss an effective method due to Huneke 10

[H2] for testing the Noetherian property of s(p) where p is a height two prime ideal of a 3-dimensional regular local ring, e.g. k[[X,Y,ZR ]]. As an application, we will show that if p is a height two prime of a 3-dimensional regular local ring R and e(R/p) = 3 then s(p) is Noetherian. In particular p is a set theoretic complete intersection. We begin withR the following basic

Proposition 6.1. Let (R, m) be a local ring, p a prime ideal of R. Then s(p) is a Noetherian ring if and only if there is k 1 so that p(k)n = p(kn) for all n 1R. ≥ ≥ Proof. Let B := (p) be Noetherian. Then the ideal B = ∞ p(n)tn is finitely s + M R n=1 r1 r2 rs (ri) generated. Let B+ = (a1t , a2t , . . . , ast ) where ai p for i = 1, 2, . . . , s. Let (k∈)n (kn) r = lcm(r1, . . . , rs) and set k = rs. We show that p = p for all n 1. Any (m) u1 ≥us element of p is an R-linear combination of monomials of the form a1 . . . as where u1r1 + ... + usrs m. If m k = rs then some uiri r. Set v = r/ri. Then ≥ ≥ ≥ u1 us u1 ui v us v a1 . . . as = (a1 . . . ai − . . . as )ai . (m r) (r) The first factor in the above expression belongs to p − and the second belongs to p . (m) (m r) (r) Thus p p − p for m k. This implies that for any positive integer ` ⊆ ≥ p(k+r`) p(k)p(r)` p(k)p(r`) ⊆ ⊆ Hence for all n 1, p(nk) p(k)n. The reverse inclusion always holds. The converse≥ has⊆ been proved in [GHNV]. We sketch a proof for the case when R is in addition a Nagata ring. Recall that a ring B is called a Nagata ring if it is Noetherian and for every prime ideal p of B, the integral closure of B/p in any finite extension field of the quotient field of B/p is a finite B/p-module. Nagata showed that finitely generated algebras over Nagata rings are Nagata rings [Mat, Theorem 72]. Nagata also proved that complete local rings are Nagata rings. Coming back to the proof of the converse, let us assume that (R, m) is a local ring which is also Nagata. For example R can be taken to be k[[X,Y,Z]] or k[X,Y,Z]m where k is a field and m is any maximal ideal. Let k 1 and p(kn) = p(k)n for all ≥ n 1. Put (p) = and ∞ p(nk)tnk = . Then = R[b tk, . . . , b tk] where s M k k 1 g ≥ R R i=. R R (k) p = (b1, b2, . . . , bg). Let L denote the quotient field of R. Then the quotient field of ,L(t), is a finite extension of that of k. Since R is Nagata, is Nagata. Thus the R R R integral closure of k in L(t) is a finite k-module. Since is an integral extension of R R R k, it follows that is a finite k-module, hence it is a Noetherian ring. R R R Theorem 6.2. (Cowsik, Vasconcelos) Let (R, m) be a local ring with infinite residue field and dimension d. Let p be a nonmaximal prime ideal of R. If s(p) is Noetherian R then there exist f1, f2, . . . , fd 1 p such that p = q(f1, f2, . . . , fd 1). − ∈ − Proof. Let k be a positive integer for which p(k)n = p(nk) for all n 1. Then depth R/In 1 for all n 1 where I = p(k). By Burch’s inequality `(≥I) d 1. Let ≥ ≥ ≤ − (f1, f2, . . . , fd 1) be a reduction of I. Then √I = p = (f1, . . . , fd 1). − q − 11

7 Associativity Formula for Multiplicities

By Cowsik-Vasconcelos theorem, a height two prime p of a local ring (R, m) of dimension 3 is a set theoretic complete intersection if the symbolic Rees algebra of p is Noetherian. Huneke [H2] has devised a very effective method to check Noetherian property of s(p). We will present Huneke’s method in the next section. It’s proof makes heavy useR of the associativity formula for multiplicities. Let q be an m-primary ideal of a local ring (R, m) of dimension d. Let M be a finitely generated R-module. Then dim M := dim R/annM d. For large n, `(M/qnM) ≤ is given by a polynomial Pq(M, n) with rational coefficients. Write Pq(M, n) as

d Pq(M, n) = e(q, M)n /d! + ... where e(q, M) is a non-negative integer and e(q, M) = 0 if and only if dim M < d. We call e(q, M), the multiplicity of M with respect to q.

Lemma 7.1. Let q be an m-primary ideal of a local ring (R, m). Let M be a finitely generated R-module. Let 0 M 0 M M 00 0 be an exact sequence of R-modules. → → → → Then e(q, M) = e(q, M 0) + e(q, M”).

Proof. We view M 0 as a submodule of M. Then

`(M/qnM) = l(M 00 /qnM 00 ) + `(M 0 /M 0 qnM). ( ) ∩ ∗ n n r r By Artin-Rees Lemma q M M 0 = q − (q M M 0 ) for some r and all n r. Hence for all n r ∩ ∩ ≥ ≥ n n n r q M 0 M 0 q M q − M 0 . ⊆ ∩ ⊆ n r n n Thus `(M 0 /q − M 0 ) `(M 0 /M 0 q M) `(M 0 /q M 0 ) , hence ≤ ∩ ≤ lim (d!/nd)`(M 0 /M 0 qnM) = e(q, M 0 ). n →∞ ∩ Multiply ( ) by d!/nd and take the limit to finish the proof. ∗

Theorem 7.2 Let (R, m) be a local ring, q an m-primary ideal and p1, p2, . . . , pr be all the minimal primes of R such that dim R = d = dim R/pi for all i. Then

r e(q, M) = e(q , R/p )`(M ). X i i pi i=1

Proof. Let M = M0 M1 ... Mk = 0 be a filtration such that Mi/Mi+1 R/Qi ⊇ ⊇ ⊇ ' for i = 0, 1, 2, . . . , k 1 and Qi Spec R. By Lemma 7.1 we get − ∈ k 1 e(q, M) = − e(q, R/Q ) X i i=0 dim R/Qi=d 12

For any prime p of R, Mp M1p ... Mkp = 0 and ⊃ ⊃ ⊃

0 if p = Qi (Mi/Mi+1)p = (R/Qi)p = ( 6 Rp/Qip if p = Qi.

Hence `Rp(Mp) is the number of times p occurs in Q0,Q1,...,Qk 1 . Thus { − } r e(q, M) = e(q, R/p )`(M ) X i pi i=1 8 Huneke’s Criterion for Noetherian property of Sym- bolic Rees Algebras

Lemma 8.1 Let (R, m) be a three dimensional regular local ring. Let f, g be a regular sequence and put I = (f, g). Then In has no embedded associated primes.

Proof. Suppose m is an associated prime of In = J then there is an h R J such ∈ \ that mh J. Set A = R[ft, gt] and B = ∞ Intn. The ring B is the integral closure of M ⊆ n=0 A in R[t]. It is easy to see that A R[u, v]/(gv fu) where the isomorphism takes u to f and g to v. Since ht mA = 2 we' get ht mB −= 2, since B is integral over A. Since mh J, (mh)tn B. Hence mBhtn B. If htn is not in B then mB an associated prime⊆ of a principal⊆ ideal. Since B is⊆ normal and ht mB = 2, we get⊆ a contradiction. Thus htn B which yields h J which is a contradiction. ∈ ∈ Remark. The above lemma is a very special case of a theorem of E. Dade [D] to the effect that in a formally equidimensional local ring the integral closure of an ideal, whose height and analytic spread coincide, has no embedded components.

Theorem 8.2. (Rees Multiplicity Theorem) Let (R, m) be a quasi-unmixed local ring, J I be m-primary ideals. Then e(I) = e(J) if and only if J is a reduction of I. ⊆ Proof. See [HIO, 19]. § Theorem 8.3. (Huneke). Let (R, m) be a 3-dimensional regular local ring with infinite residue field. Let p be a height 2 prime ideal of R. Then s(p) is a Noetherian ring if and only if there are elements f p(k), g p(`) for some Rk and ` and an x m p so that ∈ ∈ ∈ \ e(x, R/(f, g)) = k`e(x, R/p).

Proof. (Goto [GN]). Let s(p) be a Noetherian ring. Hence there is an integer k such that p(k)n = p(kn) for all Rn 1. Put I = p(k). Then depth R/In = 1 for all n 1. By Burch’s inequality `(I) =≥ 2. Let J = (f, g) be a reduction of I. Let x m≥ p. Since R/(f, g) is Cohen-Macaulay of dimension 1, e(x, R/(f, g)) = `(R/(x,∈ f, g))\ = 13 e(J, R/(x)) = e(I, R/(x)), since J is a reduction of I. To find e(I, R/(x)) we need to calculate `(R/(In, x)) = e(x, R/I n) since R/In is Cohen-Macaulay for all n. Set B = R/xR. By associativity formula we get

nk + 1 `(B/InB) = e(x, R/In) = e(x, R/p)`(R /pnkR ) = e(x, R/p) ! p p 2

Equating coefficients of n2/2 we get

e(x, R/(f, g)) = k2e(x, R/p). ( ) ∗ Conversely let f p(k) and g p(`) so that e(x, R/(f, g)) = k`e(x, R/p). Then gk, f ` p(k`) and ∈ ∈ ∈ e(x, R/(f `, gk)) = (k`)2e(x, R/p). Hence we may assume that k = `. We prove that (f, g) = J is a reduction of p(k) = I. Let := min(R/J) = min (R/J n) for all n. By the associativity formula for multiplicities F `(B/J n) = `(R/(x, J n)) = e(x, R/J n) = e(x, R/q)`(R /J nR ) X q q q ∈F Taking n large and equating the leading coefficients we obtain

e(J, B) = e(x, R/q)e(JR ,R ) X q q q ∈F e(x, R/p)e(JRp,Rp) ≥ e(x, R/p)e(IRp,Rp) ≥ = e(x, R/p)k2 = e(J, B)

Hence = p and e(JRp,Rp) = e(IRp,Rp). By Rees Multiplicity Theorem, IRp F { } (k) ⊆ JRp. Since J has no embedded components, J is p-primary, so I J. Hence `(p ) = 2. ⊆ By Theorem (3.3) of [GHNV] it follows that s(p) is Noetherian. We present a proof when R is a Nagata ring. Since J n = In Rand by Lemma 8.1, J n has no embed- n n (nk) ded components, I Rp R = I = p for all n 1. Now s(p) is integral over (nk) nk ∩n nk ≥ R p t := k = I t . Thus s(p) is integral over the Rees algebra of I with Lk R L R t as an indeterminate. Since R is Nagata, s(p) is a finite (I)-module. Hence it is Noetherian. R R

Corollary 8.4. Let (R, m) be a 3-dimensional regular local ring with R/m infinite. Let p be a height two prime with e(R/p) = 3. Then the symbolic Rees algebra s(p) is Noetherian. In particular, p is a set- theoretic complete intersection. R

Proof. We may choose x m m2 such that e(x, R/p) = 3 = `(R/(p, x)). If p m2 then ∈ \ n (n) 6⊆ p is a complete intersection and so p = p for all n 1. Hence s(p) is Noetherian. So let p m2. Since `(R/(p, x)) = `(R/(m2, x)) we get≥ (m2, x) =R (p, x). We calculate ⊆ 14

`(R/(x, p(2))) by the associativity formula for multiplicities. Since R/p(2) is Cohen- (2) 2 Macaulay, e(x, R/p ) = `(Rp/p Rp)`(R/(p, x)) = 9. On the other hand, `(R/(x, p2)) = `(R/(x, m4)) = 10. Then there is f m3 p(2) such that (p(2), x) = (x, f, m4). Let “ ” denote images in S = R/(x) and denote∈ ∩ the − ∗ 4 initial form of an element in grnS where n = m/(x). Then deg(f)∗ = 3. Choose g n ∈ such that deg(g)∗ = 4. Then `(R/(f, g, x)) = `(S/(f, g)) = deg(f)∗ deg(g)∗ = 12 = 2 e(x, R/p)2 . Hence s(p) is Noetherian. R

Remark. J. Herzog and B. Ulrich [HU] have shown that if e(R/p) 5 then p is a set-theoretic complete intersection. ≤

9 Exercises

Reductions, analytic spread and integral closures (R, m) will denote a local ring with infinite residue field k.

1. Let I = (a, b, c) be an R-ideal. Show that (an, bn, cn) is a reduction of In for all n. ¯ 2. Let I be an R-ideal and S a mult. closed subset of R. Show that IRS = IRS.

n 3. Show that if grI (R) is reduced then I is integrally closed for all n. 4. Let p be a prime ideal of a regular local ring R. Show that p(n) = p(n) for all n.

5. Show that if R is normal then all principal ideals of R are integrally clsoed.

6. Find a minimal reduction of I = (X2,Y 2,XYZ) in R = k[[X,Y,Z]].

7. Let I be a basic ideal of R. Show that mIn In = mIn. ∩

8. The elements a1, a2, . . . , an R are called analytically independent if for any homo- ∈ geneous polynomial f(X1,...,Xn) R[X1,X2,...Xn], whenever f(a1, a2, . . . , an) n ∈ ∈ mI , then all coefficients of f are in m. Put I = (a1, . . . , an). Show that a1, . . . , an are analytically independent iff F (I) k[X1,...Xn]. '

9. Show that `(I) = max n a1, . . . , an I are analytically iondependent . { | ∈ } 10. Using analytically independent elements give another proof of the fact that for an ideal I of R, there is a minimal prime q of R such that `(I + q/q) = `(I).

11. Show that ideals of the principal class have no proper reductions.

12. Let R = k[[X2,X3,Y ]],I = (X2,X3Y ). Show that `(I) = 1 and In = (X2n,X2n+1). Asymptotic prime divisors

13. Find A∗(I) for the ideals in Ex. 6 and Ex. 12. 15

2 2 14. Find A∗(I) for I = (X ,XYZ,Y ) k[[X,Y,Z]] ⊆

15. Find A∗(p) where p = p(3, 4, 5). Rees Algebras

16. Show that the integral clsoure of R[It] in R[t] is the graded ring R It¯ I2t2 ⊕ ⊕ ⊕ · · ·

17. Let I and J be ideals of a commutative ring R. Put J ∗ = JR[t] R[It]. Show that (i) ∩ R[It]/J ∗ R/J[((I +J)/J)t] (ii) If J1,J2 are R-ideals then (J1 J2)∗ = J ∗ J ∗. (iii) ' ∩ 1 ∩ 2 If J = q1 ... qn is a reduced primary decomposition of J in R then J ∗ = q∗ ... q∗ ∩ ∩ 1 ∩ ∩ n is a reduced primary decomposition of J ∗ in R[It]. Length and Multiplicities

18. Let (R, m) be a d-dimensional Cohen-Macaulay local ring and a1, a2, . . . , ad a system of parameters of R. Show that for n1, n2, . . . , nd IN, ∈

n1 nd `(R/(a1 , . . . , ad )) = n1n2 . . . nd`(R/(a1, . . . , ad)). 19. Let (R, m) be a Cohen-Macaulay local ring with infinite residue field. Prove Ab- hyankar’s inequality e(R) µ(m) dim R + 1. Show that equality holds if and only if for some (hence all) minimal≥ reduction− J of m, Jm = m2.

20. Let (R, m) be a C-M ring. Let J be such that Jm = m2. Find `(R/mn).

21. Put R = k[X,Y ]. Let 0 = an < an 1 < < a1 and 0 = b1 < b2 < < bn be − ai bi ··· n 1 ··· integers. Set I = (X Y i = 1, 2, . . . , , n). Show that `(R/I) = − ai(bi+1 bi). | Pi=1 − Cowsik-Nori-Dade Theorem and Huneke’s Theorem

22. Let R = k[[X,Y,Z]] and p = p(3, 4, 5) in this and the next two problems. Show that there is a d p(2) such that p(2) = (d, p2). ∈ 23. Let I = p(2). Show that R/In is Cohen-Macaulay for all n but I is not a complete intersection.

(2n) (2)n (2n+1) (2n) (2) 2 24. Show that p = p for all n and p = pp . Thus s(p) = R[pt, p t ]. R 3 5 3 4 2 2 25. Let p = (X Y Z ,YZ X ,XZ Y ) in R = k[[X,Y,Z]]. Show that s(p) is Noetherian by− using Huneke’s− Theorem.− R 16

Open Problems

26. Find a criterian for the Noetherian property of s(p) where p is a dimension 2 prime of a regular local ring. R

27. Let d 2 and m 1 and gcd(d, m) = 1. Put ni = d + (i 1)m for i = 1, 2, . . . , d. ≥ ≥ − Let A = k[[X1,...,Xd]] and k[[T ]] be formal power series rings over a field k. Let ni ϕ : A k[[T ]] denote the homomorphism of k-algebras defined by ϕ(Xi) = T −→ for i = 1, 2, . . . , d. Put p = kerϕ. It is proved in [GN] that s(p) is Noetherian and for d 4, it is even Cohen-Macaulay. Is it Cohen-MacaulayR for all d? ≤ 28. Find an analogue of Huneke’s theorem on Noetherian symbolic Rees algebras for a dimension one Cohen-Macaulay ideal of a regular local ring.

29. Let a1, a2, . . . , an be a sequence of positive integers so that n 1 of them are in arithmetic progression. Let p denote the kernel of the k-algebra− homomorphism ai ϕ : k[[X1,...,Xn]] k[[T ]] defined by ϕ(Xi) = T for i = 1, 2, . . . , n. Is s(p) Noetherian ? D. P.−→ Patil [P] showed that p is a set- theoretic complete intersection.R

30. Let p and q be dimension one primes of a 3-dimensional regular local ring. Is the bigraded ring p(r)q(s)trts ever Noetherian? M 1 2 r,s 0 ≥ References

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