PROJECTIVELY FULL IDEALS IN

NOETHERIAN RINGS

William Heinzer Purdue University

joint with Catalin Ciuperca, Jack Ratliff and Dave Rush

SETUP AND DEFINITIONS

Let I be a regular proper ideal of a Noetherian ring R. DEFINITION. An ideal J of R is projectively equivalent to I if there exist positive integers m and n such that I^{m} and J^{n} have the same integral closure, i.e., (I^{m})_{a} = (J^{n})_{a}. NOTATION. Let P(I) denote the set of integrally closed ideals projectively equivalent to I. FACT. The set P(I) is linearly ordered and discrete with respect to inclusion.

MORE SETUP AND DEFINITIONS

Since P(I) is discrete and since J ∈ P(I) implies (J^{n})_{a} ∈ P(I), there is naturally associated to I and P(I) a numerical semigroup, i.e., a subsemigroup S(I) of the additive semigroup of nonnegative integers ℕ_{0} that contains all sufficiently large integers. DEFINITION. The set P(I) is said to be projectively full if S(I) = ℕ_{0}, or equivalently, if every element of P(I) is the integral closure of a power of the largest element K of P(I), i.e., every element of P(I) has the form (K^{n})_{a}, for some positive integer n. If this holds, the ideal K is said to be projectively full.

MAIN GOALS

1. Describe ideals K that are projectively full.

2. Describe ideals I such that P(I) is projectively full.

3. Can things be improved by passing to an integral extension?

THEOREM. If R contains the field of rational numbers, then there exists a finite free integral extension ring A of R such that P(IA) is projectively full; and if R is an integral domain, then there also exists a finite integral extension domain B of R such that P(IB) is projectively full.

EXAMPLES

EXAMPLE 1. Let (R, M) be a Noetherian local ring having the property

a ∈ M^{i} \ M^{i+1} and b ∈ M^{j} \ M^{j+1} ⟹ ab ∉ M^{i+j+1}, then M is projectively full. Thus if the associated graded ring

G(R, M) = R/M ⊕ M/M^{2} ⊕ ⋯ ⊕ M^{n}/M^{n+1} ⊕ ⋯ is an integral domain, then M is projectively full.

EXAMPLE 2. Let k be a field and let R = k[[x^{2}, x^{3}]]. Then the maximal ideal M = (x^{2}, x^{3})R is not projectively full. The numerical semigroup S(M) is generated by 2 and 3.

DISCUSSION

With R = k[[x^{2}, x^{3}]] as in Example 2, R is not integrally closed. Things improve by passing to the integral extension R[x] = k[[x]]. For each regular proper ideal I of R, P(IR[x]) is projectively full.

Want to give an example of a normal local domain (R, M) such that M is not projectively full?

EXAMPLE 3. Let F be a field and let x, y, z, w be variables. Let

R_{0} = F [x, y]_{(x,y)} and R = \frac{R_{0}[z, w]}{(z^{2} − x^{3} − y^{3}, w^{2} − x^{3} + y^{3})}.

If 2 and 3 are units of F, then R is a 2-dim normal local domain, and the maximal ideal M = (x, y, z, w)R is not projectively full.

SOME HISTORY

The concept of projective equivalence of ideals and the study of ideals projectively equivalent to I was introduced by Samuel in

Some asymptotic properties of powers of ideals, Annals of Math 56 (1952), 11-21.

and further developed by Nagata in

Note on a paper of Samuel concerning asymptotic properties of ideals, Mem. Coll. Sci. Univ. Kyoto, Ser. A Math. 30 (1957), 165-175.

MORE HISTORY

Rather than ‘projectively equivalent’, Hartmut Göhner uses the term ‘asymptotically equivalent’ in Semifactoriality and Muhly’s condition (N) in two-dimensional local rings, J. Algebra 34 (1975), 403-429.

Göhner mentions that the expression ‘projective asymptotic equivalence’ is used by David Rees in Valuations associated with ideals (II), J. London Math. Soc. 36 (1956), 221-228. and by H. T. Muhly in On the existence of asymptotically irreducible ideals, J. London Math. Soc. 40 (1965), 99-107.

RESULTS OF REES

Let I be a regular proper ideal of a Noetherian ring R. For each x ∈ R, let v_{I}(x) = max{k ∈ ℕ | x ∈ I^{k}}, and define

\overline{v}_{I}(x) = lim_{k→∞}(\frac{v_{I}(x^{k})}{k}).

Rees established that:

(a) \overline{v}_{I}(x) is well defined;

(b) for each k ∈ ℕ and x ∈ R, \overline{v}_{I}(x) ≥ k ⟺ x ∈ (I^{k})_{a}

(c) there exist discrete valuations v_{1}, …, v_{g} defined on R, 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\}.

Describing the Rees Valuation Rings

For simplicity, assume R is a Noetherian integral domain with field of fractions F. Let t be an indeterminate, The Rees ring of R with respect to I is the graded subring

R = R[t^{−1}, It] = ⨁_{n∈ℤ} I^{n}t^{n}

of the Laurent polynomial ring R[t^{−1}, t]. The integral closure B of R[t^{−1}, It] is a Krull domain, and B_{P} is a DVR for each minimal prime P of t^{−1}B. The set Rees I of Rees valuation rings of I is precisely the set of rings V = B_{P} ∩ F, where P is a minimal prime of t^{−1}B.

THE REES INTEGERS OF I

Let (V_{1}, N_{1}), …, (V_{g}, N_{g}) be the Rees valuation rings of I. The integers (e_{1}, …, e_{g}), where IV_{i} = N_{i}^{e_{i}}, are the Rees integers of I. PROPOSITION. A sufficient condition for I to be projectively full is that gcd(e_{1}, …, e_{g}) = 1. This is not a necessary condition.

EXAMPLE 4. Let (R, M) be a 2-dimensional regular local ring with M = (x, y)R. The ideal I = (x, y^{2})R is integrally closed with unique Rees valuation ring V = R[x/y^{2}]_{MR[x/y^{2}]}. The integrally closed ideals projectively equivalent to I are precisely the powers I^{n} of I. Thus I is projectively full. The maximal ideal of V is N = yV and IV = N^{2}, so the gcd is two not one.

Projective Equivalence and Rees Valuation Rings

Recall that ideals I and J are projectively equivalent if (I^{m})_{a} = (J^{n})_{a} for some m, n ∈ ℕ. If I and J are projectively equivalent, then Rees I = Rees J. The converse holds if I or J has only one Rees valuation ring, but fails in general. Steve McAdam, Jack Ratliff and Judy Sally show in Integrally closed projectively equivalent ideals, in Commutative Algebra, MSRI Pub. 15, 1988, 391-405 that if I and J are projectively equivalent, then the Rees integers of I and J are proportional. The converse also holds: if Rees I = Rees J and the Rees integers of I and J are proportional, then I and J are projectively equivalent.

Projectively full ideals of a 2-dim RLR

EXAMPLE 4. (continued) (R, M) is a 2-dim regular local ring with M = (x, y)R. Zariski’s theory of unique factorization of complete (= integrally closed) ideals of R as finite products of simple complete ideals implies P(I) is projectively full for each nonzero proper ideal I of R. The ideal I has a unique Rees valuation ring if and only if I is a power of a simple complete ideal. If I factors as I = I_{1}^{f_{1}} ⋯ I_{g}^{f_{g}}, where I_{1}, …, I_{g} are distinct simple complete ideals, then I is projectively full if and only if gcd(f_{1}, …, f_{g}) = 1.

Projective fullness and Rees integers

EXAMPLE 4. (continued) (R, M) is a 2-dim regular local ring with M = (x, y)R, and I = I_{1}^{f_{1}} ⋯ I_{g}^{f_{g}} is the factorization of the M-primary ideal I as a product of distinct simple complete ideals. How do the integers f_{1}, …, f_{g} relate to the Rees integers of I? The simple complete M-primary ideals of R are in one-to-one correspondence with the prime divisors birationally dominating R. Thus the Rees valuation rings of I are (V_{1}, N_{1}), …, (V_{g}, N_{g}), where (V_{j}, N_{j}) corresponds to I_{j}. If I_{j}V_{j} = N_{j}^{c_{j}}, then the Rees integers of I are e_{1} = c_{1}f_{1}, …, e_{g} = c_{g}f_{g}.

EXAMPLE 4. (continued) (R, M) is a 2-dim regular local ring with M = (x, y)R. Let I = (x^{2}, xy^{2}, y^{3})R. Notice that J = (x^{2}, y^{3})R is a reduction of I and JI = I^{2}, so the reduction number r_{J}(I) = 1. Let

V = R[\frac{xy^{2}}{x^{2}} , \frac{y^{3}}{x^{2}} ]_{MR[ \frac{xy^{2}}{x^{2}} , \frac{y^{3}}{x^{2}} ]} = R[\frac{y^{2}}{x} , \frac{y^{3}}{x^{2}} ]_{MR[ \frac{y^{2}}{x} , \frac{y^{3}}{x^{2}} ]}.

One sees that V is a valuation ring with maximal ideal N = (y^{2}/x)V, and I is a simple complete ideal. The ideals of R that are contracted from V descend as follows: M = N^{2} ∩ R ⊋ (x, y^{2})R = N^{3} ∩ R ⊋ M^{2} = N^{4} ∩ R ⊋ (x, y^{2})M = N^{5} ∩ R ⊋ I = N^{6} ∩ R. The ideals in P(I) are precisely the ideals I^{m} = N^{6m} ∩ R, for m ∈ ℕ.

THE ASSOCIATED GRADED RING G(R, I)

PROPOSITION. If G(R, I) = R/I ⊕ I/I^{2} ⊕ ⋯ ⊕ I^{n}/I^{n+1} ⊕ ⋯ has a minimal prime p such that p is its own p-primary component of (0), then I has a Rees integer equal to one. Therefore I is projectively full.

More can be said using the Rees ring R = R[t^{−1}, It], and the identification G(R, I) = R/t^{−1}R. Let R′ denote the integral closure of R.

PROPOSITION. The ideal I has a Rees integer equal to one if and only if t^{−1}R′ has a minimal prime p such that t^{−1}R′_{p} = pR′_{p}.

EXAMPLE 5. An example of a 2-dim normal local domain (R, M) such that M is projectively full and the associated graded ring G(R, M) is not reduced. Let F be an algebraically closed field with char F = 0, and let R_{0} be a 2-dim regular local domain with maximal ideal M_{0} = (x, y)R_{0} and coefficient field F, e.g., R_{0} = F [x, y]_{(x,y)}, or R_{0} = F [[x, y]]. Then V_{0} = R_{0}[y/x]_{xR_{0}[y/x]} is the unique Rees valuation ring of M_{0}. Let

R = R_{0}[z], where z^{2} = x^{3} + y^{j} with j ≥ 3. It is readily checked that R is 2-dim normal local with maximal ideal M = (x, y, z)R. Notice that I = (x, y)R is a reduction of M since z is integral over I.

EXAMPLE 5 (continued). Since I = (x, y)R is a reduction of M, every Rees valuation ring of M is an extension of V_{0}. Let V be a Rees valuation ring of M and let v denote the normalized valuation with value group ℤ corresponding to V. Then v(x) = v(y) and the image of y/x in the residue field of V is transcendental over F. Since z^{2} = x^{3} + y^{j} and j ≥ 3, we have

2v(z) = v(z^{2}) = v(x^{3} + y^{j}) = 3v(x).

We conclude that v(x) = 2 and v(z) = 3. Therefore V is ramified over V_{0}. This implies that V is the unique extension of V_{0} and thus the unique Rees valuation ring of M.

EXAMPLE 5 (continued). For each positive integer n, let I_{n} = {r ∈ R | v(r) ≥ n}. Thus I_{2} = M. Since V is the unique Rees valuation ring of M, we have I_{2n} = (M^{n})_{a} for each n ∈ ℕ. To show M is projectively full, we prove that V is not the unique Rees valuation ring of I_{2n+1} for each n ∈ ℕ. Consider the inclusions

M^{2} ⊆ I_{4} ⊂ (z, x^{2}, xy, y^{2})R := J ⊆ I_{3} ⊂ M.

Since λ(M/M^{2}) = 3 and since the images of x and y in M/M^{2} are F -linearly independent, J = I_{3} and M^{2} = I_{4} = (M^{2})_{a}.

EXAMPLE 5 (continued). Since x^{3} = z^{2} − y^{j} and j ≥ 3, L = (z, y^{2})R is a reduction of I_{3} = (z, x^{2}, xy, y^{2})R. Indeed, (x^{2})^{3} ∈ L^{3} and (xy)^{3} ∈ L^{3} implies x^{2} and xy are integral over L. It follows that V is not a Rees valuation of I_{3}, for zV ≠ y^{2}V. Consider M^{3} ⊂ I_{3}M ⊆ I_{5} ⊂ I_{4} = M^{2}. Since the images of x^{2}, xy, y^{2}, xz, yz in M^{2}/M^{3} are an F -basis, it follows that I_{3}M = I_{5} and M^{3} = (M^{3})_{a} = I_{6}. Proceeding by induction, we assume M^{n+1} = (M^{n+1})_{a} = I_{2n+2}, and consider

M^{n+2} ⊂ I_{3}M^{n} ⊆ I_{2n+3} ⊂ M^{n+1} = I_{2n+2}.

EXAMPLE 5 (continued). Since the images in M^{n+1}/M^{n+2} of {x^{a}y^{b} | a + b = n + 1} ∪ {zx^{a}y^{b} | a + b = n} is an F -basis, λ(M^{n+1}/M^{n+2}) = 2n + 3, and the inequalities λ(M^{n+1}/I_{2n+3}) ≥ n + 2 and λ(I_{3}M^{n}/M^{n+2}) ≥ n + 1 imply I_{3}M^{n} = I_{2n+3} and M^{2n+2} = (M^{2n+2})_{a}. Therefore the ideal I_{2n+3} has a Rees valuation ring different from V, and thus is not projectively equivalent to M. We conclude that M is projectively full. We have also shown that M is a normal ideal.

Questions

Let (R, M) be a complete normal Noetherian local domain of altitude two.

1. What are necessary and sufficient conditions in order that M is projectively full?

2. What are necessary and sufficient conditions in order that P(I) is projectively full for each nonzero proper ideal I of R?

EXAMPLE 6. An example of a (complete) 2-dim normal local domain (R, M) such that M is not projectively full. Let F be an algebraically closed field with char F = 0, and let R_{0} be a 2-dim regular local domain with maximal ideal M_{0} = (x, y)R_{0} and coefficient field F, e.g., R_{0} = F [x, y]_{(x,y)}, or R_{0} = F [[x, y]]. Let k < i be relatively prime positive integers ≥ 2, and let

R = R_{0}[z, w], where z^{k} = x^{i} + y^{i} and w^{k} = x^{i} − y^{i}. It is readily checked that R is 2-dim normal local with maximal ideal M = (x, y, z, w)R. Also R is a free R_{0}-module of rank k^{2}.

EXAMPLE 6 (continued). With R = R_{0}[z, w] as above, we want to show that M = (x, y, z, w)R has a unique Rees valuation ring and that M is not projectively full. L = (x, y)R is a reduction of M, for z^{k} ∈ (x^{i}, y^{i})R ⊆ L^{k} and w^{k} ∈ (x^{i}, y^{i})R ⊆ L^{k} imply z and w are integral over L. Thus each Rees valuation ring V of M is an extension of the order valuation ring V_{0} = R_{0}[y/x]_{xR_{0}[y/x]} of R_{0}. To show there exists a unique Rees valuation ring of M, we observe that V as an extension of V_{0} ramifies of degree k and undergoes a residue field extension of degree ≥ k.

EXAMPLE 6 (continued). To show V ramifies of degree k over V_{0}, observe that kv(z) = v(z^{k}) = v(x^{i} + y^{i}) = iv(x) = iv(y) implies v(z) = i and v(x) = v(y) = k. Similarly, v(w) = i. Let N denote the maximal ideal of V. We have (z, w)V = N^{i} and MV = (x, y)V = N^{k}. The residue field of V_{0} is F (\overline{τ}), where \overline{τ} is the image of τ = y/x and is transcendental over F. Now w/z is a unit of V and (\frac{w}{z})^{k} = \frac{x^{i}−y^{i}}{x^{i}+y^{i}} = \frac{1−τ^{i}}{1+τ^{i}}. It follows that the residue class of w/z in V/N is algebraic of degree k over F (\overline{τ}). This proves that V is the unique Rees valuation ring of M.

EXAMPLE 6 (continued). To show M is not projectively full, notice that (z^{k}, w^{k})R = (x^{i} + y^{i}, x^{i} − y^{i})R, and since char F ≠ 2, (x^{i} + y^{i}, x^{i} − y^{i})R = (x^{i}, y^{i})R. Since (x^{i}, y^{i})R is a reduction of M^{i}, we have ((z^{k}, w^{k})R)_{a} = (M^{i})_{a}. Also ((z^{k}, w^{k})R)_{a} = ((z, w)^{k}R)_{a}. Therefore (z, w)R and M are projectively equivalent, so V is the unique Rees valuation ring of (z, w)R, and ((z, w)R)_{a} = N^{i} ∩ R. We have M = N^{k} ∩ R, M^{n}V = N^{nk} and (M^{n})_{a} = N^{nk} ∩ R, for each positive integer n. Since i is not a multiple of k, M is not projectively full.

RATIONAL SINGULARITIES

Joe Lipman in his paper Rational singularities, with applications to algebraic surfaces and unique factorization, Publ. Math. Inst. Hautes Études Sci. N^{o} 36 (1969), 195-279. extended Zariski’s theory of complete ideals in 2-dim regular local rings to 2-dim normal local rings R having a rational singularity. Lipman proved that R has unique factorization of complete ideals if and only if the completion of R is a UFD. For R having this property, it follows that P(I) is projectively full for each nonzero proper ideal I, e.g., R = F [[x, y, z]], where z^{2} + y^{3} + x^{5} = 0.

MORE ON RATIONAL SINGULARITIES

Let (R, M) be a normal local domain of altitude two. Göhner proves that if R has a rational singularity, then the set of complete asymptotically irreducible ideals associated to a prime R-divisor v consists of the powers of an ideal A_{v} which is uniquely determined by v. In our terminology, this says that if I is a nonzero proper ideal of R having only one Rees valuation ring, then P(I) is projectively full. Göhner’s proof involves choosing a desingularization f : X → Spec R such that v is centered on a component E_{1} of the closed fiber on X.

THE CLOSED FIBER 1

Let E_{2}, …, E_{n} be the other components of the closed fiber on X. Let E_{X} denote the group of divisors having the form ∑_{i=1}^{n} n_{i}E_{i}, with n_{i} ∈ ℤ, and consider

E_{X}^{+} = {D ∈ E_{X} | D ≠ 0 and (D · E_{i}) ≤ 0 for all 1 ≤ i ≤ n}, and E_{X}^{#} = {D ∈ E_{X} | D ≠ 0 and O(−D) is gen. by its sections over X}. Lipman shows E_{X}^{#} ⊆ E_{X}^{+} and equality holds if R has a rational singularity. Also, if D = ∑_{i} n_{i}E_{i} ∈ E_{X}^{+}, then negative-definiteness of the intersection matrix (E_{i} · E_{j}) implies n_{i} ≥ 0 for all i.

THE CLOSED FIBER 2

For if D ∈ E_{X}^{+} and D = A − B, where A and B are effective, then (A − B · B) ≤ 0 and (A · B) ≥ 0 imply (B · B) ≥ 0, so B = 0. Let v = v_{1}, v_{2}, …, v_{n} denote the discrete valuations corresponding to E_{1}, …, E_{n}. Associated with D = ∑_{i} n_{i}E_{i} ∈ E_{X}^{#} one defines the complete M-primary ideal I_{D} = {r ∈ R | v_{i}(r) ≥ n_{i} for 1 ≤ i ≤ n}. This sets up a one-to-one correspondence between elements of E_{X}^{#} and complete M-primary ideals that generate invertible O_{X}-ideals.

THE CLOSED FIBER 3

Lipman suggested to me the following proof that P(I) is proj. full for each complete M-primary ideal I if R has a rational singularity. Fix a desingularization f : X → Spec R such that I generates an invertible O_{X}-ideal and let D = ∑_{i} n_{i}E_{i} ∈ E_{X}^{#} be the divisor associated to I. Let g = gcd{n_{i}}. Since E_{X}^{+} = E_{X}^{#}, (1/g)D ∈ E_{X}^{#}. The ideals J ∈ P(I) correspond to divisors in E_{X}^{#} that are integral multiples of (1/g)D. Thus if K is the complete M-primary ideal associated to (1/g)D, then each J ∈ P(I) is the integral closure of a power of K, so P(I) is projectively full.

INTEGRAL EXTENSIONS

QUESTION. Let I be a nonzero ideal of a Noetherian domain R. Does there always exist a finite integral extension domain B of R such that P(IB) is projectively full? Let I = (b_{1}, …, b_{g})R be a nonzero regular ideal of the Noetherian ring R, let R_{g} = R[X_{1}, …, X_{g}] and K = (X_{1}^{c} − b_{1}, …, X_{g}^{c} − b_{g})R_{g}, where c is a positive integer. Then A = R_{g}/K is a finite free integral extension of rank c^{g} of R. Let x_{i} = X_{i} mod K. Then J = (x_{1}, …, x_{g})A is such that (IA)_{a} = (J^{c})_{a}, so IA and J are projectively equivalent.

MAIN THEOREM

THEOREM. Let I = (b_{1}, …, b_{g})R be as above and let (V, N) be a Rees valuation ring of I. Assume that b_{i}V = IV = N^{c} for each i with 1 ≤ i ≤ g, and that c is a unit of R. Then the finite free integral extension A = R_{g}/K is such that J = (x_{1}, …, x_{g})A is projectively full. Hence P(IA) is projectively full. If R is an integral domain and p is a minimal prime of A, then B = A/p is an integral extension domain of R such that P(IB) is projectively full.