PROJECTIVELY FULL IDEALS IN NOETHERIAN RINGS

Catalin Ciuperca, William J. Heinzer, Louis J. Ratliff Jr., and David E. Rush

Abstract Let R be a Noetherian commutative ring with unit 1 ≠ 0, and let I be a regular proper ideal of R. The set P(I) of integrally closed ideals projectively equivalent to I is linearly ordered by inclusion and discrete. There is naturally associated to P(I) a numerical semigroup S(I); we have S(I) = IN if and only if every element of P(I) is the integral closure of a power of the largest element J of P(I). If this holds, the ideal J and the set P(I) are said to be projectively full. If I is invertible and R is integrally closed, we prove that P(I) is projectively full. We investigate the behavior of projectively full ideals in various types of ring extensions. We prove that a normal ideal I of a local ring (R, M) is projectively full if I ⊈ M^{2} and both the associated graded ring G(M) and the fiber cone ring F (I) are reduced. We present examples of normal local domains (R, M) of altitude two for which the maximal ideal M is not projectively full.

1 INTRODUCTION.

All rings in this paper are commutative with a unit 1 ≠ 0. Let I be a regular proper ideal of the Noetherian ring R (that is, I contains a regular element of R and I ≠ R). The concept of projective equivalence of ideals and the study of ideals projectively equivalent to I was introduced by Samuel in [19] and further developed by Nagata in [12]. Making use of interesting work of Rees in [17], McAdam, Ratliff, and Sally in [11, Corollary 2.4] prove that the set P(I) of integrally closed ideals projectively equivalent to I is linearly ordered by inclusion (and discrete). They also prove that if I and J are projectively equivalent, then the set Rees I of Rees valuation rings of I is equal to the set Rees J of Rees valuation rings of J and the values of I and J with respect to these Rees valuation rings are proportional [11, Proposition 2.10]. We observe in [1] that the converse also holds and further develop the connections between projectively equivalent ideals and their Rees valuation rings. For this purpose, we define in [1] the ideal I to be projectively full if the set P(I) of integrally closed ideals projectively equivalent to I is precisely the set {(I^{n})_{a}} consisting of the integral closures of the powers of I. If there exists a projectively full ideal J that is projectively equivalent to I, we say that P(I) is projectively full. As described in [1], there is naturally associated to I and to the projective equivalence class of I a numerical semigroup S(I). One has S(I) = IN, the semigroup of nonnegative integers under addition, if and only if P(I) is projectively full.

Our goal in the present paper is to build on the work in [11] and [1] by further developing the concept of projectively full ideals and examining the numerical semigroup S(I). In Section 2 we present several results relating P(I) and P(IA), for certain R-algebras A. In Section 3, these results are applied to explore the relationship between the projective fullness of P(I) and P(IA), for certain R-algebras A. Several methods are given for obtaining projectively full ideals. We prove that an integrally closed complete intersection ideal of a local ring is projectively full. We also prove that if I is a proper invertible ideal of an integrally closed Noetherian domain, then P(I) is projectively full. In Section 4 we present classes of examples of projectively full ideals. For instance, we prove that if R_{P} is a regular local domain, then P I is projectively full for all regular ideals I ⊈ P. In Section 5 we present a family of examples of integrally closed local domains (R, M) of altitude two for which M, and therefore P(M), is not projectively full.

Our notation is as in [13] and [6]. A ring is said to be integrally closed if it is integrally closed in its total quotient ring. In particular, a ring that is equal to its total quotient ring is integrally closed. If we are given a ring homomorphism from a ring R to a ring A, then we say that A with respect to this homomorphism is an R-algebra.

We thank the referee for helpful suggestions for revising this paper.

2 PROJECTIVELY EQUIVALENT IDEALS.

In this section we prove several elementary results about projectively equivalent ideals. For this, we need the following definitions. (Throughout, IN denotes the set of nonnegative integers, and IN_{+} (resp., Q_{+}, IR_{+}) denotes the set of positive integers (resp., rational numbers, real numbers).)

Definition 2.1 Let I be an ideal in a Noetherian ring R.

(2.1.1) The Rees ring R(R, I) of R with respect to I is the graded subring R(R, I) = R[u, tI] of R[u, t], where t is an indeterminate and u = 1/t. (2.1.2) R′ denotes the integral closure of R in its total quotient ring. (2.1.3) If R is local with maximal ideal M, then a(I) denotes the analytic spread of I (so a(I) = altitude(R(R, I)/((u, M)R(R, I))) (see (2.1.1))). (2.1.4) I_{a} denotes the integral closure of I in R, so I_{a} = {b ∈ R | b satisfies an equation of the form b^{n} + i_{1}b^{n−1} + ⋯ + i_{n} = 0, where i_{k} ∈ I^{k} for k = 1, …, n}. The ideal I is said to be integrally closed in case I = I_{a}, and I is normal in case (I^{i})_{a} = I^{i} for all i ∈ IN_{+}. (2.1.5) An ideal J in R is a reduction of I in case J ⊆ I and JI^{n} = I^{n+1} for some n ∈ IN. (2.1.6) An ideal J in R is projectively equivalent to I in case (I^{i})_{a} = (J^{j})_{a} for some i, j ∈ IN_{+}.

Concerning (2.1.6), Samuel introduced projectively equivalent ideals in 1952 in [19], and a number of properties of projective equivalence can be found in [4], [5], [8], [9], [10], [11], [15], [16].

Remark 2.2 Let R be a Noetherian ring. Then (2.2.1) The relation “I is projectively equivalent to J” is an equivalence relation on I = {I | I is an ideal of R}. (2.2.2) [11, (2.1)(b)]: If I and J are ideals in R and if i, j, k, l ∈ IN_{+} with \frac{i}{j} = \frac{k}{l}, then (I^{i})_{a} = (J^{j})_{a} if and only if (I^{k})_{a} = (J^{l})_{a}. (2.2.3) Assume that I and J are projectively equivalent in R and let K be an ideal in R. Then (I + K)/K and (J + K)/K are projectively equivalent in R/K. (2.2.4) Let A be an R-algebra. If I, J are projectively equivalent in R, then IA, JA are projectively equivalent in A.

Concerning (2.2.4), it is not true in general that integral closedness of ideals is preserved under a faithfully flat ring extension. This need not be true even if (R, M) is a regular local domain and A = R̂ is the M-adic completion of R [3]. Thus there exists a regular local domain R and an integrally closed ideal I of R such that P(IR̂) ≠ {JR̂ | J ∈ P(I)} (see (2.4.3)).

With the preceding paragraph in mind, we note in (2.8) below that P(I), Rees I (see (2.3)), and d(I) (see (2.4.4)) behave nicely when passing to: R[X]; R(X); R_{S} (for certain multiplicatively closed subsets S of R); and, R/K (where K ⊆ Rad(R), the nilradical of R), and we also show that d(I) ≤ d(IA) for certain types of R-algebras A. For this, we first define, for a regular proper ideal I in a Noetherian ring R, the Rees valuation rings of I, the set P(I), and the positive integer d(I).

Definition 2.3 Let I be a regular proper ideal in a Noetherian ring R, for each x ∈ R let v_{I}(x) = max{k ∈ IN | x ∈ I^{k}} (as usual, I^{0} = R and v_{I}(x) = ∞ in case x ∈ I^{k} for all k ∈ IN), and let \overline{v}_{I}(x) = lim_{k→∞}(\frac{v_{I}(x^{k})}{k}). Rees shows in [17] that: (a) \overline{v}_{I}(x) is well defined; (b) for each k ∈ IN and x ∈ R, \overline{v}_{I}(x) ≥ k if and only if x ∈ (I^{k})_{a} (as usual, (I^{0})_{a} = R); and, (c) there exist valuations v_{1}, …, v_{g} defined on R (with values in IN ∪ {∞}) and positive integers e_{1}, …, e_{g} such that, for each x ∈ R, \overline{v}_{I}(x) = min\{\frac{v_{i}(x)}{e_{i}} | i = 1, …, g\}. (These v_{i} and e_{i} are described as follows: let z_{1}, …, z_{d} be the minimal prime ideals z in R such that z + I ≠ R, for i = 1, …, d, let R_{i} = R/z_{i}, let F_{i} be the quotient field of R_{i}, let R_{i} = R(R_{i}, (I + z_{i})/z_{i}), let p_{i,1}, …, p_{i,h_{i}} be the (height one) prime divisors of uR_{i}′ (see (2.1.2)), let w_{i,j} be the valuation of the discrete valuation ring W_{i,j} = R_{i}′_{p_{i,j}}, let e_{i,j} = w_{i,j}(u), let V_{i,j} = W_{i,j} ∩ F_{i}, and define v_{i,j} on R by v_{i,j}(x) = w_{i,j}(x + z_{i}). Then v_{1}, …, v_{g} are the valuations v_{1,1}, …, v_{d,h_{d}} resubscripted and e_{1}, …, e_{g} are the corresponding e_{i,j} resubscripted, and Rees I = {(V_{1}, N_{1}), …, (V_{g}, N_{g})}, where V_{i} is the valuation ring of the valuation v_{i}; note that IV_{i} = N_{i}^{e_{i}}V_{i}.

Definition 2.4 Let I be a regular proper ideal in a Noetherian ring R. (2.4.1) For α ∈ IR_{+} let I_{α} = \{x ∈ R | \overline{v}_{I}(x) ≥ α\} (see (2.3)). (2.4.2) W(I) = \{α ∈ IR_{+} | \overline{v}_{I}(x) = α for some x ∈ R\}. (2.4.3) U(I) = {α ∈ W(I) | I_{α} is projectively equivalent to I} (see (2.4.1), (2.4.2), and (2.1.6)), and P(I) = {I_{α} | α ∈ U(I)}. (2.4.4) d(I) is the smallest positive integer d such that, for all J ∈ P(I), (J^{d})_{a} = (I^{j})_{a} for some j ∈ IN_{+}. (See [11, Theorem 2.8].)

Remark 2.5 Let R be a Noetherian ring and let I be a regular proper ideal in R.

(2.5.1) Concerning (2.4.1), for each α ∈ IR_{+} the ideal I_{α} is an integrally closed ideal (= (I_{α})_{a}) in R, I_{n} = (I^{n})_{a} for all n ∈ IN_{+}, and for all k ∈ IN_{+} and for all I_{α} ∈ P(I) it holds that (I_{α}^{k})_{a} = I_{kα}, by [11, (2.1)(g), (2.1)(c) and (2.5)]. (2.5.2) The sets W(I) and U(I) of (2.4.2) and (2.4.3) are discrete subsets of Q_{+}, by [11, (1.1) and (2.8)]. (2.5.3) For the set P(I) of (2.4.3), P(I) = {J | J is an integrally closed ideal in R that is projectively equivalent to I}, and P(I) is linearly ordered by inclusion, by [11, (2.4)]. (2.5.4) Concerning (2.4.4), it is shown in [11, (2.8) and (2.9)] that there exists a unique smallest positive integer d(I) that is a common divisor (but not necessarily the greatest common divisor) of the integers e_{1}, …, e_{g} of (2.3) such that, for each ideal J in R that is projectively equivalent to I, (J^{d(I)})_{a} = (I^{j})_{a} for some j ∈ IN_{+}. Also, d(I)α ∈ IN_{+} for all α ∈ U(I). Further, if H, J ∈ P(I) and if d(H), d(J) are the corresponding unique positive integers for H, J, then (H^{d(J)})_{a} = (J^{d(H)})_{a}, by [1, (4.8.3)]. Finally, there exist n^{∗}(I) ∈ IN_{+} such that \{α ∈ U(I) | α ≥ n^{∗}(I)\} = \{n^{∗}(I) + \frac{k}{d(I)} | k ∈ IN\} (in fact, each large n ∈ IN_{+} is a suitable choice for n^{∗}(I)), by [11, (2.8)].

In Theorem 2.6 (together with its corollary) we show that d(I) ≤ d(IA) for certain standard types of R-algebras A.

Theorem 2.6 Let I be a regular proper ideal in a Noetherian ring R and let A be a Noetherian ring that is an R-algebra having the property that: (a) IA is regular and proper; and,

(b) if H ⊊ J in P(I), then (HA)_{a} ⊊ (JA)_{a} in P(IA). Then d(IA) ≥ d(I). If d(IA) > d(I), then d(IA) = kd(I) for some k ∈ IN_{+}.

Proof. By the last sentence in (2.5.4) let n be a large enough positive integer so that, for all integers m ≥ n, there are exactly d = d(I) ideals in P(I) between (I^{m+1})_{a} and (I^{m})_{a} (counting one endpoint) and there are exactly d′ = d(IA) ideals in P(IA) between (I^{m+1}A)_{a} and (I^{m}A)_{a} (counting one endpoint). Let H_{0}, H_{1}, …, H_{d} in P(I) such that (I^{n+1})_{a} = H_{d} ⊊ ⋯ ⊊ H_{0} = (I^{n})_{a}. Then each (H_{i}A)_{a} ∈ P(IA) (by (2.2.4)) and (I^{n+1}A)_{a} = (H_{d}A)_{a} ⊊ ⋯ ⊊ (H_{0}A)_{a} = (I^{n}A)_{a}, by (b). It follows that d′ ≥ d.

Now assume that d′ > d. Then to show that d′ = kd for some k ∈ IN_{+} it suffices to show that there are exactly \frac{d′}{d} − 1 ideals in P(IA) that are strictly between (H_{i+1}A)_{a} and (H_{i}A)_{a} for i ∈ {0, 1, …, d − 1}.

For this, it may clearly be assumed that d ≥ 2, so it suffices to show that if H_{i+1} ⊊ H_{i} ⊊ H_{i−1} are three consecutive ideals in P(I) between (I^{n+1})_{a} and (I^{n})_{a}, then the number of ideals in P(IA) between (H_{i+1}A)_{a} and (H_{i}A)_{a} is the same as the number of ideals in P(IA) between (H_{i}A)_{a} and (H_{i−1}A)_{a}.

For this, recall that H_{j} = I_{n+ \frac{j}{d}} (see (2.5.4)) and similarly (since (I^{n+1}A)_{a} = (IA)_{n+1} ⊆ (H_{i+1}A)_{a} ⊊ (H_{i}A)_{a} ⊊ (H_{i−1}A)_{a} ⊆ (IA)_{n} = (I^{n}A)_{a}) it follows that, for j ∈ {i − 1, i, i + 1}, (H_{j}A)_{a} = (IA)_{n+ \frac{h_{j}}{d′}} for some integer h_{j} between j and d′. Therefore there exist exactly h_{i+1} − h_{i} ideals in P(IA) between (H_{i+1}A)_{a} and (H_{i}A)_{a} (counting one endpoint), and there exist exactly h_{i} − h_{i−1} ideals in P(IA) between (H_{i}A)_{a} and (H_{i−1}A)_{a} (counting one endpoint), so it suffices to show that h_{i+1} − h_{i} = h_{i} − h_{i−1}.

For this, it follows from (2.5.1) that, for all m ∈ dIN_{+}, (H_{j}^{m})_{a} = (I_{n+ \frac{j}{d}} ^{m})_{a} = (I^{nm+ \frac{j}{d} m})_{a} and, for all m ∈ d′IN_{+}, ((H_{j}A)^{m})_{a} = (((IA)_{n+ \frac{h_{j}}{d′}})^{m})_{a} = ((IA)^{nm+ \frac{h_{j}}{d′} m})_{a} = ((I^{nm+ \frac{h_{j}}{d′} m})_{a}A)_{a}. Therefore with m = dd′ we get ((I^{ndd′+jd′})_{a}A)_{a} = ((H_{j}^{dd′})_{a}A)_{a} = ((H_{j}A)^{d′d})_{a} = ((I^{ndd′+h_{j}d})_{a}A)_{a}. Therefore it follows from (b) that (I^{ndd′+jd′})_{a} = (I^{nd′d+h_{j}d})_{a}, so we get: (i) (i+1)d′ = dh_{i+1} ∈ IN_{+} (for j = i + 1); (ii) id′ = dh_{i} ∈ IN_{+} (for j = i); and, (iii) (i − 1)d′ = dh_{i−1} ∈ IN_{+} (for j = i − 1). It follows by subtracting (ii) from (i) and (iii) from (ii) that h_{i+1} − h_{i} = \frac{d′}{d} = h_{i} − h_{i−1} ∈ IN_{+}, so the number of ideals in P(IA) between (H_{i+1}A)_{a} and (H_{i}A)_{a} is the number of ideals in P(IA) between (H_{i}A)_{a} and (H_{i−1}A)_{a}. Therefore d(IA) = d′ = kd = kd(I), where k = h_{i} − h_{i−1} ≥ 2.

Corollary 2.7 Let I be a regular proper ideal in a Noetherian ring R. Then (a) and (b) of Theorem 2.6 hold for the following types of R-algebras A:

1. A is a Noetherian integral extension ring of R such that IA is regular.

2. A is a faithfully flat Noetherian extension ring of R.

3. A = R_{S} for some multiplicatively closed subset S of R such that IR_{S} ≠ R_{S}. Therefore d(IA) = kd(I) for some positive integer k (possibly k = 1) for such rings A.

Proof. For (1) and (2) it is well known that (JA)_{a} ∩R = J_{a} for all regular proper ideals J in R, so (b) holds for such rings A. Also, (a) holds for the rings A in (1) by hypothesis, and (a) holds for the rings A in (2) by faithful flatness (since I is regular in R).

For the rings R_{S} as in (3), by hypothesis IR_{S} is proper, so (a) holds by flatness. For (b), let d = d(I) and let n and H_{0}, …, H_{d} be as in the first paragraph of the proof of (2.6). Suppose that H_{i+1}R_{S} = H_{i}R_{S} for some i ∈ {0, 1, …, d − 1}. (Note that JR_{S} = (JR_{S})_{a} if J is an ideal in R such that J = J_{a}.) Let H = H_{i+1} and J = H_{i}. Since H, J ∈ P(I), there exist h, j ∈ IN_{+} such that (H^{d})_{a} = (I^{h})_{a} and (J^{d})_{a} = (I^{j})_{a} (by (2.5.4)). Therefore (I^{h}R_{S})_{a} = H^{d}R_{S} = J^{d}R_{S} (since H_{i+1}R_{S} = H_{i}R_{S}) = (I^{j}R_{S})_{a}, hence h = j (since IR_{S} is a regular proper ideal). Therefore, since (H^{d})_{a} = (I^{h})_{a} and (J^{d})_{a} = (I^{j})_{a}, it follows that (H^{d})_{a} = (J^{d})_{a}, so H_{a} = J_{a}, hence H_{i+1} = H_{i}, and this is a contradiction. It follows that the ideals H_{i}R_{S} are distinct (so (b) holds) and are ideals in P(IR_{S}) between (I^{n+1}R_{S})_{a} and (I^{n}R_{S})_{a}, hence d(IR_{S}) ≥ d(I).

The last statement is clear from (2.6) and what has already been shown.

It would be interesting to know whether Corollary 2.7 holds in general for a Noetherian ring A that is a flat R-algebra such that IA ≠ A.

We next consider classes of R-algebras A for which d(IA) = d(I). For these algebras, P(I) and Rees I also extend nicely. The proofs of these results ((2.8.1) - (2.8.4)) are straightforward, so they are omitted.

Proposition 2.8 Let I be a regular proper ideal in a Noetherian ring R. (2.8.1) Let K be an ideal in R such that K ⊆ Rad(R). Then P((I +K)/K) = {(J +K)/K | J ∈ P(I)}, Rees (I + K)/K = Rees I, and d((I + K)/K) = d(I). (2.8.2) Let P_{1}, …, P_{k} be the centers in R of the Rees valuations of I, and let S be a multiplicatively closed subset of the nonzero elements of R. Assume that S ∩ P_{i} = ∅ if and only if i = 1, …, h (for some integer h with 1 ≤ h ≤ k). Then Rees IR_{S} = {(V, N) ∈ Rees I | N ∩ R ∈ {P_{1}, …, P_{h}}}, and d(IR_{S}) ≥ d(I). Moreover, if h = k, then P(IR_{S}) = {JR_{S} | J ∈ P(I)} and d(IR_{S}) = d(I). (2.8.3) Let X_{1}, …, X_{g} be independent indeterminates, and let A = R[X_{1}, …, X_{g}]. Then P(IA) = {JA | J ∈ P(I)}, Rees IA = {V [X_{1}, …, X_{g}]_{N[X_{1},...,X_{g}]} | (V, N) ∈ Rees I}, and d(IA) = d(I). (2.8.4) Let X_{1}, …, X_{g} be independent indeterminates. As in [13, pp. 17-18], let A = R(X_{1}, …, X_{g}) denote the quotient ring of the polynomial ring R[X_{1}, …, X_{g}] with respect to the multiplicatively closed set of polynomials whose coefficients generate the unit ideal of R. Then P(IA) = {JA | J ∈ P(I)}, Rees IA = {V [X_{1}, …, X_{g}]_{N[X_{1},...,X_{g}]} | (V, N) ∈ Rees I}, and d(IA) = d(I)

Remark 2.9 Let I be a regular proper ideal in a Noetherian ring R. Let A be either R[X_{1}, …, X_{g}] or R(X_{1}, …, X_{g}) (g ≥ 1), and let J be an ideal in A. If there exist i, j ∈ IN_{+} such that either J^{j} = I^{i}A or (J^{j})_{a} = (I^{i}A)_{a}, then there exists an ideal G in R such that (G^{j})_{a} = (I^{i})_{a} and ((GA)^{j})_{a} = ((G^{j})_{a})A = (J^{j})_{a}, so (J^{j})_{a} is the integral closure of the extension of the j-th power of an ideal in R.

Proof. If either J^{j} = I^{i}A or (J^{j})_{a} = (I^{i}A)_{a}, then IA, J are projectively equivalent in A, so J_{a} ∈ P(IA) (by (2.5.3)). But by (2.8.3) (if A = R[X_{1}, …, X_{g}]) or by (2.8.4) (if A = R(X_{1}, …, X_{g})) P(IA) = {HA | H ∈ P(I)}. It follows that J_{a} = GA, where G = J_{a} ∩ R. It therefore follows that ((I^{i})_{a})A = (J^{j})_{a} = ((J_{a})^{j})_{a} = ((GA)^{j})_{a} = ((G^{j})_{a})A, so (J^{j})_{a} = ((G^{j})_{a})A and (G^{j})_{a} = (I^{i})_{a}.

In Proposition 2.10, we consider projective equivalence for invertible ideals in an integrally closed Noetherian ring. Concerning the hypothesis of (2.10), we remark that an integrally closed local ring R that contains a regular proper principal ideal is an integral domain. Moreover, if R is an integrally closed Noetherian ring with Rad(R) = (0), then R is a finite product of Noetherian integrally closed domains.

Proposition 2.10 Let R be an integrally closed Noetherian ring and let I be a regular proper ideal of R.

1. If I is invertible, then every ideal projectively equivalent to I is invertible.

2. If R is local and I = bR is principal, then every ideal J in R that is projectively equivalent to I is principal and invertible.

Proof. Since a regular ideal I of a Noetherian ring R is invertible if and only if IR_{M} is principal for each maximal ideal M of R and since projective equivalence behaves well with respect to localization, to prove item 1 it suffices to prove item 2. Thus we may assume (R, M) is an integrally closed local ring and I = bR is a regular proper principal ideal (so R is an integral domain). Let J be an ideal in R that is projectively equivalent to bR, so (J^{i})_{a} = (b^{m}R)_{a} for some i, m ∈ IN_{+}. Then a(J) = 1 (see (2.1.3)), since a(J) = a((J^{i})_{a}) = a((b^{m}R)_{a}) = a(bR) = 1. Assume temporarily that R/M is infinite and let xR be a minimal reduction of J. Then xR = (xR)_{a}, since R is integrally closed, and xR ⊆ J ⊆ (xR)_{a}, so J = xR is principal. Now, if R/M is finite, then T = R[X]_{MR[X]} is an integrally closed local domain with infinite residue field and JT is projectively equivalent to bT, by (2.2.4), so JT is principal. Since a minimal generating set for J is a generating set for JT, it can be reduced to a minimal (one element) generating set for JT, so J is principal (since J = JT ∩ R). Since J is principal, J is invertible.

Remark 2.11 Without the assumption that R is integrally closed, easy examples show that Proposition 2.10 fails. For example, if (R, M) is a local domain of altitude one that is not integrally closed and I = xR is a reduction of M, then M is projectively equivalent to I and M is not invertible. For a specific example, let t be an indeterminate over a field k and consider the subring R = k[[t^{2}, t^{3}]] of the formal power series ring k[[t]]. Then I = t^{2}R is a reduction of M = (t^{2}, t^{3})R, so M is projectively equivalent to I.

3 PROJECTIVELY FULL IDEALS.

Projectively full ideals are introduced in [1, Section 4]. It is observed in [1, (4.12)] that P(I) is projectively full for every nonzero proper ideal I in a regular local domain of altitude two; see also [11, (3.6)]). In this section we develop basic properties of projectively full ideals. We then determine various classes of ideals I for which either I or P(I) is projectively full. Concerning the basic properties, our main results show that: for certain R-algebras A, if IA is projectively full, then I is projectively full (see (3.2)); the converse holds for A as in (2.8) (by (3.5)); and, the converse need not hold, even if A is a finite free integral extension domain of R (see (4.2.1)). Concerning ideals I where either I or P(I) is projectively full, we show that: the integrally closed complete intersection ideals of a local ring are projectively full (see (3.6) and (3.7)); and P(I) is projectively full for an invertible ideal I in an integrally closed Noetherian ring (see (3.8)).

We begin with the following remark.

Remark 3.1 Let I be a regular proper ideal in a Noetherian ring R. (3.1.1) It is immediate from the relevant definitions that for each J ∈ P(I) (see (2.3.3)) we have {(J^{i})_{a} | i ∈ IN_{+}} ⊆ P(I), and P(I) is projectively full if and only if there exists J ∈ P(I) such that {(J^{i})_{a} | i ∈ IN_{+}} = P(I). It is clear that if such an ideal J exists, then J must be the largest element in the linearly ordered (discrete) set P(I). The numerical semigroup S(I) = d(I)U(I)∪{0} (where U(I) is as in (2.3.3)) is the semigroup of nonnegative integers under addition, if and only if P(I) is projectively full. (3.1.2) It follows from (2.5.4) and (3.1.1) that P(I) is projectively full if and only if d(K) = 1, where K is the largest element in P(I); cf [1, (4.11)]. Also, I is projectively full if and only if d(I) = 1.

Proposition 3.2 is an immediate corollary of (2.7) and (3.1.2).

Proposition 3.2 Let I be a regular proper ideal in a Noetherian ring R and let A be one of the following types of R-algebras:

1. A is a Noetherian integral extension ring of R such that IA is regular.

2. A is a faithfully flat Noetherian extension ring of R.

3. A = R_{S} for some multiplicatively closed subset S of R such that IR_{S} ≠ R_{S}.

If IA is projectively full, then I is projectively full.

Proof. If IA is projectively full, then d(IA) = 1, by (3.1.2), so d(I) = 1, by (2.7), so I is projectively full, by (3.1.2).

Remark 3.3 (3.3.1) The converses of (3.2)(1) and (3.2)(2) are not true; in (4.2.1) below we give specific examples where A is a finite free integral extension and I is a projectively full ideal such that IA is not projectively full. (3.3.2) The converse of (3.2)(3) is not true; in fact, it may happen that:

1. I is projectively full, but IR_{S} is not projectively full.

2. P(I) is projectively full, but P(IR_{S}) is not projectively full.

Proof. (of 3.3)(2) For (1), let (R, M = (x, y, z)R) be a regular local domain of altitude three, let p = zR, let P = (x, y)R, and let I = pP^{2}. Then I is projectively full, by (4.1.3) below, p and P are the centers in R of two of the Rees valuation rings of I, and IR_{P} = P^{2}R_{P} is not projectively full. (However, in this example P(IR_{P}) = {P^{i}R_{P} | i ∈ IN_{+}} is projectively full.)

For (2), let A be a Noetherian ring having a regular proper ideal J such that P(J) is not projectively full. For example, let A be the normal local domain of [1, Example 4.14]. Let X be an indeterminate over A and let R = A[X]. Then the ideal I = XJR is projectively full by (4.1.4), so P(I) is projectively full. Let S be the multiplicative system generated by X. Then IR_{S} = JR_{S}, and P(JR_{S}) is not projectively full.

In the case where A is a faithfully flat Noetherian extension of R, or the case where A = R_{S} and IA ≠ A, it would be interesting to know if P(IA) is projectively full implies that P(I) is projectively full.

Remark 3.4 There are often many ideals that localize to the same ideal in a localization. With this in mind, if JR_{S} = IR_{S} and IR_{S} is projectively full, then so are both I and J, by (3.2)(3). So assume, for example, that (R, M) is a regular local domain, let P ≠ M be a nonzero prime ideal in R, and let b ∈ R − P. Then for all positive integers n it holds that every ideal between b^{n}P and P is projectively full (since b^{n}P R_{P} = P R_{P} is projectively full).

In Proposition 3.5 we consider extension rings A of R for which the converse of (3.2) holds. This result is an immediate corollary of (2.8) and (3.1.2).

Proposition 3.5 Let I be a regular proper ideal in a Noetherian ring R. (3.5.1) Let K be an ideal in R such that K ⊆ Rad(R). Then I (respectively P(I)) is projectively full in R if and only if (I + K)/K (respectively P((I + K)/K)) is projectively full in R/K. (3.5.2) Let P_{1}, …, P_{k} be the centers in R of the Rees valuations of R and let S be a multiplicatively closed subset of R such that S ∩ (P_{1} ∪ ⋯ ∪ P_{k}) = ∅. Then I (respectively P(I)) is projectively full in R if and only if IR_{S} (respectively P(IR_{S})) is projectively full in R_{S}. (3.5.3) Let X_{1}, …, X_{g} be indeterminates and let A = R[X_{1}, …, X_{g}]. Then I (respectively P(I)) is projectively full in R if and only if IA (respectively P(IA)) is projectively full in A. (3.5.4) Let X_{1}, …, X_{g} be indeterminates, and let A = R(X_{1}, …, X_{g}). Then I (respectively P(I)) is projectively full in R if and only if IA (respectively P(IA)) is projectively full in A.

Proof. For (3.5.1), I is projectively full if and only if d(I) = 1 (by (3.1.2)) if and only if d((I +K)/K) = 1 (by (2.8.1)) if and only if (I +K)/K is projectively full (by (3.1.2)). Also, P(I) is projectively full if and only if there exists J ∈ P(I) such that d(J) = 1 (by (3.1.2)) if and only if d((J + K)/K) = 1 (by (2.8.1)) if and only if P((I + K)/K) is projectively full (by (3.1.2), since (J + K)/K ∈ P((I + K)/K)).

The proof of (3.5.2) (resp., (3.5.3), (3.5.4)) is similar using (2.8.2) (resp., (2.8.3), (2.8.4)) in place of (2.8.1).

In Proposition 3.6 (and its corollary) we show that the integrally closed complete intersection ideals of a local ring as classified by Goto are projectively full.

Proposition 3.6 Let (R, M) be a local ring and I a normal ideal in R with I ⊈ M^{2}. If both the associated graded ring G(M) = ⨁_{n≥0} M^{n}/M^{n+1} and the fiber cone ring F (I) = ⨁_{n≥0} I^{n}/MI^{n} are reduced, then I is projectively full.

Proof. Since G(M) is reduced, the maximal ideal M is also normal. For it is shown in [18, Theorem 2.1] that G(M) = R(R, M)/uR(R, M), so G(M) is reduced if and only if uR(R, M) is the intersection of its minimal prime divisors if and only if pR(R, M)_{p} = uR(R, M)_{p} for all prime divisors p of uR(R, M); it follows that each such R(R, M)_{p} is a discrete valuation ring, so uR(R, M) = (uR(R, M))_{a}, by [14, Theorem 2.10], so M^{n} = u^{n}R(R, M) ∩ R = (u^{n}R(R, M))_{a} ∩ R = M^{n}_{a} for all n ∈ IN_{+}, hence M is normal. Let J be an integrally closed ideal such that (J^{t})_{a} = I^{s} for some positive integers t, s. Then we must have s ≥ t. If not, then I^{s} ⊆ (J^{s+1})_{a} ⊆ M^{s+1}. Choose a ∈ I \ M^{2}. If a′ denotes the image of a in M/M^{2} ⊆ G(M), then (a′)^{s} = 0, hence a′ = 0, contradicting the choice of a.

In particular, since (J^{t})_{a} ⊆ I^{t}, we have J ⊆ I. Let k be the positive integer such that J ⊆ I^{k} and J ⊈ I^{k+1}. We will prove that J = I^{k}.

We first show that J ⊈ MI^{k}. Assume that J ⊆ MI^{k}. Then I^{s} = (J^{t})_{a} ⊆ (M^{t}I^{kt})_{a}. This implies that kt ≤ s − t, otherwise I^{s} ⊆ (M^{t}I^{s−t+1})_{a} ⊆ M^{s+1}, which, as shown above, is not true. But then (J^{t})_{a} = I^{s} ⊆ I^{(k+1)t} and hence J ⊆ I^{k+1}, contradicting the choice of k.

We now consider the fiber cone ring F (I) = ⨁_{n≥0} I^{n}/MI^{n}. Let x ∈ J \ MI^{k}. The image of x in (J + MI^{k}/MI^{k}) ⊆ (I^{k}/MI^{k}) is nonzero and, since F (I) is reduced, we have x^{t} ∈ J^{t} \ MI^{kt}. This shows that J^{t} ⊈ MI^{kt}, and since J^{t} ⊆ I^{s}, we get s ≤ kt. On the other hand, since J ⊆ I^{k}, we have I^{s} = (J^{t})_{a} ⊆ I^{kt}, and therefore s ≥ kt. In conclusion, s = kt and from (J^{t})_{a} = I^{kt} we obtain I^{k} ⊆ J_{a} = J.

In [2], Goto described the M-primary integrally closed complete intersection ideals in a local ring (R, M). He proves that such ideals I exist only when the ring R is regular, in which case there exist regular parameters x_{1}, …, x_{d} and a positive integer n such that I = (x_{1}^{n}, x_{2}, …, x_{d})R. Moreover, all the powers of such an ideal are integrally closed, i.e., the ideal is normal. An immediate corollary of Proposition 3.6 shows that the ideals of this type are also projectively full.

Corollary 3.7 Let (R, M) be a regular local domain of altitude d ≥ 2 and let x_{1}, …, x_{d} be a regular system of parameters. Then every ideal of the form I = (x_{1}^{n}, x_{2}, …, x_{d})R is projectively full.

Proof. Under these assumptions, both G(M) = R(R, M)/uR(R, M) and F (I) = R(R, M)/(u, M)R(R, M) are polynomial rings in d variables with coefficients in R/M, and therefore reduced. Also, I is normal, so the conclusion follows from (3.6).

In Proposition 3.8, we show that the projective equivalence class of each proper invertible ideal in an integrally closed Noetherian ring is projectively full.

Proposition 3.8 Let R be an integrally closed Noetherian ring and let I be a nonzero proper ideal of R.

1. If I is invertible, then P(I) is projectively full.

2. If R is local and I is principal (or equivalently invertible), say I = bR, then there exists x ∈ R such that P(bR) = {x^{i}R | i ∈ IN_{+}}.

Proof. Assume that I is invertible. By Proposition 2.10 every ideal projectively equivalent to I is invertible. Let K be the largest element of P(I) and let H ∈ P(I). Since H is an arbitrary element of P(I), to prove that P(I) is projectively full it suffices to prove that H is a power of K. By the linear order of P(I), there exists n ∈ IN_{+} such that K^{n+1} ⊆ H ⊆ K^{n}. If H = K^{n} we are done. If H ⊊ K^{n}, then H = JK^{n} for some invertible ideal J such that K ⊆ J ⊊ R. Since H and K are projectively equivalent and since invertible ideals of an integrally closed ring are integrally closed, there exist positive integers h and k such that H^{h} = K^{k}. Therefore K^{k} = H^{h} = J^{h}K^{nh}. If k ≤ nh, then multiplying this equation by K^{−k} gives a contradiction to the fact that J is a proper ideal. Therefore k > nh. Multiplying by K^{−nh}, we obtain K^{k−nh} = J^{h}. Thus K and J are projectively equivalent. Since K is the largest ideal projectively equivalent to H, we have K = J and H = K^{n+1}. This proves item 1. Since an invertible ideal of a local domain is principal, item 2 is an immediate consequence of item 1.

In connection with Proposition 3.8, a question we have considered, but not resolved, is whether P(I) is always projectively full if I is a divisorial ideal of an integrally closed Noetherian domain. A complicating factor here is that the integral closure of a power of the divisorial ideal I may fail to be divisorial. We discuss this in Remark 3.9.

Remark 3.9 Let R be an integrally closed Noetherian domain. Invertible ideals of R are divisorial, and a divisorial ideal I of R is uniquely representable as the intersection of symbolic powers of the height-one prime ideals that contain it, say

I = P_{1}^{(e_{1})} ∩ P_{2}^{(e_{2})} ∩ ⋯ ∩ P_{g}^{(e_{g})}.

Let v_{i} denote the normalized valuation associated to the valuation domain R_{P_{i}}. Then v_{i}(I) = e_{i}. The valuations v_{i} and positive integers e_{i} are a subset of the valuations and associated positive integers mentioned in Definition 2.3. If I is invertible, then this subset is all the valuations of Definition 2.3, and Rees I = {R_{P_{1}}, …, R_{P_{g}}} since I is invertible if and only if IR_{P} has analytic spread one for every prime ideal P of R that contains I.

Notice that (I^{n})_{a} is divisorial for all n ∈ ℕ_{+} if and only if Rees I = {R_{P_{1}}, …, R_{P_{g}}}. A prime ideal P is the center of a Rees valuation ring of I if the analytic spread of IR_{P} is equal to the height of P, and the converse holds if R_{P} is quasi-unmixed [7, Prop. 4.1]. If I ⊂ P and P is of height 2, then R_{P} is Cohen-Macaulay and thus unmixed, so P is the center of a Rees valuation ring of I if and only if IR_{P} has analytic spread 2. Thus if R has altitude two, then the Rees valuation rings of a divisorial ideal I of R are all centered on height-one primes of R if and only if I is invertible. However, there exists a normal local domain (R, M) of altitude 3 that has a divisorial height-one prime ideal P such that P has analytic spread 2 (so P is not invertible) and yet R_{P} is the unique Rees valuation ring of P (so (P^{n})_{a} = P^{(n)} is divisorial for all n ∈ IN_{+}). For a specific example, let k be a field and let R = k[[x, y, z, w]], where xy = zw. Then P = (x, z)R is a height-one prime of R and P has analytic spread 2. Since the localization of R at any nonmaximal prime ideal is a regular local domain, if P ⊂ Q with Q a prime of R of height 2, then P R_{Q} is principal. Therefore R_{P} is the unique Rees valuation ring of P.

In connection with Remark 3.9, it seems natural to ask:

Question 3.10 If I is a divisorial ideal of an integrally closed Noetherian domain and if the integral closure of I^{n} is divisorial for every n ∈ ℕ_{+}, does it follow that P(I) is projectively full?

Remark 3.11 Without the assumption that R is integrally closed, easy examples show that Proposition 3.8 fails. Indeed, if (R, M) is a local domain of altitude one such that the integral closure of R is a valuation domain, then all the M-primary ideals of R are projectively equivalent, and given an arbitrary numerical semigroup S, it is possible to construct a local domain (R, M) of altitude one such that the integral closure of R is a valuation domain and such that S(M) = S. Let b_{1} < b_{2} < ⋯ < b_{r} be positive integers having greatest common divisor 1, and let S = ⟨b_{1}, b_{2}, …, b_{r}⟩ denote the numerical semigroup determined by b_{1}, …, b_{r}. Let t be an indeterminate over the field k and let R be the subring k[[t^{b_{1}}, t^{b_{2}}, …, t^{b_{r}}]] of the formal power series ring k[[t]]. Then R is local with maximal ideal M = (t^{b_{1}}, …, t^{b_{r}})R and the integral closure of R is the valuation domain k[[t]]. The integrally closed M-primary ideals of R are precisely the M-primary ideals that are contracted from k[[t]]. If I is M-primary, then I_{a} = Ik[[t]]∩R and Ik[[t]] = t^{b}k[[t]], where b ∈ S. Thus the integrally closed M-primary ideals are in one-to-one correspondence with the elements of S = ⟨b_{1}, … b_{r}⟩. Therefore S(M) = S. We conclude that every numerical semigroup S is realizable as S(M) for a local domain (R, M) of altitude one. In particular, M is projectively full if and only if S = IN, or, equivalently, if and only if R = k[[t]].

In Remark 3.12 we obtain a partial extension of Proposition 3.8 to integrally closed regular principal ideals in Noetherian rings that are not integrally closed.

Remark 3.12 Let R be a Noetherian ring and let I = bR be an integrally closed regular proper principal ideal. Then the following hold: (3.12.1) For each ideal H in R that is projectively equivalent to I, H^{n} is principal for infinitely many n ∈ IN_{+}, and H^{n} = (H^{n})_{a} for all large n ∈ IN_{+}. (3.12.2) If the largest ideal in P(I) is principal, say xR, then P(I) is projectively full and each ideal projectively equivalent to I is a power of xR. (3.12.3) If c is a regular element in R such that Rad(cR) = Rad(bR), then bR = (bR)_{a} if and only if cR = (cR)_{a}. Therefore if b is a regular element in R such that bR = (bR)_{a}, then the conclusion of (3.12.1) holds for all principal ideals cR such that Rad(cR) = Rad(bR).

Proof. For (3.12.1), let H be an ideal in R that is projectively equivalent to I. Then (H^{h})_{a} = b^{d}R for some d, h ∈ IN_{+}. (It follows from [14, Theorem 2.10] that bR = (bR)_{a} if and only if R_{p} is integrally closed for all prime divisors p of bR; it follows from this (and the fact that bR and b^{i}R have the same prime divisors for all i ∈ IN_{+}) that if bR = (bR)_{a}, then b^{i}R = (b^{i}R)_{a} for all i ∈ IN_{+}.) Therefore H^{h} ⊆ (H^{h})_{a} = b^{d}R, so there exists an ideal J in R such that H^{h} = b^{d}J. Therefore (b^{d}R)_{a} = b^{d}R = (H^{h})_{a} = (b^{d}J)_{a}, hence R = b^{d}R : b^{d}R = (b^{d}J)_{a} : b^{d}R = J_{a}, so J = R and H^{h} = b^{d}R. Therefore H^{hn} = b^{dn}R is integrally closed and principal for all n ∈ IN_{+}. It therefore follows from [7, (11.15)] that H^{i} = (H^{i})_{a} for all large i ∈ IN_{+}.

For (3.12.2), assume that xR is the largest ideal in P(I) and let H be an ideal in R that is projectively equivalent to I. To complete the proof it must be shown that H = x^{i}R for some i ∈ IN_{+}.

For this, by the linear order of P(I) there exists n ∈ IN_{+} such that x^{n+1}R ⊆ H ⊆ x^{n}R. If H = x^{n}R, we are done. If H ⊊ x^{n}R, then H = x^{n}J for some ideal J in R such that xR ⊆ J ⊊ R. Since H are xR are projectively equivalent, (H^{h})_{a} = x^{i}R for some h, i ∈ IN_{+}. Therefore (x^{hn}J^{h})_{a} = x^{i}R. If i ≤ hn, then (x^{hn−i}J^{h})_{a} = (x^{hn}J^{h})_{a} : x^{i}R = x^{i}R : x^{i}R = R, and this contradicts J ≠ R. Therefore i > hn, so (J^{n})_{a} = (x^{hn}J^{n})_{a} : x^{hn}R = x^{i}R : x^{hn}R = x^{i−hn}R, so J is projectively equivalent to xR. Since J ⊇ xR and xR is the largest ideal in P(I), it follows that J = xR, hence H = x^{n}J = x^{n+1}R.

(3.12.3) follows as in the parenthetical part of the proof of (3.12.1).

4 EXAMPLES OF PROJECTIVELY FULL IDEALS.

We use Remark 3.1.2 to obtain several classes of examples of projectively full ideals.

Example 4.1 (4.1.1) If the integer e_{i} = 1 for some Rees valuation ring V_{i} of I (see (2.3)), then I is projectively full. Moreover, if J is a regular proper ideal in R that is not contained in the center P_{i} in R of V_{i}, then IJ is projectively full. (4.1.2) Let P be a prime ideal in a Noetherian ring R such that R_{P} is a regular local domain. Then P is projectively full. Moreover, P I is projectively full for all regular ideals I in R such that I ⊈ P. In particular, if (R, M) is a regular local domain, then M^{n}P is projectively full for all n ∈ IN_{+} and for all nonzero prime ideals P ≠ M. (4.1.3) Let x be a regular parameter in a regular local domain R and let I be an ideal in R such that ht(I) ≥ 2. Then xI is projectively full. (4.1.4) Let R be a Noetherian domain and let X be an indeterminate. Then XIR[X] is projectively full in R[X] for every nonzero ideal I in R. (4.1.5) Let R be a Noetherian domain, let X be an indeterminate, and let I be a regular proper ideal in R that is not projectively full. Therefore d := d(I) > 1. For each n > 1 in IN_{+} it holds that X^{n}IR[X] is projectively full if and only if n and d are relatively prime. In particular, X^{n}IR[X] is projectively full for all n ∈ IN_{+} if and only if I is projectively full in R.

Proof. It is noted in (2.4.4) that the integer d = d(I) is a common divisor of the integers e_{1}, …, e_{g} of (2.3). Therefore, since I is projectively full if and only if d(I) = 1 (by (3.1.2)), the first statement in (4.1.1) is clear.

For the second statement in (4.1.1), it is shown in [1, (3.6)] that for all regular proper ideals I and J in R one has Rees I ∪ Rees J ⊆ Rees IJ. Therefore V_{i} is a Rees valuation ring of IJ, and if J ⊈ P_{i}, then the integer e_{i} of V_{i} for IJ is also one (since (IJ)V_{i} = IV_{i} = N^{e_{i}} = N, where N is the maximal ideal of V_{i}), so IJ is projectively full by the first statement of this remark.

For (4.1.2), the integer e of the order valuation ring V of P is one, so the first two statements follow immediately from the second statement in (4.1.1), and the third statement is a special case of the second statement.

For (4.1.3), if x is a regular parameter in a regular local domain R, then R_{xR} is a regular local domain, so it follows from (4.1.2) that if ht(I) > 1, then xI is projectively full.

For (4.1.4), P = XR[X] is a prime ideal such that R[X]_{P} is a regular local domain. Therefore, since IR[X] ⊈ XR[X] for all nonzero ideals I of R, it follows from (4.1.2) that XIR[X] is projectively full.

For (4.1.5), assume first that e > 1 is a divisor of d = d(I), say d = eq. Let n^{∗}(I) ∈ IN_{+} such that {n^{∗}(I) + (i/d) | i ∈ IN} ⊆ U(I) (see (2.5.4)). Then n^{∗}(I) + (q/d)) ∈ U(I), so the ideal H := I_{n^{∗}(I)+(q/d)} ∈ P(I), by (2.4.3), hence (H^{d})_{a} = (I^{(n^{∗}(I))d+q})_{a}, by [11, (2.3)(b)]. However, ((n^{∗}(I))d+q)/d) = (n^{∗}(I)e+1)/e, so (H^{e})_{a} = (I^{n^{∗}(I)e+1})_{a}, by (2.5.1). Therefore, for all h ∈ IN_{+} it holds that

((X^{(n^{∗}(I)e+1)h}H)^{e})_{a} = ((X^{eh(n^{∗}(I)e+1)}H^{e}))_{a} = ((X^{eh})^{(n^{∗}(I)e+1)}H^{e}))_{a} =

((X^{eh}I)^{n^{∗}(I)e+1})_{a}, so X^{(n^{∗}(I)e+1)h}H is projectively equivalent to X^{eh}I, and it is clear that (X^{(n^{∗}(I)e+1)h}H)_{a} is not the integral closure of any power of X^{eh}I, so X^{eh}I is not projectively full.

Conversely, assume that d ∈ IN_{+} is such that X^{d}I is not projectively full, so it must be shown that d = d_{i}e for some divisor d_{i} > 1 of d = d(I) and for some e ∈ IN_{+}. By hypothesis there exists an ideal J in R[X] that is projectively equivalent to X^{d}I such that J_{a} ≠ ((X^{d}I)^{n})_{a} for all n ∈ IN_{+}. Now (J^{j})_{a} = ((X^{d}I)^{n})_{a} = X^{nd}(I^{n})_{a}, for some j, n ∈ IN_{+}, and by [11, (2.1)(b)] it may be assumed that j, n are relatively prime (and j > 1, by the preceding sentence). Now J^{j} ⊆ (J^{j})_{a} = X^{nd}(I^{n})_{a} ⊆ X^{nd}R[X], hence J^{j} = X^{nd}K for some ideal K in R[X] such that K ⊈ XR[X] (since IR[X] ⊈ XR[X]). Let m ∈ IN such that J ⊆ X^{m}R[X] and J ⊈ X^{m+1}R[X], so J = X^{m}H for some ideal H in R[X]. Then X^{mj}H^{j} = J^{j} = X^{nd}K. Therefore mj = nd (= nd(I)), since K ⊈ XR[X] and H^{j} ⊈ XR[X] (since H ⊈ XR[X] and XR[X] is a prime ideal). Therefore K = H^{j}, so J^{j} = X^{mj}H^{j}, so X^{mj}(H^{j})_{a} = (J^{j})_{a} = X^{nd}(I^{n})_{a}, so (H^{j})_{a} = (I^{n})_{a}. Therefore by (2.9), if we let G = H_{a} ∩ R, then H_{a} = GR[X], (G^{j})_{a}R[X] = (H^{j})_{a}, and (G^{j})_{a} = (I^{n})_{a} (so G is projectively equivalent to I (in R)). Therefore (G^{d(I)})_{a} = (I^{i})_{a} for some i ∈ IN_{+}, by (2.5.4).

Since (G^{j})_{a} = (I^{n})_{a} and (G^{d(I)})_{a} = (I^{i})_{a}, it follows from [11, (2.1)(b)] that ji = nd(I). Therefore nd(I) = mj (resp., ji = nd(I)) and n, j are relatively prime, so it follows that m = en (resp., d(I) = jf) for some e (resp., f) ∈ IN_{+}. Therefore j is a divisor of d(I), and since nd = mj = enj it follows that d = je for some divisor j of d(I) and for some e ∈ IN_{+}, and it was noted in the preceding paragraph that j > 1.

The final statement is clear from what has already been shown.

It follows from (4.1.2) that every nonzero prime ideal of a regular local domain is projectively full. If (R, M) is a regular local domain and dim R = 2, then P(I) is projectively full for every nonzero proper ideal I of R [1, (4.12)]. It would be interesting to know whether this is also true when dim R ≥ 3.

Remark 4.2 (4.2.1) A projectively full ideal may fail to extend to a projectively full ideal in a finite free integral extension domain. (4.2.2) Concerning (4.1.1), there exist regular ideals bA and J in a Noetherian domain A and positive integers d > 1 such that: bA has a Rees valuation ring with integer e = 1; b^{d−1}J, and b^{d+1}J are projectively full; and, b^{d}J is not projectively full. Therefore: (a) the product of an ideal H that has a Rees valuation with integer e = 1 and an ideal which is projectively full and is contained in H need not be projectively full; and, (b) the product of an ideal H that has a Rees valuation with integer e = 1 and an ideal which is not projectively full and is contained in H may be projectively full. (4.2.3) Concerning (4.1.2), it is often the case for a height-one prime ideal P of an integrally closed Noetherian domain R that the rank-one discrete valuation domain V = R_{P} is not the only Rees valuation ring of P.

Proof. For (4.2.1), let I be a nonzero ideal in a Noetherian domain R that is not projectively full (so d = d(I) > 1), so (X^{d}I)R[X^{d}] is projectively full (by (4.1.4)). However, (4.1.5) shows that (X^{d}I)R[X] is not projectively full in the free (of degree d) integral extension domain R[X] of R[X^{d}].

For (4.2.2), let R and I be as in (4.1.5), let d > 1 be a divisor of d(I) such that d − 1 and d + 1 are relatively prime to d(I), let b = X, let A = R[X], and let J = IA. Then V = A_{bA} is a regular local domain, so bA has V as a Rees valuation ring with e = 1. Also, it follows from (4.1.5) that b^{d−1}J is projectively full in A, b^{d}J is not projectively full in A, and b^{d+1}J is projectively full in A. The last statement (concerning (a) and (b)) clearly follows from this.

For (4.2.3), let P be a height-one prime ideal of an integrally closed local domain (R, M) of altitude 2. Then V = R_{P} is the unique Rees valuation ring of P if and only if P is invertible if and only if the analytic spread a(P) of P is one. With this in mind, let x, y be independent indeterminates over a field k and let R be the subring k[[x^{2}, xy, y^{2}]] of the formal power series ring k[[x, y]]. Then R is a normal local domain and P = (x^{2}, xy)R is a height-one prime ideal of R such that a(P) = 2. Therefore P is projectively full and has more than one Rees valuation ring.

One problem we have not been able to solve is: given a nonzero ideal in a Noetherian domain R, does there always exist a finite integral extension domain A of R such that P(IA) is projectively full? In Proposition 4.3 we give a “logical” candidate for A and show that, at least, for each J ∈ P(I) it holds that (JA)_{a} is the power of some fixed ideal in P(IA).

In Proposition 4.3, for elements b_{1}, …, b_{g} of a Noetherian ring R, we let R[b_{1}^{1/k}, …, b_{g}^{1/k}] denote an integral extension ring of R generated by elements b_{1}^{1/k}, …, b_{g}^{1/k} that are k-th roots of b_{1}, …, b_{g}, respectively. This integral extension ring of R can be obtained in several ways. If R is an integral domain, the extension A = R[b_{1}^{1/k}, …, b_{g}^{1/k}] can be constructed to also be an integral domain. On the other hand, one can also construct R[b_{1}^{1/k}, …, b_{g}^{1/k}] so that it is a finite free R-module of rank gk. In any case, we note that if R is local with maximal ideal M, then A is local with maximal ideal (M, b_{1}^{1/k}, …, b_{g}^{1/k})A.

Proposition 4.3 Let I be a regular proper ideal in a Noetherian ring R and assume that P(I) is not projectively full. Let K be the largest ideal in P(I), let b_{1}, …, b_{g} be regular elements in A that generate K, and let d(K) = k (so k > 1). Let A = R[b_{1}^{1/k}, …, b_{g}^{1/k}] and let H = (b_{1}^{1/k}, …, b_{g}^{1/k})A. Then for each ideal J ∈ P(K) = P(I) it holds that (JA)_{a} = (H^{n})_{a} for some n ∈ IN_{+}.

Proof. Note first that H^{[k]} = ((b_{1}^{1/k})^{k}, (b_{2}^{1/k})^{k}, …, (b_{g}^{1/k})^{k})A = KA, and it is clear that H^{[k]} is a reduction of H^{k}, so (KA)_{a} = (H^{k})_{a}. Now let J ∈ P(K), so (J^{k})_{a} = (K^{n})_{a} for some n ∈ IN_{+}, by (2.5.4). Therefore ((JA)^{k})_{a} = ((J^{k})_{a}A)_{a} = ((K^{n})_{a}A)_{a} = (((KA)_{a})^{n})_{a} = (((H^{k})_{a})^{n})_{a} = (H^{kn})_{a}. Thus ((JA)^{k})_{a} = (H^{kn})_{a}, so (JA)_{a} = (H^{n})_{a}.

Concerning (4.3), the only relation between d(K) and d(H) we have been able to determine is d(K) ≤ kd(H). This follows from: d(K) ≤ d(KA) (by (2.7)(1)) = d(H^{k}) (since KA = H^{h}) = kd(H) (by [1, (4.8.3)]).

For a regular proper ideal J = (b_{1}, …, b_{g})R in a Noetherian ring R, we present in Example 4.4 a construction for obtaining a finite integral extension ring A of R[X] such that P(KA) is projectively full, where K = (b_{1}X^{k}, …, b_{g}X^{k})R[X] and k = d(J).

Example 4.4 Let J be a regular proper ideal in a Noetherian ring R, assume that J is not projectively full and is the largest ideal in P(J), let d(J) = k (so k > 1, by (3.1.2)), let b_{1}, …, b_{g} be regular elements in R that generate J, and let B = R[b_{1}^{1/k}, …, b_{g}^{1/k}]. Let X be an indeterminate, let K = (b_{1}X^{k}, b_{2}X^{k}, …, b_{g}X^{k})R[X], let A = R[X, b_{1}^{1/k}X, …, b_{g}^{1/k}X], and let H = (b_{1}^{1/k}X, …, b_{g}^{1/k}X)A. Then K is not projectively full in R[X], d(K) = k, A is obtained from R[X] by adjoining the k-th root b_{i}^{1/k}X = (b_{i}X^{k})^{1/k} of each generator b_{i}X^{k} of K to R[X], and H is projectively full in A and is projectively equivalent to KA.

Proof. By (4.1.5), K is not projectively full and d(K) = k. Let C = B[X], so C is a finite integral extension ring of A and HC = (b_{1}^{1/k}X, …, b_{g}^{1/k}X)C is projectively full (by (4.1.4), since (b_{1}^{1/k}, …, b_{g}^{1/k})B is an ideal in B). Therefore H is projectively full in A, by (3.2)(1), and H is projectively equivalent to KA, by (4.3).

5 NON-PROJECTIVELY FULL MAXIMAL IDEALS.

In Sections 3 and 4 a number of examples of regular ideals I of a Noetherian ring are constructed for which I or P(I) is projectively full. The main result in this section gives a family of integrally closed local domains (R, M) of altitude two for which the maximal ideal M is not projectively full.

Example 5.1 Let x, y, Z, W be independent indeterminates, let F be a field whose characteristic is not 2, and let (R_{0}, M_{0}) denote the regular local domain F [x, y]_{(x,y)} of altitude two. Let k < i ≤ j be positive integers that are units in F and set R = R_{0}[z, w] = R_{0}[Z, W]/(Z^{k} − x^{i} − y^{j}, W^{k} − x^{i} + y^{j})R_{0}[Z, W]. Then the following hold:

(i) R is an integrally closed Cohen-Macaulay local domain (with maximal ideal M = (x, y, z, w)R) of altitude two. (ii) If j = i and if i, k are relatively prime, then M has a unique Rees valuation ring V. (iii) If j = i and if i, k are relatively prime, then M = (x, y, z, w)R is not projectively full, d(M) = k, and ((z, w)R)_{a} = M_{i/k}.

Proof. We first show that R is a Cohen-Macaulay local ring of altitude two. For this, let K = (Z^{k} −x^{i} −y^{j}, W^{k} −x^{i} +y^{j})R_{0}[Z, W], let z = Z +K and w = W +K in R_{0}[Z, W]/K, and let R = R_{0}[z, w], so z^{k} = x^{i} + y^{j} and w^{k} = x^{i} − y^{j} are in R_{0}, but possibly R is not an integral domain (since we do not yet know that K is a prime ideal). Also, it is clear that R is an integral extension ring of R_{0} and that M = (x, y, z, w)R is a maximal ideal in R, and since z^{k} = x^{i} + y^{j} ∈ (x, y)R_{0} = M_{0} and w^{k} = x^{i} − y^{j} ∈ (x, y)R_{0} = M_{0}, by integral dependence it follows that every maximal ideal in R contains (x, y, z, w)R, hence R is local. Further, K is a height two (not necessarily prime) ideal in the locally regular UFD R_{0}[Z, W], so R is a free (of degree k^{2}) integral extension ring of the altitude two regular local domain R_{0} (with {z^{m}w^{n} | m = 0, …, k − 1 and n = 0, …, k − 1} as a free basis), so altitude(R) = 2 and R is Cohen-Macaulay, by [13, (25.16)], so R is a Cohen-Macaulay local ring of altitude two. In particular, R satisfies (S_{i}) for all i ∈ IN (that is, the maximum length of a prime sequence in R_{p} is ht(p) for all p ∈ Spec(R).

Since R is (S_{2}), to show that R is integrally closed it suffices (by [6, (23.8)]) to show that R satisfies (R_{1}) (that is, R_{p} is a regular local domain for all height one prime ideals p in R). We do this in the next four paragraphs.

Let p be a height one prime ideal in R, let P be the preimage of p in R_{0}[Z, W], and let T = R_{0}[Z, W]_{P}. Then T is a regular local domain of altitude three and f = Z^{k} − x^{i} − y^{j}, g = W^{k} − x^{i} + y^{j} are in Q = P T. Also, f_{Z} = kZ^{k−1}, f_{W} = 0, g_{Z} = 0, and g_{W} = kW^{k−1}, and the determinant (= k^{2}Z^{k−1}W^{k−1}) of the two by two matrix consisting of these four polynomials is in Q if and only if either: (a) Z ∈ Q; or, (b) W ∈ Q (since k is a unit in R_{0}). That is, if and only if either: (a′) z ∈ p; or, (b′) w ∈ p. Therefore by [6, (30.4)], R_{p} is a regular local domain except, possibly, in cases (a′) or (b′).

To handle cases (a′) and (b′), note first that if z ∈ p, then z^{k} = x^{i} + y^{j} ∈ p, so xy ∉ p (since, otherwise, the height two M-primary ideal (x, y)R is contained in the height one prime ideal p). Similarly, if w ∈ p, then w^{k} = x^{i} − y^{j} ∈ p, so xy ∉ p. Therefore if q is a height one prime ideal in R that contains either x or y, then R_{q} is a regular local domain (by the preceding paragraph).

Next (as just above) let T be the altitude three regular local domain R_{0}[Z, W]_{P} and let Q = P T be the maximal ideal of T (so (f = Z^{k} − x^{i} − y^{j}, g = W^{k} − x^{i} + y^{j})T ⊆ Q). Also, f_{x} = −ix^{i−1}, f_{y} = −jy^{j−1}, g_{x} = −ix^{i−1}, g_{y} = jy^{j−1}, and the two by two determinant (= −2ijx^{i−1}y^{j−1}) consisting of these four elements is in Q if and only if either: (c) x ∈ Q; or,

(d) y ∈ Q (since 2ij is a unit in R_{0}). Modulo K, it follows that either: (c′) x ∈ p; or, (d′) y ∈ p. Therefore by [6, (30.4)], R_{p} is a regular local domain except, possibly, in cases (c′) or (d′). However, the preceding paragraph shows that R_{p} is a regular local domain in both cases (c′) and (d′).

It follows that R_{p} is a regular local domain for all height one prime ideals p in R, so R is normal (by (R_{1}), (S_{2}) (see [6, Theorem 23.8])). Therefore R is an integrally closed reduced local ring, so R is a local domain (by [6, (9.11)]), hence K is a prime ideal and R is an integrally closed Cohen-Macaulay local domain. This completes the proof of (i).

For (ii), assume that j = i and that i, k are relatively prime, and notice that z^{k} ∈ (x^{i}, y^{i})R ⊆ (x, y)^{k}R and w^{k} ∈ (x^{i}, y^{i})R ⊆ (x, y)^{k}R, so (x, y)R is a reduction of M = (x, y, z, w)R. It follows that each Rees valuation ring of M is an extension of the order valuation ring V_{0} = R_{0}[y/x]_{xR_{0}[y/x]} = R_{0}[x/y]_{yR_{0}[x/y]} of R_{0}. (Notice that xV_{0} = yV_{0} is the maximal ideal N_{0} of V_{0} and that if we let t = y/x and \overline{t} = t + N_{0} in the field V_{0}/N_{0}, then V_{0}/N_{0} = (R_{0}/M_{0})(\overline{t}) and \overline{t} is transcendental over R_{0}/M_{0}.)

With this in mind, it follows from [20, Theorem 19, p. 55] that k^{2} = [R : R_{0}] ≥ ∑i=1^{g} e_{i}f_{i}, where V_{1}, …, V_{g} are the extensions (to the quotient field of R) of V_{0}, e_{i} is the ramification index of N_{0} in V_{i} (so N_{0}V_{i} = N_{i}^{e_{i}}, where N_{i} is the maximal ideal of V_{i}), and f_{i} is the relative degree [(V_{i}/N_{i}) : (V_{0}/N_{0})]. It will now be shown that g = 1 and that e_{1} = k = f_{1}

For this, if v is the valuation of any of the valuation rings V ∈ {V_{1}, …, V_{g}}, then v(x^{i}) = iv(x), v(y^{i}) = iv(y), and kv(z) = v(z^{k}) = v(x^{i} + y^{i}) = iv(x) (since v is an extension of the order valuation v_{0}). Therefore kv(z) = iv(x) and, similarly, kv(w) = iv(x). Since k, i are relatively prime (by hypothesis), it follows that v(z) = v(w) ≥ i and v(x) ≥ k. Also, v(x) = v(N_{0}) (since xV_{0} = N_{0}, as noted above), so the ramification index e of N_{0} in V is at least k.

Also, as noted in the preceding paragraph, v(z) = v(w), so \frac{w}{z} is a unit in V, and (\frac{w}{z})^{k} = \frac{w^{k}}{z^{k}} = \frac{x^{i}−y^{i}}{x^{i}+y^{i}}. As above, let t = \frac{y}{x}, so t is a unit in V_{0} whose residue class \overline{t} in V_{0}/N_{0} is transcendental over R_{0}/M_{0} and \frac{x^{i}−y^{i}}{x^{i}+y^{i}} = \frac{1−t^{i}}{1+t^{i}} ∈ V_{0}, It follows that the residue class of \frac{w}{z} in V/N is algebraic of degree k over V_{0}/N_{0}, so [V/N : V_{0}/N_{0}] ≥ k.

Therefore, by the preceding two paragraphs, for each (V_{i}, N_{i}) ∈ {(V_{1}, N_{1}), …, (V_{g}, N_{g})} it holds that N_{0}V_{i} = N_{i}^{e_{i}} ⊆ N_{i}^{k} (so e_{i} ≥ k) and f_{i} = [V_{i}/N_{i} : V_{0}/N_{0}] ≥ k. But k^{2} = [R : R_{0}] ≥ _{∑i=1}^{g} e_{i}f_{i}, so it follows that g = 1 and that e_{1} = k = f_{1}. In what follows, we denote (V_{1}, N_{1}) by (V, N), e_{1} by e, and f_{1} by f. After normalizing v, we have v(x) = v(y) = k (so v(M) = k) and v(z) = v(w) = i.)

For (iii), assume that j = i and that i, k are relatively prime. To see that M is not projectively full, note that M^{k} ⊇ (z^{k}, w^{k})R = (x^{i} + y^{i}, x^{i} − y^{i})R (since j = i), and (x^{i} + y^{i}, x^{i} − y^{i})R = (2x^{i}, 2y^{i})R = (x^{i}, y^{i})R (since 2 is a unit in R), and (x^{i}, y^{i})R is a reduction of M^{i}. It follows that ((z^{k}, w^{k})R)_{a} = (M^{i})_{a}, so ((z, w)^{k}R)_{a} = (M^{i})_{a}. Therefore (z, w)R and M are projectively equivalent and ((z, w)R)_{a} = M_{i/k}, by [11, (2.3)], so ((z, w)R)_{a} is not the integral closure of any power of M (since k and i are relatively prime), hence M is not projectively full.

By the preceding paragraph i/k ∈ U(M) (see (2.4.3)). Therefore, since i, k are relatively prime and since d(M)U(M) ⊆ IN_{+} (by (2.5.4)), it follows that d(M) is a multiple of k. On the other hand, d(M) is a divisor of the integer e associated to the Rees valuation ring (V, N) of M, by (2.5.4). By (2.3), this integer e is given by MV = N^{e}, so v(M) = e. However, v(M) = k, by the second preceding paragraph, so it follows that d(M) is a divisor of k. Therefore d(M) = k.

In the next remark we note (with brief indications of proofs) several properties of two rings related to the rings R[z, w] of (5.1).

Remark 5.2 With notation as in (5.1), let A = R[M/x] = R[y/x, z/x, w/x] and let R = R[u, tM] = R[u, tx, ty, tz, tw] (where t is an indeterminate and u = 1/t). Also, with the assumptions as in (5.1.2), let (V, N) be the unique Rees valuation ring of M. Then the following hold: (5.2.1) xA′ is N ∩ A′-primary, A is Cohen-Macaulay, xA is p′-primary, where p′ = N ∩ A, p′ = (M, z/x, w/x)A, and A is not integrally closed. Moreover, if i − k is a unit in R_{0}, then A_{p′} is not integrally closed, but A_{p} is a regular local domain for all height one prime ideals p ≠ p′. (5.2.2) uR′ is primary, R is Cohen-Macaulay, uR is p^{∗}-primary, where p^{∗} = (u, M, tz, tw)R, and R is not integrally closed. Moreover, if i − k is a unit in R_{0}, then R_{p^{∗}} is not integrally closed, but R_{p} is a regular local domain for all height one prime ideals p ≠ p^{∗}.

Proof. (We only sketch the proofs.)

xA′ is N ∩ A′-primary (by [11, (2.9)]) and uR′ is primary (since M has a unique Rees valuation ring).

A (resp., R) is a free (of degree k^{2}) integral extension domain of the locally regular UFD A_{0} = R_{0}[M_{0}/x] = R_{0}[y/x] (resp., R_{0} = R_{0}[u, tM_{0}] = R_{0}[u, tx, ty]) (since (z/x)^{k} = x^{i−k} + y^{i−k}(y/x)^{k} and (w/x)^{k} = x^{i−k} − y^{i−k}(y/x)^{k} imply (f_{1}, g_{1})A_{0}[Z, W] = (Z^{k} − x^{i−k} − y^{i−k}(y/x)^{k}, W^{k} − x^{i−k} + y^{i−k}(y/x)^{k})A_{0}[Z, W] is a height two prime ideal) (resp., (tz)^{k} = u^{i−k}(tx)^{i} + u^{i−k}(ty)^{i} and (tw)^{k} = u^{i−k}(tx)^{i} − u^{i−k}(ty)^{i}) imply (f_{2}, g_{2})R_{0}[Z, W] = (Z^{k} − u^{i−k}(tx)^{i} − u^{i−k}(ty)^{i}, W^{k} − u^{i−k}(tx)^{i} + y^{i−k}(ty)^{i})R_{0}[Z, W] is a height two prime ideal)), so it follows from [13, (25.16)] that A (resp., R) is Cohen-Macaulay.

Since A (resp., R) is Cohen-Macaulay and xA′ (resp., uR′) has a unique (height one) prime divisor, it follows from the structure of A (resp., R) that xA (resp., uR) has a unique prime divisor, say p′ (resp., p^{∗}). And it then follows that p′ = N ∩ A = (M, z/x, w/x)A (resp., p^{∗} = (u, M, tz, tw)R).

Using the determinant of f_{1Z} = kZ^{k−1}, f_{1W} = 0, g_{1Z} = 0, and g_{1W} = kW^{k−1} and then of f_{1x} = −(i − k)x^{i−k−1}, f_{1τ} = −ky^{i−k}τ^{k−1}, g_{1x} = −(i − k)x^{i−k−1}, and g_{1τ} = ky^{i−k}τ^{k−1} (with τ = y/x), it follows from [6, (30.4)] that A_{p} is a regular local domain for all p ≠ p′. However, A_{p′} is not a regular local domain, since, otherwise, it would follow that V/N = V_{0}/N_{0} (where (V_{0}, N_{0}) is the order valuation ring of R_{0}), and this contradicts the fact (shown in the proof of (5.1)(ii)) that w/z is algebraic of degree k over V_{0}/N_{0}. It therefore follows that A is not integrally closed.

It is clear that analogous statements hold for B = R[M/y] = R[x/y, z/y, w/y] and y in place of A and x. With this in mind, since R[1/(tx)] = A[tx, 1/(tx)] (resp., R[1/(ty)] = B[ty, 1/(ty)]), and since uA[tx, 1/(tx)] = xA[tx, 1/(tx)] (resp., uA[ty, 1/(ty)] = yA[ty, 1/(ty)]), it follows that R is not integrally closed, that R_{p^{∗}} is not integrally closed, and that R_{p} is a regular local domain for all height one prime ideals p ≠ p^{∗}.

In Example 5.1 M has only one Rees valuation ring. In the final remark in this paper we consider the ideals in P(I) in the case where I has only one Rees valuation ring.

Remark 5.3 Let I be a proper ideal in a Noetherian domain R and assume that I has only one Rees valuation ring, say (V, N). Then P(I) ⊆ {N^{i} ∩ R | i ∈ IN}, so if P = N ∩ R is the center of V on R, then each ideal in P(I) is a P -primary valuation ideal (that is, it is contracted from a valuation overring of R). Assume that I is maximal in P(I). If IV = N^{e}, then I = N^{e} ∩ R and I is projectively full if and only if each J ∈ P(I) has the property that JV = N^{ne} for some n ∈ IN. In general, the inclusion P(I) ⊆ {N^{i} ∩ R | i ∈ IN} need not be an equality. For example, if (R, M) is a regular local domain of altitude two and M = (x, y)R, then I = (x, y^{2})R has a unique Rees valuation ring. To see this one can apply [1, (2.9)] or [11, (3.1)]. Notice that MR[x/y^{2}] = yR[x/y^{2}] is a height-one prime ideal and V = R[x/y^{2}]_{yR[x/y^{2}]} is a valuation domain. Also, MR[y^{2}/x] is a height-one prime ideal and R[y^{2}/x]_{MR[y^{2}/x]} = V. Thus V is the unique Rees valuation ring of I = (x, y^{2})R. The ideal I is projectively full (by (3.7)) and I ⊊ M = N ∩R. Also, IV = N^{2}, so P(I) = {N^{2n} ∩R}_{n=1}^{∞}.

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Department of Mathematics, University of Missouri, Columbia, Missouri 65211-0001 E-mail address: catalin@math.missouri.edu

Department of Mathematics, Purdue University, West Lafayette, Indiana 47909-1395 E-mail address: heinzer@math.purdue.edu

Department of Mathematics, University of California, Riverside, California 92521-0135 E-mail address: ratliff@math.ucr.edu

Department of Mathematics, University of California, Riverside, California 92521-0135 E-mail address: rush@math.ucr.edu