NORMAL IDEALS OF GRADED RINGS

Sara Faridi

Mathematics Department, University of Michigan, Ann Arbor, MI 48109 email: faridi@math.lsa.umich.edu

Abstract. For a graded domain R = k[X_{0}, ..., X_{m}]/J over an arbitrary domain k, it is shown that the ideal generated by elements of degree ≥ mA, where A is the least common multiple of the weights of the X_{i}, is a normal ideal.

1. Introduction

In the theory of resolution of singularities, one wishes to be able to blow up a singular variety along a closed subscheme and obtain a smooth variety birational to the original one. One question that comes up is the existence of such resolutions; it is known for varieties over fields of characteristic zero [H], and is conjectured in the prime characteristic case. Another question is describing the closed subschemes that give smooth blowups. Translating this problem into the language of algebra, one is interested in ideals I of a ring R such that Proj R[It] is a smooth variety, where R[It] = ⊕_{n∈N}I^{n}t^{n} is the Rees ring of R along I. This leads one to focus on the Rees ring and its properties. In particular one would like to know when the Rees ring is normal. If R is a normal domain, R[It] is normal if and only if I is a normal ideal, where by normal ideal we mean an ideal all of whose positive powers are integrally closed. Normal ideals have been studied in different cases, and in some special cases necessary and sufficient conditions have been given for an ideal to be normal; see for example [G], [O], and [HS].

Typeset by AMS-TEX

Key words and phrases. Rees Ring, Normal Ideal.

I would like to thank my advisor Professor Karen Smith for all her help and many fruitful discussions during the preparation of this paper.

SARA FARIDI

In general given a ring, finding an ideal whose blowup is regular has proven to be a difficult problem. The same is true even for finding normal ideals. In this paper we will construct a normal ideal I for a general graded domain R = k[X_{0}, ..., X_{m}]/J which depends only on the weights of the variables X_{0}, ..., X_{m}, and is thus very simple to construct. We will begin by recalling some necessary definitions. For more on the construction of Rees rings see [E].

Let R be any N-graded domain, which is a quotient of a polynomial ring k[X_{0}, ..., X_{m}] modulo a homogeneous ideal J, where k = R_{0} is an arbitrary domain and X_{0}, ..., X_{m} are variables of positive weights A_{0}, ..., A_{m}. In practice, k is usually a field. Let m = (X_{0}, ...., X_{m}) be the irrelevant ideal of R. By abuse of notation, by X_{0}, ..., X_{m} we will mean the images of the variables X_{0}, ..., X_{m} in R. Throughout this paper R_{≥α} refers to the ideal of R generated by the elements of degree at least α in the graded ring R.

We begin our search for normal ideals by reviewing some basic facts about them.

Definition 1. For Noetherian domains S ⊆ T the integral closure of S in T, denoted by \overline{S}, is defined as all x ∈ T that satisfy an equation of the form x^{n} + a_{1}x^{n−1} + ... + a_{n−1}x + a_{n} = 0 where a_{i} ∈ S, for i = 1, ..., n.

Definition 2. For a Noetherian domain S and an ideal J of S, the integral closure of J in S, denoted by \overline{J}, is defined as

all x ∈ S that satisfy an equation of the form x^{n} + a_{1}x^{n−1} + ... + a_{n−1}x + a_{n} = 0 where a_{i} ∈ J^{i}, i = 1, ..., n; or equivalently,

all x ∈ S for which there exists a nonzero c ∈ S such that cx^{n} ∈ J^{n} for all positive integers n (see [Ho]).

Note. These definitions also hold in the nondomain case; we just have to choose c not in any minimal prime of S in the second statement.

Here are a few well-known facts that we shall use:

Theorem 1. If S is a normal domain, then S[It] is normal iff I^{n} is integrally closed for every positive integer n.

Theorem 2. In an N-graded domain (S, n), for any positive integer α, the n-primary ideal I = S_{≥α} is integrally closed.

Using Theorems 1 and 2, we will be looking for an (m-primary) ideal I of the form R_{≥α}, with the property that I^{n} = R_{≥nα} for all integers n ≥ 1. Note that from Lemma (2.1.6) in [EGA] one can deduce that such an α always exists. Theorem 3 gives an explicit value for α.

2. Main Theorem

Theorem 3. Let R be a graded domain, which is a quotient of a polynomial ring k[X_{0}, ..., X_{m}] modulo a homogeneous ideal J, where k is an arbitrary domain and X_{0}, ..., X_{m} are variables of positive weights A_{0}, ..., A_{m}. Let A be the least common multiple of A_{0}, ..., A_{m}. Then the ideal I = R_{≥mA} is a normal ideal. In particular, if R is normal, the Rees ring R[It] is normal.

We will show that for all positive integers p, I^{p} = R_{≥pmA}. By Theorem 2, this will complete the proof.

Inductive Step. Let R, I, and A be as in Theorem 3. For p ≥ 2, if I^{p−1} = R_{≥(p−1)mA} then I^{p} = R_{≥pmA}.

It is obvious that I^{p} ⊆ R_{≥pmA}. We have to show that the other inclusion holds. Also, observe that elements in R_{≥pmA} are sums of monomials in the X_{i} with coefficients in k, whose degrees are larger than or equal to pmA, and therefore we notice that R_{≥pmA} is generated by such monomials. Hence it is enough to show that every monomial of k[X_{0}, ..., X_{m}] which lies in R_{≥pmA} also lies in I^{p}.

Lemma. With notation as above, let X_{0}^{c_{0}}...X_{m}^{c_{m}} ∈ R_{≥pmA}, where c_{0}, ..., c_{m} are nonnegative integers, and let a_{0}, ..., a_{m} be positive integers such that a_{i}A_{i} = A for all i. Assume I^{p−1} = R_{≥(p−1)mA}. Fix n such that 1 ≤ n ≤ m, and suppose that a_{i} ≤ c_{i} for 0 ≤ i < n. For each i smaller than n, let k_{i} be the unique positive integer such that k_{i}a_{i} ≤ c_{i} < (k_{i} + 1)a_{i}. Then:

1) if k_{0} + ... + k_{n−1} ≤ m − 1, then for some a_{j} (n ≤ j ≤ m), a_{j} ≤ c_{j};

2) if k_{0} + ... + k_{n−1} ≥ m, then X_{0}^{c_{0}}...X_{m}^{c_{m}} ∈ I^{p}.

Proof of Lemma. 1) If c_{i} < a_{i} for all n ≤ i ≤ m, then

pmA ≤ c_{0}A_{0} + ... + c_{m}A_{m}

< (k_{0} + 1)a_{0}A_{0} + ... + (k_{n−1} + 1)a_{n−1}A_{n−1} + a_{n}A_{n} + ... + a_{m}A_{m}

= (k_{0} + ... + k_{n−1})A + nA + (m − n + 1)A
≤ (m − 1)A + nA + (m − n + 1)A
= 2mA.

It follows that p < 2, contrary to the hypothesis.

2) Choose nonnegative integers s_{0}, ..., s_{n−1} such that s_{i} ≤ k_{i} for all 0 ≤ i ≤ n − 1, and s_{0} + ... + s_{n−1} = m. We see that

X_{0}^{c_{0}}...X_{m}^{c_{m}} = X_{0}^{s_{0}a_{0}}...X_{n−1}^{s_{n−1}a_{n−1}} · X_{0}^{c_{0}−s_{0}a_{0}}...X_{n−1}^{c_{n−1}−s_{n−1}a_{n−1}}X_{n}^{c_{n}}...X_{m}^{c_{m}}.

Now

deg X_{0}^{s_{0}a_{0}}...X_{n−1}^{s_{n−1}a_{n−1}}
= s_{0}A + ... + s_{n−1}A
= mA,

which implies that X_{0}^{s_{0}a_{0}}...X_{n−1}^{s_{n−1}a_{n−1}} ∈ I. On the other hand

deg X_{0}^{c_{0}−s_{0}a_{0}}...X_{n−1}^{c_{n−1}−s_{n−1}a_{n−1}}X_{n}^{c_{n}}...X_{m}^{c_{m}}
= deg X_{0}^{c_{0}}...X_{m}^{c_{m}} − mA
≥ pmA − mA
= (p − 1)mA.

Therefore X_{0}^{c_{0}−s_{0}a_{0}}...X_{n−1}^{c_{n−1}−s_{n−1}a_{n−1}}X_{n}^{c_{n}}...X_{m}^{c_{m}} ∈ I^{p−1}, so X_{0}^{c_{0}}...X_{m}^{c_{m}} ∈ I^{p}. ■

Proof of Inductive Step. We have X_{0}^{c_{0}}...X_{m}^{c_{m}} ∈ R_{≥pmA}, where c_{0}, ..., c_{m} are nonnegative integers, and we want to show that it lies in I^{p} as well. Let a_{0}, ..., a_{m} be positive integers such that a_{i}A_{i} = A for all i.

If c_{i} ≥ ma_{i} for any i, say if c_{0} ≥ ma_{0}, then we will get

X_{0}^{c_{0}}...X_{m}^{c_{m}} = X_{0}^{ma_{0}} · X_{0}^{c_{0}−ma_{0}}...X_{m}^{c_{m}}.

Now, deg X_{0}^{ma_{0}} = ma_{0}A_{0} = mA, and so X_{0}^{ma_{0}} ∈ I. On the other hand,

deg X_{0}^{c_{0}−ma_{0}}...X_{m}^{c_{m}} = deg X_{0}^{c_{0}}...X_{m}^{c_{m}} − mA ≥ pmA − mA = (p − 1)mA.

So X_{0}^{c_{0}−ma_{0}}...X_{m}^{c_{m}} ∈ I^{p−1} by our assumption. Therefore X_{0}^{c_{0}}...X_{m}^{c_{m}} ∈ I^{p}, and we are done.

So let us look at the case where c_{i} < ma_{i} for all i.

Now, we cannot have c_{i} < a_{i} for all i, because in that case we get

pmA ≤ c_{0}A_{0} + ... + c_{m}A_{m} < a_{0}A_{0} + ... + a_{m}A_{m} = (m + 1)A,

and so

pm < m + 1 ⇒ p < 1 + \frac{1}{m} ≤ 2,

which is a contradiction.

So without loss of generality let c_{0} ≥ a_{0}, and let k_{0} be the positive integer (0 < k_{0} < m) such that k_{0}a_{0} ≤ c_{0} < (k_{0} + 1)a_{0}.

Hereafter, we proceed by induction: Having chosen k_{0}, ..., k_{n−1}, if k_{0} + ... + k_{n−1} ≥ m, then X_{0}^{c_{0}}...X_{m}^{c_{m}} ∈ I^{p} by the above lemma, and hence we are done.

However, if k_{0} + ... + k_{n−1} ≤ m − 1, by part 1 of the lemma there exists a j ≥ n for which a_{j} ≤ c_{j}. We can without loss of generality assume that j = n, and repeat the same cycle. Thus we are reduced to considering the case where a_{i} ≤ c_{i} for i = 0, ..., m. Let the k_{0}, ..., k_{m} be the unique positive integers for which k_{i}a_{i} ≤ c_{i} < (k_{i} + 1)a_{i}, (0 ≤ i ≤ m). We can easily see that k_{0} + ... + k_{m} ≥ m, because otherwise

pmA ≤ c_{0}A_{0} + ... + c_{m}A_{m}
< (k_{0} + 1)a_{0}A_{0} + ...(k_{m} + 1)a_{m}A_{m}
= (k_{0} + ... + k_{m})A + (m + 1)A
≤ (m − 1)A + (m + 1)A
= 2mA,

which implies that p < 2, contrary to our assumption. It follows then from part 2 of the same lemma that X_{0}^{c_{0}}...X_{m}^{c_{m}} ∈ I^{p}. This completes the proof of the inductive step. ■

Conclusion of Proof of Theorem 3. We need I^{p} to be integrally closed for all p ≥ 1. By Theorem 2, it suffices to have I^{p} = R_{≥pmA} for p ≥ 1. We prove this by induction. The case p = 1 is the definition of I. If I^{k} = R_{≥kmA} for all 1 ≤ k ≤ p − 1, then from the inductive step it follows that I^{p} = R_{≥pmA}. If R is normal, it follows from Theorem 1 that R[It] is a normal ring. ■

3. Examples

Example 1. Let k be a field, and R = k[x, y, z]/(x^{2} + y^{3} − z^{5}), where x, y, z have weights 15,10, and 6, respectively. the least common multiple of these variables is 30, and therefore by theorem 3,

\begin{aligned}I & = R_{≥60} = (x^{4}, x^{3}y^{2}, x^{3}yz, x^{3}z^{3}, x^{2}y^{3}, x^{2}y^{2}z^{2}, x^{2}yz^{4}, x^{2}z^{5}, xy^{5}, \\ & xy^{4}z, xy^{3}z^{3}, xy^{2}z^{5}, xyz^{6}, xz^{8}, z^{10}, yz^{9}, y^{2}z^{7}, y^{3}z^{5}, y^{4}z^{4}, y^{5}z^{2}, y^{6}) \end{aligned}

is a normal ideal for this ring.

Example 2. In general, if k is a field, and R = k[x, y, z]/(x^{a} + y^{b} + z^{c}) is a domain, where the variables x, y, z have weights bc, ac, and ab, respectively, the ideal I = R_{≥2abc} will be a normal ideal of R.

Example 3. For the polynomial ring R = k[x, y, z] over a field k, one can find many normal ideals by assigning different weights to the variables x, y and z. For example, if we set deg x = 1, deg y = 1 and deg z = 2 we find that the ideal I = R_{≥4} = (x^{4}, x^{3}y, x^{2}y^{2}, x^{2}z, xy^{3}, xyz, y^{4}, y^{2}z, z^{2}) is a normal ideal.

Remark 1. The method described in Example 3 above for finding normal ideals of polynomial rings is not as interesting in the case of two variables. It is a theorem due to Zariski (see [Z]), that the set of integrally closed ideals in a regular local ring of dimension two is closed under multiplication. Therefore, in the ring R = k[x, y], all integrally closed ideals are normal. In particular, all ideals of the form I = R_{≥α}, where α is a positive integer, are normal (all powers of I are integrally closed).

Remark 2. Looking at Theorems 1 and 2, it is natural to wonder whether for an N-graded normal ring R and all positive integers α, one can say that \overline{(R_{≥α})^{n}} = R_{≥nα} for n ∈ N. The answer in general is no. For example, consider the ring R = k[x, y] where k is a field, deg x = 2, and deg y = 3. Let I = (R_{≥7})^{3}. We will check if \overline{I} is equal to R_{≥21}. Since R_{≥7} = (x^{4}, yx^{2}, xy^{2}, y^{3}), we can calculate I = (x^{12}, x^{10}y, x^{8}y^{2}, x^{6}y^{3}, x^{5}y^{4}, x^{4}y^{5}, x^{3}y^{6}, x^{2}y^{7}, xy^{8}, y^{9}). If Γ is the set of pairs (a, b) corresponding to generators x^{a}y^{b} of I, we plot the points in Γ on the R^{2} plane, and we see that there are no pairs of positive integers (c, d) in the convex hull of the region Γ+R^{2}^{+}, besides those in Γ+R^{2}^{+} itself. This implies that I is an integrally closed ideal (see [E]). On the other hand x^{11} ∈ R_{≥21}, but x^{11} does not belong to I = \overline{I}. Therefore R_{≥21} cannot be the integral closure of I. (Note that it also follows from Zariski’s theorem mentioned in Remark 1 that I is an integrally closed ideal.)

Remark 3. One might ask if all ideals of the form I = R_{≥α}, α ∈ N, in a normal graded ring are normal. The answer is no. A counterexample is the ring R = k[x, y, z]/(x^{2} + y^{3}z + z^{4}), where x, y and z have degrees 2, 1 and 1, respectively. The ideal I = R_{≥1} = (x, y, z) is integrally closed, but I^{2} = (x^{2}, xy, xz, y^{2}, yz, z^{2}) is not: (x)^{2} + (y^{2})(yz) + (z^{2})^{2} = 0, and hence x ∈ \overline{I^{2}}, but x is not in I^{2}. This, by the way, is an example of a ring whose test ideal is not normal, given by Hara and Smith in [HSm]. For more on test ideals and tight closure theory in general, see [HH].

References

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[EGA] Grothendieck, A., Èlèments de Gèomètrie Algèbrique II, I.H.E.S. Publ. Math. 4 (1965).

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[H] Hironaka, H., Resolution of Singularities of an Algebraic Variety Over a Field of Characteristic Zero, Ann. Math. 79 (1964), 109-326.

[HH] Hochster, M. and Huneke, C., Tight Closure, Invariant Theory, and the Briancon-Skoda Theorem, J. Amer. Math. Soc. 3 (1990), 31-116.

[Ho] Hochster, M., The Tight Integral Closure of a Set of Ideals, To appear in J. Algebra.

[HS] Huneke, C. and Sally, J.D., Birational Extensions in Dimension Two and Integrally Closed Ideals, J. Algebra 115 (1988), 481-500.

[HSm] Hara, N. and Smith, K.E., The Strong Test Ideal, preprint.

[O] O’Carroll, L., A Generic Approach to the Structure of Certain Normal Ideals, Compositio Math. 104 (1996), 217-225.

[Z] Zariski, O. and Samuel, P., Commutative Algebra, Vol. 2,, Van Nostrand, Princeton, 1960.