Taut-Level Hilbert Rings IV

Taut-Level Hilbert Rings IV

Kazunori FUJITA

INTRODUCTION. In this paper, all rings are commutative with identity and finite dimensional. In [ 1] an integral domain A is said to be a Ko-domain if some integral extension domain B of A is a Krull domain, and by [ 1, Proposition 1] , a domain A is a Ko-domain if and only if the integral closure of A is a Krull domain. A result [7, Theorem 2.7] states that if A is an integral domain, G is a torsion-free abelian group which has ace on cyclic subgroups, and M is a submonoid of G such that A [M] is gr-noetherian, then A [M] is a Ko-domain. In [6, p.6] a ring A is said to be taut-level if ht(p) +dim(A!p) =dim(A) for any prime ideal pin A. In [2] it was shown that a taut-level noetherian ring A is a Hilbert ring if and only if A [X] is taut-level, and it was shown that if A is a taut-level noethericµi Hilbert ring, then A [X1, ·· ·, Xn] is taut-level. Moreover, in [1] a ring A is said to be a strong Ko-ring if Alp is a Ko-domain for any prime ideal p in A. In this pal?er, we prove that a taut-level strong Ko-ring A is a Hilbert ring if and only if A [X] is taut-level, and we also show that for fields L~ K, the ring K

+

^xL [x] is a Ko-domain if and only if L is algebraic over K, where xis an indeterminate.

LEMMA 1. Let A be a Ko-domain, and let q be a prime ideal in A such that ht (q)

~ 2. Then 1 p E Ht1 (A) I q ~ p ^f is an infinite set, where Ht1 (A) is the set of height one prime ideals in A.

PROOF. Set U = 1 p E Ht1 (A) I q ~ p ^{f .} Let p1, · · ·, pn be elements of U. Then q

¥

^{pi U}

· · · U pn. Let a be an element of q-pi U · · · U pn. The height of any minimal prime ideal of aA is one by [ 1, Theorem 3] . Let pn+1 be a minimal prime ideal of aA such that q

~ p. Then pn+1 EU and pn+1 =i= pi (i= 1, · · ·, n). Therefore U is an infinite set.

LEMMA 2. Let A be a Ko-domain, and let Ube an infinite subset of Hti (A). Then np=O.

pEU

PROOF. The number of minimal prime ideals of aA is finite for any non-unit a ==I= 0 of A by [1, Lemma 8]. Hence n p=0 holds.

pEU

LEMMA 3. Let A be a Ko-domain, and let P be a non-zero prime ideal in A [X]

such that PnA=0. Let Ube an infinite subset of Ht1 (A). Then there exist an element q of U and QESpec(A[X]) such that QnA=q, Q~P.

PROOF. Let K be the quotient field of A. Since P==l=0 and PnA=0, there exists f(X) EA [X] such that PK[X] f(X) K[X]. Write

f(X)

=ax r+bX r-l+···+c

where a==l=0, b, ···, c EA. We have n q=0 by Lemma 2. Therefore there exists q E·u

j

^EU

such that aft.q. Then B=A[l/a, X IPA[l/a, X] is integral over A[l/a], hence there exists a prime ideal Nin B such that NnA[lla] =qA[l!a].Let Q= ^(f)-1(N) nA[X], where ^(fJ: A[l/a, X]~B be the natural A-algebra homomorphism. Then Q~P and Q

nA=q.

We say that a ring is a strong Ko-ring if Alp is a Ko-domain for any prime ideal p in

A.

LEMMA 4. Let A be a strong Ko-ring. Then ( 1) A is a strong S-ring.

(2) ht (pA [X]) = ht (p) for any prime ideal P in A.

(3) ht(p) =ht(PnA) +1 for any prime ideal Pin A[X] such that P;} (PnA)A[X].

(4) dim (A [X]) =dim (A)+ 1.

PROOF. Alp is a Ko-domain for any prime ideal in A. Thus Alp is S-domain by [ 1, Corollary 4]. Therefore A is a strong S-ring. So (2), (3) and (4) are immediate from [3, Theorem 39].

LEMMA 5. Let A be a strong Kq-domain. If A is a Hilbert ring, then dim(A[X]IP)

= dim (A) holds for any non-zero prime ideal P in A [X] such that P nA

=

0.

PROOF. We prove the assertion by induction on n=dim (A). If n . 0, then A is a field, hence the assertion is obvious. Assume that n

=

l. Suppose that P is a maximal ideal in A [X]. Then PnA =0 is maximal in A because A is a Hilbert ring. Therefore

1 ~dim(A [X]IP) <dim(A [X]) =dim (A)+ 1 =2

and hence dim(A[X]IP) =1. Now we assume that n=dim(A) L2. Let OCp1Cp2C···

Cpn be a saturated chain of prime ideals of length n in A. Then the set U= 1 q E Spec (A)

I

^{pz:)q:)0 f} is an infinite set by Lemma 1, and so by Lemma 3, there exist an element q of U and QESpec(A[X]) such that QnA=q, Q:)P. Since dim(Alq) =n -1, dim(A[X]IQ) =dim(Alq) holds by our induction. Then dim(A[X]IP) Ldim(A [X]IQ) +l=dim(Alq) +l=n. On the other hand, A[X]IP is algebraic over A, so that dim(A[X]IP)~dim(A) =n holds. Thus we have dim(A[X]IP) =dim(A).

LEMMA 6. Let A be a strong Ko-ring.

If

A is a Hilbert ring, then dim(A[X]IP) = dim(Al(PnA) holds for any prime ideal Pin A[X] such that P~ (PnA)A[X].

PROOF. The Proof is immediate by applying Lemma 5 to a strong Ko-domain Al(P nA).

THEOREM 7. Let A be a taut-level strong Ko-ring. Then the following statements are equivalent.

( 1) A is Hilbert ring.

(2) A [X] is taut-level.

PROOF. (1)

•

(2) Let P be a prime ideal in A[X] and set p=PnA. Since A is taut-level, ht(p) +dim(Alp) =dim(A). Assume at first that P=pA[X]. Then ht(P) = ht(p), dim(A[X]) =dim(A) +1 and dim(A[X]IP) =dim( (Alp) [X]) =dim(AIP) +1 hold by Lemma 4. Therefore we obtain that ht(P) +dim(A[X]IP) =dim(A[X]).

Next, assume that P~pA[X]. Then dim(A[X]IP) =dim(Alp) by Lemma 6. Since ht(P) =dim(Ap[X]) =ht(p) +1, we see that ht(P) +dim(A[X]IP) =ht(p) +dim (Alp)+ 1 =dim (A) + 1 = dim (A [X]). So ht (P) +dim (A [X] IP) = dim (A [X]). Thus A [X] is taut-level.

(2)

• (

1) Let M be a maximal ideal in A [X]. Since A [X] is taut-level, ht (M) = dim (A [X]). We shall show that p= MnA is a maximal id~al in A. Since M~pA [X], Lemma 4 yields that ht (M) = ht (p) + 1 ~ dim (A) + 1. Therefore ht (p) = dim (A) , so p is a maximal ideal A. Thus A is Hilbert ring.

Let B--;;> A be integral domains such that B is integral over A. Then B is said to be almost finite over A if the quotient field of B is finite over that of A [5, p. 30].

COROLLARY 8. If a taut-level Hilbert domain A zs almost finite over some noetherian domain, then A [X] is taut-level.

PROOF. A is a strong Ko-domain by [ 1, Proposition 1]. Therefore A

[X]

is taut-level by Theorem 7.

LEMMA 9. Let L-:JK be fields and let x be an indeterminate. Then the integral closure of K+xL[x] is K'+xL[x], where K' is an algebraic closure of Kin L.

PROOF. The proof is straightforward, and we omit it.

THEOREM 10. Let L-:JK be fields and let x be an indeterminate. Then K+xL[x] zs a Ko-domain if and only if Lis algebraic over K.

PROOF. Assume at first that L is not algebraic over K. Let K' be the algebraic closure of K in L. Then by Lemma 9, B

=

^K'

+

^xL [x] is the integral closure of K

+

xL[x]. Since P=xL[x] is a height one prime ideal in B, ·BP is a one dimensional normal local domain. Let t be any element of L- K'. We see easily that t and lit are not contained in BP Therefore BP is not a valuation domain. Thus B is not a Krull domain, and hence K+xL[x] is not a Ko-domain by [1, Proposition I]. Conversely, assume that L is algebraic over K. Then L[x] is the integral closure of K+xL[x].

Therefore K+xL[x] is a Ko-domain.

PROPOSITION 11. Let L-:J K be fields and let x1, · · ·, xn be indeterminates. If L is algebraic over K, then· (K

+

(x1, · · ·, xn) L [x1, · · ·, xn]) [X] is taut-level.

PROOF. Let A=K+ (x1, ···, xn)L[x1, ···, xnJ. Then K[x1, ···, xn] CACL[x1, ···, xn], and L [x1, ···, xn] is integral over K[x1, ···, xn]. Therefore A is a taut-level strong Ko-domain and is a Hilbert ring. Hence A [X] is taut-level by Theorem 7.

LEMMA 12. Let A be strong Ko-ring, and let p, P be prime ideals of A such that pCP, ht(Plp) =2. Set U=

1

qESpec(A)lpCqCP

l.

Lets be an element of A-p.

Then there exists an element q of U which does not contains.

PROOF. Since Alp is a Ko-domain,

n

^qlp

=

0 by Lemma 2. Therefore s tE q for some

qEU

elements q of U.

PROPOSITION 13. Let A be a strong Ko-ring. If A is a Hilbert ring, then dim (As) = dim (A) holds for any non-zero divisor s of A.

PROOF. Let p0Cp1 Cp2C ··· Cpn-1 Cpn be a saturated chain of prime ideals of length n in A, where n=dim (A). Since s is a non-zero divisor, po does not contain s. Then there exists a prime ideal q1 in A such that p0Cq1 Cp2 ands <f.q1 by Lemma 12. So we can find inductively a saturated chain of prime ideals p0Cq1 Cq2C ··· Cqn-1 Cpn in A such that s<f.qj (j=l, 2, ···, n-1). Since A is a Hilbert ring, we have

n

^M=qn-1

MEE

where E is the set of maximal ideals in A containing qn-1. Therefore there exists an element M of E such that s<f.M. Thus dim(As) ² n, and hence dim(As) =dim(A).

PROPOSITION 14. Let A be a taut-level strong Ko-ring.

If

A is a Hilbert ring, then As is taut-level for any non-zero divisor s of A.

PROOF. Let ps be a prime ideal in As, where p E Spec (A). Since A is taut-level, ht(p) +dim(A/p) =dim(A). Proposition 13 implies that dim(As) =dim(A) and dim ((A/p)s)=dim(A/p) hold. Therefore ht(ps)+dim(As!ps)=dim(As). Hence As is taut- level.

DEFINITION. Let A be a ring. We say that A is a stably strong Ko-ring if A [X1, ···, Xn] is a strong Ko-ring for any positive integer n.

As direct consequence of Theorem 7 and Proposition 14, we have :

THEOREM 15. Let A be a stably strong Ko-ring. If A is a Hilbert ring, then A [X1,

···, Xn] and A[X1, X1-1

, ···, Xn, Xn-1

] are taut-level.

PROOF. A [X1, · · ·, Xn] is a taut-level ring by Theorem 7. Therefore A [X1, · · ·, Xn] is a taut-level strong Ko-ring, so Proposition 14 shows that A [X1, ·· ·, Xn, 1/ (X1, ···, Xn)]

is also taut-level.

References

[ 1] K. Fujita, Minimal prime ideals of a finitely generated ideal, Hiroshima Math. J.

9(1979), 491-502.

[2] K. Fujita, Taut-level Hilbert rings ill, Mem. Fae. Kagawa Univ. II, 53 (2003), 15-17.

[3] I. Kaplansky, Commutative Rings, Allyn and Bacon, Boston, 1970.

[ 4] J. L. Mott and M. Zafrullah , Unruly Hilbert domains, Canad. Math. Bull. 33 ( 1990) , 106-109.

[5] M. Nagata, Local Rings, Interscience, New York, 1962.

[6] L. J. Ratliff, Chain Conjectures in Ring Theory, Springer-Verlag, 1978.

[7] D. E. Rush, Noetherian properties in monoid rings, J. Pure Appl. Algebra 185 (2003), 259-278.