Quadratic transform of complete ideals in regular

local rings

Clare D’Cruz SPIC Mathematical Institute, 92-G. N. Chetty Road Chennai 600 017, India

clare@smi.ernet.in

1 Introduction

Throughout this paper, (R, m) will denote a regular local ring with infinite residue field. The integral closure of an ideal, denoted by \overline{I} is defined as \overline{I} = \{x ∈ R|x^{n} + a_{1}x^{n−1} + ⋯ + a_{n} = 0; a_{i} ∈ I^{i}\}. An ideal I is complete if I = \overline{I}.

The theory of complete ideals in two-dimensional regular local rings was first studied by Zariski [ZS]. Zariski showed that the product of complete ideals is complete. Moreover, every complete ideal can be uniquely factorised as a product of simple complete ideals. An elegant treatment for Zariski’s theory was given by Huneke in [H1]. Zariski’s theory does not hold true for complete ideals in regular local rings of dimension greater than two ([C], [H2], [L], [T]).

In a two dimensional regular local ring quadratic transform of a complete ideal is always complete. This result does not hold true in higher dimension. In this short note we present a class of m-primary complete ideals whose quadratic transform is complete.

It is known that if I is any m-primary ideal in a regular local ring of dimension two, then μ(I) ≤ 1+o(I), where o(I) is the m-adic order of I. If I is complete then equality holds [R, Lemma 3.1]. In this paper we show that if I is an m-primary complete ideal in a regular local ring of dimension d, then μ(I) ≥ \binom{o(I) + d − 1}{d − 1}. This is done in section two.

In section three we consider complete ideals I which are minimally generated by \binom{o(I) + d − 1}{d − 1} elements. We show that in this case m^{n}I is a complete ideal for all n ≥ 0 (Proposition 3.2). As a consequence, the quadratic transform of I is complete (Theorem 3.4).

2 Complete ideals and m-full ideals

Let (R, m) be a local ring. An ideal I is m-full if mI : x = I for some x ∈ m [W]. The relationship between complete ideals and m-full ideals was observed by Rees.

Lemma 2.1 (Rees) Let (R, m) be a local ring with infinite residue field. If I is complete, then I is m-full or I = \sqrt{0}.

Proof: See Theorem 2.4 in [G].

If I is m-full and J ⊇ I is any ideal then μ(I) ≥ μ(J) [G]. Hence it is possible to give a lower bound on the number of generators of a complete ideal.

Lemma 2.2 Let (R, m) be a regular local ring of dimension d ≥ 2. Let I be an mprimary complete ideal and let r = o(I). Then μ(I) ≥ \binom{r + d − 1}{d − 1} and equality holds if and only if there exists an element x ∈ m\m^{2} such that m^{r+1} = mI+xm^{r}.

Proof: Put J = m^{r} in Lemma 2.2 of [G].

3 Quadratic transform

In this section we prove our main result. We first recall a few basic facts.

A quadratic transform T of R is the ring obtained by localizing S = R[m/x] at a prime ideal of height at least two containing mS. The transform of an ideal I in S is defined to be the ideal x^{−r}IS where r = o(I) and will be denoted by I^{S}. We define I^{T} := I^{S}T.

Lemma 3.1 Let I be an m-primary complete ideal in (R, m). Then there exists x ∈ m \ m^{2} such that \overline{mI} : x = I.

Proof: Put J = \overline{mI}. Since I and J are complete ideals, they can be expressed as an intersection of IV (resp. JV) over all valuation rings V in the quotient field K of R which contain R. As the length of R/I and R/J are finite, there exist valuation rings V_{1}, …, V_{n} such that I = IV_{1} ∩ ⋯ ∩ IV_{n} ∩ R and J = JV_{1} ∩ ⋯ ∩ JV_{n} ∩ R. Choose x ∈ m \ m^{2} such that xV_{i} = mV_{i} for all i = 1, …, n.

Now suppose a is in J : x. Then xa is in J so that xaV_{i} ⊆ JV_{i} = xIV_{i} for all i = 1, …, n. Hence a is contained in IV_{1} ∩ ⋯ IV_{n} ∩ R = I. Thus \overline{mI} : x ⊆ I. The other inclusion is obvious.

From now on r will denote the m-adic order of I.

Proposition 3.2 Let I be an m-primary complete ideal in (R, m). If μ(I) = \binom{r + d − 1}{d − 1}, then for all n ≥ 1, m^{n}I is complete,

Proof: We induct on n. Suppose n = 1. Since μ(I) = \binom{r + d − 1}{d − 1}, by Lemma 2.2, there exists x ∈ m \ m^{2} such that mI + xm^{r} = m^{r+1}. Hence

(m^{r+1}, x) = (mI, x) ⊆ (\overline{mI}, x) ⊆ (m^{r+1}, x).

Therefore \overline{mI} ⊆ mI + xR which gives \overline{mI} = mI + x(\overline{mI} : x). By Lemma 3.1, \overline{mI} : x = I. This proves that mI is complete. Our assumption on I implies that m^{r+2} = m^{2}I +xm^{r+1} and consequently μ(mI) = \binom{r + d}{d − 1} (Lemma 2.2). If n > 1,

then by induction hypothesis m^{n−1}I is complete and μ(m^{n−1}I) = \binom{n + r + d − 2}{d − 1}. We are now reduced to the case n = 1.

Before we prove the main theorem we define contracted ideals. An ideal I is contracted if there exists S = R[m/x] such that I = IS ∩R. In a two-dimensional regular local ring an m-primary complete ideal is always contracted [ZS]. We prove this result for any dimension d.

Lemma 3.3 Let I be an m-primary complete ideal in (R, m). Then there exists x ∈ m \ m^{2} such that I = IS ∩ R, where S = R[m/x].

Proof: Since I is complete and the length of R/I is finite, there exist valuation rings V_{1}, …, V_{n} such that I = IV_{1} ∩ ⋯ ∩ IV_{n} ∩ R. Since R/m is infinite, there exists x ∈ m \ m^{2} such that xV_{i} = mV_{i} for all i = 1, …, n. Let S = R[m/x]. We claim that I = IS ∩ R. Suppose a ∈ IS ∩ R. Then a = b/x^{n} for some n with b ∈ mI^{n}. This gives ax^{n} ∈ m^{n}I. Hence ax^{n}V_{i} = m^{n}IV_{i} = x^{n}IV_{i} for all i = 1, …, n. Therefore a ∈ IV_{1} ∩ ⋯ ∩ IV_{n} ∩ R = I.

Theorem 3.4 Let I be an m-primary complete ideal in (R, m). Assume that μ(I) = \binom{r + d − 1}{d − 1}. Suppose I is contracted from S = R[m/x]. Then I^{S} is complete.

Proof: Proposition 3.2 implies that m^{n}I is complete for all n ≥ 1. Hence I^{S} is complete [H1, Proposition 3.4].

Let I be an m-primary complete ideal in R and S = R[m/x]. It is known that if ℓ_{R}(S/I^{S}) is finite, then this length is equal to ℓ_{R}(m^{r+n+1}/Im^{n+1}S ∩ R) for all n large [K, Proposition 2.16]. In addition, if μ(I) = \binom{r + d − 1}{d − 1}, then we have:

Theorem 3.5 Suppose I is an m-primary complete ideal in (R, m). Suppose μ(I) = \binom{r + d − 1}{d − 1}. Assume that I is contracted from S = R[m/x]. Then ℓ_{R}(S/I^{S}) is finite and ℓ_{R}(S/I^{S}) = ℓ(m^{r+n}/Im^{n}) for all n ≥ 0.

Proof: Since m^{n}I is complete for all n ≥ 0, the map

\frac{m^{r+n}}{m^{n}I} ⟶ \frac{S}{I^{S}}

is injective. Proposition 3.2 implies that m^{r+n+1} = xm^{r+n} + Im^{n+1} for all n ≥ 0. Hence the above map is surjective.

4 Remarks

Remark 4.1 Let R be a regular local ring of dimension d ≥ 3. Theorem 3.4 does not hold true if μ(I) > \binom{r + d − 1}{d − 1} ([T, Example 1.3]). In this example, even mI is not complete.

Remark 4.2 A natural question one can ask is: Suppose I_{1} and I_{2} are complete ideals in a regular local ring of dimension d ≥ 3. Assume μ(I_{j}) = \binom{o(I_{j}) + d − 1}{d − 1}, (j = 1, 2). Then is I_{1}I_{2} complete? This question does not have an affirmative answer ([T, Example 2.9]). This example also reveals that if I is an mprimary complete ideal in (R, m), μ(I) = \binom{r + d − 1}{d − 1} and I = IS ∩ R then

μ(I^{S}) ≠ \binom{o(I^{s}) + d − 1}{d − 1}.

Acknowledgements: The author is very grateful to J. K. Verma for introducing the subject and for his encouragement. The author also wishes to thank W. Heinzer, J. Lipman and V. Kodiyalam for their valuable suggestions.

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