Vanishing and Non-negativity of the First Normal Hilbert Coefficient

Abstract

Let $(R,\mathfrak {m})$ be a Noetherian local ring such that $\widehat{R}$ is reduced. We prove that, when $\widehat{R}$ is $S_2$, if there exists a parameter ideal $Q\subseteq R$ such that $\bar{e}_1(Q)=0$, then R is regular and $\nu (\mathfrak {m}/Q)\le 1$. This leads to an affirmative answer to a problem raised by Goto-Hong-Mandal [Goto, S., Hong, J., Mandal, M.: The positivity of the first coefficients of normal Hilbert polynomials. Proc. Amer. Math. Soc. 139(7), 2399–2406 (2011)]. We also give an alternative proof (in fact a strengthening) of their main result. In particular, we show that if $\widehat{R}$ is equidimensional, then $\bar{e}_1(Q)\ge 0$ for all parameter ideals $Q\subseteq R$, and in characteristic $p>0$, we actually have $e_1^*(Q)\ge 0$. Our proofs rely on the existence of big Cohen-Macaulay algebras.

1 Introduction

Let $(R,\mathfrak {m})$ be a Noetherian local ring of dimension d such that $\widehat{R}$ is reduced and let $I\subseteq R$ be an $\mathfrak {m}$-primary ideal. Then for $n\gg 0$, $\ell (R/\overline{I^{n+1}})$ agrees with a polynomial in n of degree d, and we have integers $\overline{e}_0(I),\dots , \overline{e}_d(I)$ such that

$$\ell (R/\overline{I^{n+1}})=\overline{e}_0(I)\left( {\begin{array}{c}n+d\\ d\end{array}}\right) - \overline{e}_1(I)\left( {\begin{array}{c}n+d-1\\ d-1\end{array}}\right) + \cdots + (-1)^d\overline{e}_d(I).$$

These integers $\overline{e}_i(I)$ are called the normal Hilbert coefficients of I.

It is well-known that $\overline{e}_0(I)$ is the Hilbert-Samuel multiplicity of I, which is always a positive integer. In this paper, we are interested in the first coefficient $\overline{e}_1(I)$. It was proved by Goto-Hong-Mandal [10] that when $\widehat{R}$ is unmixed, $\overline{e}_1(I)\ge 0$ for all $\mathfrak {m}$-primary ideals $I\subseteq R$ (which answers a question posed by Vasconcelos [30]). They proposed a further problem in [10, Section 3] regarding the vanishing of $\overline{e}_1(I)$ and the regularity of the normalization of R. Since any $\mathfrak {m}$-primary ideal I is integral over a parameter ideal when the residue field is infinite, to study $\overline{e}_1(I)$ we may assume that $I=Q$ is a parameter ideal (i.e., it is generated by a system of parameters). In this paper, we prove the following main result which will lead to an affirmative answer to the question proposed in [10]. This theorem is also a generalization of the main result of [27].

Theorem 1.1

(Theorem 3.7) Let $(R,\mathfrak {m})$ be a Noetherian local ring such that $\widehat{R}$ is reduced and $S_2$. If $\overline{e}_1(Q)=0$ for some parameter ideal $Q\subseteq R$, then R is regular and $\nu (\mathfrak {m}/Q)\le 1$.

In [7], it was shown that when R has characteristic $p>0$, for $n\gg 0$, $\ell (R/(I^{n+1})^*)$ also agrees with a polynomial of degree d and one can define the tight Hilbert coefficients $e_0^*(I),\dots , e_d^*(I)$ in a similar way (see Section 2 for more details). It is easy to see that $\overline{e}_1(I)\ge e_1^*(I)$. We strengthen the main result of [10] in characteristic $p>0$ by showing that $e_1^*(Q)\ge 0$ for any parameter ideal $Q\subseteq R$ under mild assumptions.

Theorem 1.2

(Corollary 3.3) Let $(R,\mathfrak {m})$ be an excellent local ring of characteristic $p>0$ such that $\widehat{R}$ is reduced and equidimensional. Then we have $e^*_1(Q)\ge 0$ for all parameter ideals $Q\subseteq R$.

Our proofs of both theorems rely on the existence of big Cohen-Macaulay algebras. In fact, we show that the tight Hilbert coefficients $e_i^*(I)$ is a special case of what we call the BCM Hilbert coefficients $e_i^B(I)$ associated to a big Cohen-Macaulay algebra B, and the latter can be defined in arbitrary characteristic. In this context, we will show in Theorem 3.1 that $\overline{e}_1(Q)\ge e_1^B(Q)\ge 0$ for all parameter ideals $Q\subseteq R$ when B satisfies some mild assumptions. This recovers and extends the main result of [10] in arbitrary characteristic.

Throughout this article, all rings are commutative with multiplicative identity 1. We will use $(R,\mathfrak {m})$ to denote a Noetherian local ring with unique maximal ideal $\mathfrak {m}$. We refer the reader to [4, Chapter 1-4] for basic notions such as Cohen-Macaulay rings, regular sequence, Euler characteristic, integral closure, and the Hilbert-Samuel multiplicity. We refer the reader to [29, Section 07QS] for the definition and basic properties of excellent rings. The paper is organized as follows. In Section 2 we collect the definitions and some basic results on big Cohen-Macaulay algebras and variants of Hilbert coefficients. In Section 3 we prove our main results and we propose some further questions.

2 Preliminaries

Recall that an element x in a ring R is integral over an ideal $I\subseteq R$ if it satisfies an equation of the form $x^n+ a_1x^{n-1}+\cdots + a_{n-1}x + a_n=0$, where $a_k\in I^k$. The set of all elements integral over I forms an ideal and is denoted by $\overline{I}$, called the integral closure of I. An ideal $I\subseteq R$ is called integrally closed if $I=\overline{I}$. It is well-known that an element $x\in R$ is integral over I if and only if the image of x in $R/\mathfrak {p}$ is integral over $I(R/\mathfrak {p})$ for all minimal primes $\mathfrak {p}$, see [21, Proposition 1.1.5].

Suppose that R is a Noetherian ring of prime characteristic $p>0$. The tight closure of an ideal $I\subseteq R$, introduced by Hochster–Huneke, is defined as follows:

$$\begin{aligned}I^* := \{x\in R \, | \, \text {there exists}\, c\in R-\cup _{\mathfrak {p}\in \text {Min}(R)}\mathfrak {p}\, \text { such that }\, cx^{p^e}\in I^{[p^e]} \text { for all }\, e\gg 0\}.\end{aligned}$$

An ideal $I\subseteq R$ is called tightly closed if $I=I^*$. In general, tight closure is always contained in the integral closure, that is, $I^*\subseteq \overline{I}$ (see [15, Proposition on page 58]). Similar to integral closure, an element $x\in R$ is in the tight closure of I if and only if the image of x in $R/\mathfrak {p}$ is in the tight closure of $I(R/\mathfrak {p})$ for all minimal primes $\mathfrak {p}$, see [15, Theorem on page 49].

Let R be a Noetherian complete local domain and let $I\subseteq R$ be an ideal. The solid closure of I, denoted by $I^\bigstar$, consists of those element $x\in R$ such that there exists an R-algebra S such that ${\text {Hom}}_R(S, R)\ne 0$ and such that $x\in IS$. One can define solid closure of ideals in more general rings, see [16, Definition 1.2], but we will only need this notion for complete local domains. It was shown in [16, Theorem 5.10] that solid closure is contained in the integral closure, i.e., $I^\bigstar \subseteq \overline{I}$. If R has prime characteristic $p>0$, then solid closure agrees with tight closure $I^\bigstar = I^*$, see [16, Theorem 8.6].

2.1 Big Cohen-Macaulay Algebras

Let $(R,\mathfrak {m})$ be a Noetherian local ring. An R-algebra B, not necessarily Noetherian, is called balanced big Cohen-Macaulay over R if every system of parameters of R is a regular sequence on B and $\mathfrak {m} B\ne B$. Balanced big Cohen-Macaulay algebras exist, in equal characteristic, this is due to Hochster-Huneke [18], and in mixed characteristic, this is proved by André [1] (see also [2, 3, 12]). In this article, we need to compare the closure operation induced by a balanced big Cohen-Macaulay algebra with integral closure. We begin with the following result.

In what follows, when $R\rightarrow S$ is a (not necessarily injective) homomorphism of rings, $IS\cap R$ should be interpreted as the contraction of IS to R. That is, those elements of R whose image in S are contained in IS.

Lemma 2.1

Let $(R,\mathfrak {m})$ be a Noetherian local ring. Then the following conditions are equivalent:

• (1) $\widehat{R}$ is equidimensional.

• (2) There exists a balanced big Cohen-Macaulay R-algebra B such that

  

• (3) There exists a balanced big Cohen-Macaulay R-algebra B such that $I^B\subseteq \overline{I}$ for all $I\subseteq R$.

Proof

Since $(3)\Rightarrow (2)$ is obvious, we only need to show $(1)\Rightarrow (3)$ and $(2)\Rightarrow (1)$. Suppose that $\widehat{R}$ is equidimensional and let $P_1,\dots ,P_n$ be the minimal primes of $\widehat{R}$. Let $B_i$ be any balanced big Cohen-Macaulay algebra over $\widehat{R}/P_i$. Since $\widehat{R}$ is equidimensional, each system of parameters of $\widehat{R}$ is also a system of parameters of $\widehat{R}/P_i$ and thus $B_i$ is a balanced big Cohen-Macaulay algebra over $\widehat{R}$. It follows that $B:= \prod _{i=1}^{n}B_i$ is a balanced big Cohen-Macaulay algebra over $\widehat{R}$.

Claim 2.2

$(I\widehat{R})^B= IB\cap \widehat{R} \subseteq \overline{I\widehat{R}}$.

Proof of the Claim

Since integral closure can be checked after modulo each minimal prime, it suffices to show that $(I\widehat{R})^B\cdot (\widehat{R}/P_i)\subseteq \overline{I(\widehat{R}/P_i)}$. It is easy to see (by our construction of B) that

$$(I\widehat{R})^B\cdot (\widehat{R}/P_i) = (I(\widehat{R}/P_i))^{B_i}.$$

Since $B_i$ is a solid algebra over the complete local domain $\widehat{R}/P_i$ by [16, Corollary 2.4], we have

$$(I(\widehat{R}/P_i))^{B_i} \subseteq (I(\widehat{R}/P_i))^\bigstar \subseteq \overline{I(\widehat{R}/P_i)},$$

where the second inclusion follows from [16, Theorem 5.10].$\square$

By the claim above, we have

$$I^B\subseteq (I\widehat{R})^B \cap R \subseteq \overline{I\widehat{R}} \cap R=\overline{I},$$

where the last equality follows from [21, Proposition 1.6.2].

We next assume there exists a balanced big Cohen-Macaulay R-algebra B that satisfies ($\dagger$). We first note that $\widehat{B}$ (the $\mathfrak {m}$-adic completion of B) is still a balanced big Cohen-Macaulay algebra over $\widehat{R}$ by [4, Corollary 8.5.3]. If I is an $\mathfrak {m}$-primary ideal, then we have $R/I \cong \widehat{R}/I\widehat{R}$ and $B/IB \cong \widehat{B}/I\widehat{B}$ (see [29, Tag 05GG]). It follows that $(I\widehat{R})^{\widehat{B}} = (I^B)\widehat{R} \subseteq \overline{I} \widehat{R} = \overline{I\widehat{R}}$ (where the last equality follows from [21, Lemma 9.1.1]). Thus without loss of generality, we may replace R by $\widehat{R}$ and B by $\widehat{B}$ to assume that R is complete. Suppose that R is not equidimensional. Let $P_1, \ldots , P_n$ be all the minimal primes of R such that $\dim (R/P_i) = \dim (R)$, and $Q_1, \ldots , Q_m$ be all the minimal primes of R such that $\dim (R/Q_j) < d$. We pick $y \in Q_1 \cap \cdots \cap Q_m \setminus P_1\cup \cdots \cup P_n$. Then y is a parameter element in R, and thus y is a nonzerodivisor on B, since B is balanced big Cohen-Macaulay. Since $y \cdot (P_1 \cap \cdots \cap P_n) \subseteq \sqrt{0}$, there exists t such that $y^t \cdot (P_1 \cap \cdots \cap P_n)^t=0$. It follows that $(P_1 \cap \cdots \cap P_n)^tB = 0$. Hence

$$(P_1 \cap \cdots \cap P_n)^t \subseteq \mathfrak {m}^k B \cap R \subseteq \overline{\mathfrak {m}^k}$$

for all k by ($\dagger$). Thus $(P_1 \cap \cdots \cap P_n)^t\subseteq \cap _k \overline{\mathfrak {m}^k} = \sqrt{0}$ by [21, Exercise 5.14], which is a contradiction. $\square$

Remark 2.3

In the proof of Lemma 2.1, we have proved the fact that when $(R,\mathfrak {m})$ is a Noetherian complete local domain, then every balanced big Cohen-Macaulay algebra B satisfies ($\dagger$). We suspect that when $(R,\mathfrak {m})$ is Noetherian, complete, reduced and equidimensional, then every balanced big Cohen-Macaulay algebra B such that ${\text {Supp}}(\widehat{B})={\text {Spec}}(R)$ satisfies ($\dagger$).

2.2 Hilbert Coefficients

Let $(R, \mathfrak {m})$ be a Noetherian local ring of dimension d and let $I\subseteq R$ be an $\mathfrak {m}$-primary ideal. Then for all $n \gg 0$ we have

$$\ell (R/I^{n+1}) = e_0(I) \left( {\begin{array}{c}n+d\\ d\end{array}}\right) - e_1(I) \left( {\begin{array}{c}n+d-1\\ d-1\end{array}}\right) + \cdots + (-1)^d e_d(I),$$

where $e_0(I), \dots , e_d(I)$ are all integers, and are called the Hilbert coefficients of I.

Now suppose that $R \oplus \overline{I}t \oplus \overline{I^2}t^2 \oplus \cdots$ is module-finite over the Rees algebra R[It]. For instance, by a famous result of Rees (see [21, Corollary 9.2.1]), this is the case when $\widehat{R}$ is reduced. Then one can show that for all $n\gg 0$, $\ell (R/\overline{I^{n+1}})$ agrees with a polynomial in n and one can write

$$\ell (R/\overline{I^{n+1}}) = \overline{e}_0(I) \left( {\begin{array}{c}n+d\\ d\end{array}}\right) - \overline{e}_1(I) \left( {\begin{array}{c}n+d-1\\ d-1\end{array}}\right) + \cdots + (-1)^d \overline{e}_d(I),$$

where the integers $\overline{e}_0(Q), \dots , \overline{e}_d(Q)$ are called the normal Hilbert coefficients. It is well-known that $e_0(I)=\overline{e}_0(I)$ agrees with the Hilbert-Samuel multiplicity e(I, R) of I.

We also recall the tight Hilbert coefficients studied in [7]. Again, we suppose that $\widehat{R}$ is reduced and R has characteristic $p>0$. Then we have

$$\ell (R/(I^{n+1})^*) = e_0^*(I) \left( {\begin{array}{c}n+d\\ d\end{array}}\right) - e_1^*(I) \left( {\begin{array}{c}n+d-1\\ d-1\end{array}}\right) + \cdots + (-1)^d e_d^*(I)$$

for all $n\gg 0$, and the integers $e_0^*(I),\dots ,e_d^*(I)$ are called the tight Hilbert coefficients, see [7] for more details.

Now if B is a balanced big Cohen-Macaulay R-algebra that satisfies ($\dagger$), then we know that $R\oplus I^Bt \oplus (I^2)^Bt^2 \oplus \cdots$ is an R-algebra that is also module-finite over R[It]: the fact that it is an R-algebra follows from the fact that $(I^a)^B(I^b)^B\subseteq (I^{a+b})^B$ for all a, b (i.e., $\{(I^n)^B\}_n$ form a graded family of ideals), and that it is module-finite over R[It] follows because by ($\dagger$), it is an R[It]-submodule of $R \oplus \overline{I}t \oplus \overline{I^2}t^2 \oplus \cdots$, and the latter is module-finite over R[It] (note that R[It] is Noetherian). Based on the discussion above, one can show that for all $n\gg 0$, $\ell (R/(I^{n+1})^B)$ also agrees with a polynomial in n, and we write

$$\ell (R/(I^{n+1})^B) = e_0^B(I) \left( {\begin{array}{c}n+d\\ d\end{array}}\right) - e_1^B(I) \left( {\begin{array}{c}n+d-1\\ d-1\end{array}}\right) + \cdots + (-1)^d e_d^B(I)$$

for all $n\gg 0$ (see [19] for more general results). We call the integers $e_0^B(I),\dots , e_d^B(I)$ the BCM Hilbert coefficients with respect to B. It is easy to see that $e_0^B(I)=e(I, R)$ is still the Hilbert-Samuel multiplicity of I, and that we always have $\overline{e}_1(I)\ge e_1^B(I)\ge e_1(I)$ by comparing the coefficients of $n^{d-1}$ and noting that $I^n\subseteq (I^n)^B\subseteq \overline{I^n}$ for all n by ($\dagger$).

Remark 2.4

We point out that when $(R,\mathfrak {m})$ is excellent and $\widehat{R}$ is reduced and equidimensional of characteristic $p>0$, the tight Hilbert coefficient is a particular case of BCM Hilbert coefficient. This follows from the fact that under these assumptions, there exists a balanced big Cohen-Macaulay algebra B such that $I^*=I^B$ for all $I\subseteq R$ (and any such B will satisfy ($\dagger$), since tight closure is contained in the integral closure [20, Theorem 1.3]). When R is a complete local domain this is proved in [15, Theorem on page 250]. In general, one can take such a $B_i$ for each complete local domain $\widehat{R}/P_i$, where $P_i$ is a minimal prime of $\widehat{R}$, and let $B=\prod B_i$. Since R is excellent, $I^*\widehat{R}=(I\widehat{R})^*$ (see [20, Proposition 1.5]) and as tight closure can be checked after modulo each minimal prime, it follows that $I^*\widehat{R}=(I\widehat{R})^B$ and thus $I^*=I^B$.

Throughout the rest of this article, we will be mainly working with parameter ideals, i.e., ideals generated by a system of parameters. As we mentioned in the introduction, this will not affect the study of $\overline{e}_1(I)$, since we can often enlarge the residue field and replace I by its minimal reduction.

3 The Main Results

In this section we prove our main results that $e_1^B(Q)$ (and hence $\overline{e}_1(Q)$) is always nonnegative for a parameter ideal Q, and that $\overline{e}_1(Q)=0$ for some parameter ideal Q implies R is regular.

3.1 Non-negativity of $\bar{e}_1(Q)$ and $e_1^B(Q)$

Theorem 3.1

Let $(R,\mathfrak {m})$ be a Noetherian local ring such that $\widehat{R}$ is reduced and equidimensional. Let B be any balanced big Cohen-Macaulay R-algebra that satisfies ($\dagger$). Then for all parameter ideals $Q\subseteq R$ we have

$$\overline{e}_1(Q) \ge e_1^B(Q) \ge 0 \ge e_1(Q).$$

Remark 3.2

$\overline{e}_1(Q) \ge 0$ was the main theorem of [10, Theorem 1.1], and $0 \ge e_1(Q)$ was first proved in full generality in [26, Theorem 3.6]. Our method gives alternative proofs, and is inspired by some work of Goto [9] (in fact the proof that $e_1(Q)\le 0$ via this method is due to Goto [9], see also [13, Theorem 1.1] for a generalization).

Corollary 3.3

Let $(R,\mathfrak {m})$ be an excellent local ring of characteristic $p>0$ such that $\widehat{R}$ is reduced and equidimensional. Then we have $e^*_1(Q)\ge 0$ for all parameter ideals $Q\subseteq R$.

Proof

This follows from Theorem 3.1 and Remark 2.4.$\square$

Proof of Theorem 3.1

Let $Q = (x_1, \ldots , x_d) \subseteq R$. Set $S = R[[y_1, \ldots , y_d]]$ and $\mathfrak {q} = (y_1 - x_1, \ldots ,$ $y_d - x_d) \subseteq S$. For all $n \ge 0$ we have $y_1, \ldots , y_d$ is a system of parameters on $S/\mathfrak {q}^{n+1}$, and that

$$R/Q^{n+1} = S/(\mathfrak {q}^{n+1} + (y_1, \ldots , y_d)).$$

We next note that

$$\begin{aligned} e_0(Q) = e(Q, R)= & \chi (x_1, \ldots , x_d; R)\\= & \chi (x_1, \ldots , x_d, y_1, \ldots , y_d;S)\\= & \chi (y_1, \ldots , y_d, y_1-x_1, \ldots , y_d - x_d;S)\\= & \chi (y_1, \ldots , y_d; S/\mathfrak {q})\\= & e(y_1, \ldots , y_d; S/\mathfrak {q}), \end{aligned}$$

where the equalities on the second and the fourth lines follow from the fact that $y_1,\dots ,y_d$ and $y_1 - x_1, \ldots , y_d - x_d$ are both regular sequences on S. Now since $S/\mathfrak {q}^{n+1}$ has a filtration by $\left( {\begin{array}{c}n+d\\ d\end{array}}\right)$ copies of $S/\mathfrak {q}$, by the additivity formula for multiplicity (see [21, Theorem 11.2.3]) we have

$$e(y_1, \ldots , y_d; S/\mathfrak {q}^{n+1}) = \left( {\begin{array}{c}n+d\\ d\end{array}}\right) e(y_1, \ldots , y_d; S/\mathfrak {q}).$$

Putting these together, we have

$$\left( {\begin{array}{c}n+d\\ d\end{array}}\right) e_0(Q) = \left( {\begin{array}{c}n+d\\ d\end{array}}\right) e(y_1, \ldots , y_d; S/\mathfrak {q}) = e(y_1, \ldots , y_d; S/\mathfrak {q}^{n+1}), $$

and

$$\ell (R/Q^{n+1}) = \ell \left( \frac{S/\mathfrak {q}^{n+1}}{(y_1, \ldots , y_d)S/\mathfrak {q}^{n+1}} \right) .$$

Since $y_1,\dots , y_d$ is a system of parameters of $S/\mathfrak {q}^{n+1}$, we have

$$\ell \left( \frac{S/\mathfrak {q}^{n+1}}{(y_1, \ldots , y_d)S/\mathfrak {q}^{n+1}} \right) \ge e(y_1, \ldots , y_d; S/\mathfrak {q}^{n+1}).$$

It follows that

$$\ell (R/Q^{n+1}) \ge \left( {\begin{array}{c}n+d\\ d\end{array}}\right) e_0(Q),$$

and thus $e_1(Q) \le 0$ (note that this does not require any assumption on $\widehat{R}$).

It remains to show that $e_1^B(Q) \ge 0$ (since $\overline{e}_1(Q)\ge e_1^B(Q)$ always holds, see the discussion in Section 2.2). Since $e_0^B(Q)=e_0(Q)$, it is enough to show that

$$\begin{aligned} \ell (R/(Q^{n+1})^B) \le \left( {\begin{array}{c}n+d\\ d\end{array}}\right) e_0(Q) \end{aligned}$$

(1)

for any balanced big Cohen-Macaulay algebra B. Below we will prove a slightly stronger result. Recall that for a parameter ideal $(z_1,\dots ,z_d)$ of R, the limit closure is defined as $(z_1,\dots ,z_d)^{\lim _R}:= \bigcup _t (z_1^{t+1},\dots ,z_d^{t+1}): (z_1z_2\cdots z_d)^{t}$. The limit closure does not depend on the choice of the elements $z_1,\dots , z_d$ (i.e., it only depends on the ideal $(z_1,\dots ,z_d)$). This is because $(z_1,\dots ,z_d)^{\lim _R}/(z_1,\dots ,z_d)$ is the kernel of the natural map $R/(z_1,\dots , z_d)\rightarrow H_{\mathfrak {m}}^d(R)$.

Claim 3.4

Set $\varLambda _{n+1}= \{(\alpha _1,\dots , \alpha _d)\in \mathbb {N}^d \, | \, \alpha _i\ge 1 \text { and } \sum _{i=1}^{d} \alpha _i = 1+n \}$ and for each $\alpha =(\alpha _1,\dots ,\alpha _d) \in \varLambda _{n+1}$, set $Q(\alpha )=(x_1^{\alpha _1},\dots ,x_d^{\alpha _d})$. Then we have

$$\ell \left( R/(\bigcap _{\alpha \in \varLambda _{n+1}} Q(\alpha )^{\lim _R})\right) \le \left( {\begin{array}{c}n+d\\ d\end{array}}\right) e_0(Q).$$

Proof of the Claim

Recall that we have already proved that

$$\left( {\begin{array}{c}n+d\\ d\end{array}}\right) e_0(Q) = e(y_1, \ldots , y_d; S/\mathfrak {q}^{n+1}).$$

Moreover, we always have (for example, see [23, Theorem 9])

$$e(y_1, \ldots , y_d; S/\mathfrak {q}^{n+1}) \ge \ell \left( \frac{S/\mathfrak {q}^{n+1}}{(y_1, \ldots , y_d)^{\lim _{S/\mathfrak {q}^{n+1}}}} \right) .$$

Therefore it is enough to prove that

$$\begin{aligned} \ell \left( \frac{S/\mathfrak {q}^{n+1}}{(y_1, \ldots , y_d)^{\lim _{S/\mathfrak {q}^{n+1}}}} \right) \ge \ell \left( R/(\bigcap _{\alpha \in \varLambda _{n+1}} Q(\alpha )^{\lim _R})\right) . \end{aligned}$$

(2)

Consider $z \in S$ whose image in $S/\mathfrak {q}^{n+1}$ is contained in $(y_1, \ldots , y_d)^{\lim _{S/\mathfrak {q}^{n+1}}}$. This means there exists some $t \ge 1$ such that

$$\begin{aligned} (y_1 y_2\cdots y_d)^{t} z\in & (y_1^{t+1}, \ldots , y_d^{t+1}, (y_1 - x_1, \ldots , y_d - x_d)^{n+1})\\\subseteq & (y_1^{t+1}, \ldots , y_d^{t+1}, (y_1 - x_1)^{\alpha _1}, \ldots , (y_d - x_d)^{\alpha _d}) \end{aligned}$$

for each $\alpha =(\alpha _1,\dots ,\alpha _d)\in \varLambda _{n+1}$. This implies

$$z \in (y_1, \ldots , y_d, (y_1 - x_1)^{\alpha _1}, \ldots , (y_d - x_d)^{\alpha _d})^{\lim _S} = (y_1, \ldots , y_d, x_1^{\alpha _1}, \ldots , x_d^{\alpha _d})^{\lim _S}.$$

But since $S = R[[y_1, \ldots , y_d]]$, it is straightforward to check that

$$(y_1, \ldots , y_d, x_1^{\alpha _1}, \ldots , x_d^{\alpha _d})^{\lim _S} = (x_1^{\alpha _1}, \ldots , x_d^{\alpha _d})^{\lim _R}S + (y_1, \ldots , y_d)S.$$

Thus if the image of z is contained in $(y_1, \ldots , y_d)^{\lim _{S/\mathfrak {q}^{n+1}}}$, then after modulo $(y_1, \ldots , y_d)S$, $\overline{z} \in (x_1^{\alpha _1}, \ldots , x_d^{\alpha _d})^{\lim _R}$ for each $(\alpha _1,\dots ,\alpha _d)\in \varLambda _{n+1}$, i.e., $\overline{z} \in \bigcap _{\alpha \in \varLambda _{n+1}} Q(\alpha )^{\lim _R}$. It follows that the natural surjection

$$S/\mathfrak {q}^{n+1} \xrightarrow {\mod (y_1, \ldots , y_d)S} R/Q^{n+1}$$

induces a surjection

$$\frac{S/\mathfrak {q}^{n+1}}{(y_1, \ldots , y_d)^{\lim _{S/\mathfrak {q}^{n+1}}}} \twoheadrightarrow \frac{R}{\bigcap _{\alpha \in \varLambda _{n+1}} Q(\alpha )^{\lim _R}}.$$

This clearly establishes (2) and completes the proof of the claim.$\square$

Finally, since $x_1,\dots , x_d$ is a regular sequence on B, we have $Q(\alpha )^{\lim _R}\subseteq Q(\alpha )^B$ for each $\alpha$. It follows that $\bigcap _{\alpha \in \varLambda _{n+1}} Q(\alpha )^{\lim _R} \subseteq \bigcap _{\alpha \in \varLambda _{n+1}} Q(\alpha )^B$. Now if $x \in \bigcap _{\alpha \in \varLambda _{n+1}} Q(\alpha )^B$, then we have $x \in \big (\bigcap _{\alpha \in \varLambda _{n+1}} Q(\alpha )B \big ) \cap R$. But since $x_1,\dots ,x_d$ is a regular sequence on B, it is not hard to check that $\bigcap _{\alpha \in \varLambda _{n+1}} Q(\alpha )B = Q^{n+1}B$ (see [28, Remark 3.3] or [11]) and thus $x\in Q^{n+1}B \cap R = (Q^{n+1})^B$. Therefore we have $\bigcap _{\alpha \in \varLambda _{n+1}} Q(\alpha )^B = (Q^{n+1})^B$. Putting these together, we have

$$\bigcap _{\alpha \in \varLambda _{n+1}} Q(\alpha )^{\lim _R} \subseteq \bigcap _{\alpha \in \varLambda _{n+1}} Q(\alpha )^B = (Q^{n+1})^B.$$

Therefore by Claim 3.4, we have

$$\ell (R/(Q^{n+1})^B) \le \left( {\begin{array}{c}n+d\\ d\end{array}}\right) e_0(Q)$$

as wanted.$\square$

Remark 3.5

With notation as in Theorem 3.1, we do not know whether we have

$$\ell \left( \frac{S/\mathfrak {q}^{n+1}}{(y_1, \ldots , y_d)^{\lim _{S/\mathfrak {q}^{n+1}}}}\right) = \ell \left( \frac{R}{\bigcap _{\alpha \in \varLambda _{n+1}} Q(\alpha )^{\lim _R}}\right) .$$

Remark 3.6

With notation as in Claim 3.4, fix a generating set $(x_1,\dots ,x_d)$ of Q, one may try to define $(Q^n)^{\lim } := \bigcap _{\alpha \in \varLambda _{n}} Q(\alpha )^{\lim }$ and call this the limit closure of $Q^n$. However, it is not clear to us whether this is independent of the choice of the generators $x_1,\dots ,x_d$. It is also not clear to us (even when fixing the generators $(x_1,\dots ,x_d)$ of Q) whether $\{(Q^n)^{\lim }\}_n$ form a graded family of ideals, i.e., we do not know whether $(Q^a)^{\lim } (Q^b)^{\lim } \subseteq (Q^{a+b})^{\lim }$ for all a, b.

3.2 Vanishing of $\bar{e}_1(Q)$

In this subsection we prove our main result. Recall that for a finitely generated R-module M, we use the notation $\nu (M)$ to denote its minimal number of generators.

Theorem 3.7

Let $(R,\mathfrak {m})$ be a Noetherian local ring such that $\widehat{R}$ is reduced and $S_2$. If $\overline{e}_1(Q)=0$ for some parameter ideal $Q\subseteq R$, then R is regular and $\nu (\mathfrak {m}/Q)\le 1$.

Proof

We first note that if R is Cohen-Macaulay, then by [19, Corollary 4.9], Q is integrally closed.¹ But then by the main result of [8], R is regular and $\nu (\mathfrak {m}/Q)\le 1$.

We may assume that R is complete. We use induction on $d:=\dim (R)$. If $d\le 2$, then R is Cohen-Macaulay and we are done by the previous paragraph. Now suppose that $d\ge 3$ and we have established the theorem in dimension $<d$. Let $Q=(x_1,\dots ,x_d)$, $R^\prime =$ $R[t_1,\dots ,t_d]_{\mathfrak {m} R[t_1,\dots ,t_d]}$, and $x=t_1x_1+ \cdots +t_dx_d$.

Claim 3.8

We have $R^{\prime \prime }:=\widehat{R^\prime }/x\widehat{R^\prime }$ is reduced, equidimensional, and $S_2$ on the punctured spectrum. Moreover, we have $\overline{e}_1(QR^{\prime \prime })=0$.

Proof

This is essentially contained in [10, Proof of Theorem 1.1] under the assumption that R is (complete and) normal. The key ingredient is [22, Theorem 2.1]. Since [22] does not require the normal assumption, the same proof as in [10] works in our setup. For the ease of the reader (and also because the $S_2$ on the punctured spectrum conclusion is not stated in [10]), we give a complete and self-contained argument here.

First of all, since $R^\prime$ is $S_2$ and $R_0$, we know that $R^\prime /xR^\prime$ is $S_1$ and $R_0$ (see [25, Lemma 10]), so $R^\prime /xR^\prime$ and thus $R^{\prime \prime }$ is reduced (as $R^\prime /xR^\prime$ is excellent). $R^{\prime \prime }$ is clearly equidimensional since $\widehat{R^\prime }$ is so and x is a parameter in $\widehat{R^\prime }$. To see $R^{\prime \prime }$ is $S_2$ on the punctured spectrum, it is enough to show $R^\prime /xR^\prime$ is $S_2$ on the punctured spectrum (as $R^\prime /xR^\prime$ is excellent). Now we use a similar argument as in [25, Lemma 10] (the idea follows from [14]): every non-maximal $P^\prime \in {\text {Spec}}(R^\prime /xR^\prime )$ corresponds to a prime ideal of $R^\prime$ that contracts to a non-maximal $P\in {\text {Spec}}(R)$, thus $(R^\prime /xR^\prime )_{P^\prime }$ is a localization of $R_P[t_1,\dots ,t_d]/(t_1x_1+\cdots + t_dx_d)$, but at least one $x_i$ is invertible in $R_P$ (say $x_1$ is invertible) so the latter is isomorphic to $R_P[t_2,\dots , t_d]$, which is $S_2$ as $R_P$ is $S_2$, thus $R^\prime /xR^\prime$ is $S_2$ on the punctured spectrum as wanted.

It remains to show that $\overline{e}_1(QR^{\prime \prime })=0$. By [21, Corollary 6.8.13], we have a short exact sequence

$$0\rightarrow R^\prime /\overline{Q^n} \xrightarrow {\cdot x} R^\prime /\overline{Q^{n+1}} \rightarrow R^\prime /(x, \overline{Q^{n+1}})\rightarrow 0.$$

Since $\overline{e}_1(Q)=0$, for $n\gg 0$ we have

$$\ell (R^\prime /\overline{Q^{n+1}})=\overline{e}_0(Q)\cdot \left( {\begin{array}{c}n+d\\ d\end{array}}\right) + \overline{e}_2(Q)\cdot \left( {\begin{array}{c}n+d-2\\ d-2\end{array}}\right) + o(n^{d-2}),$$

$$\ell (R^\prime /\overline{Q^n})=\overline{e}_0(Q)\cdot \left( {\begin{array}{c}n+d-1\\ d\end{array}}\right) + \overline{e}_2(Q)\cdot \left( {\begin{array}{c}n+d-3\\ d-2\end{array}}\right) + o(n^{d-2}).$$

It follows that

$$\begin{aligned} \ell (R^\prime /(x, \overline{Q^{n+1}})) = \overline{e}_0(Q)\cdot \left( {\begin{array}{c}n+d-1\\ d-1\end{array}}\right) + o(n^{d-2}). \end{aligned}$$

(3)

We next show that for all $n\gg 0$, $\overline{Q^n}(R^\prime /xR^\prime )= \overline{Q^n(R^\prime /xR^\prime )}$. Once this is proved, we will have $\overline{Q^n}R^{\prime \prime }= \overline{Q^nR^{\prime \prime }}$ for all $n\gg 0$ by [21, Lemma 9.1.1] and thus (3) will tell us that

$$\ell (R^{\prime \prime }/\overline{Q^{n+1}R^{\prime \prime }})= \overline{e}_0(Q)\left( {\begin{array}{c}n+d-1\\ d-1\end{array}}\right) +o(n^{d-2}).$$

Since x is a general element of Q, we have $\overline{e}_0(Q)=e(Q, R^\prime )=e(QR^{\prime \prime }, R^{\prime \prime })=\overline{e}_0(QR^{\prime \prime })$ and so the above equation implies that $\overline{e}_1(QR'')=0$ as wanted.

To show $\overline{Q^n}(R^\prime /xR^\prime )= \overline{Q^n(R^\prime /xR^\prime )}$ for $n\gg 0$, let $\mathcal {R}^\prime$ denote the integral closure of $R^\prime [Qt, t^{-1}]$ inside $R^\prime [t, t^{-1}]$. Concretely, $\mathcal {R}^\prime$ is the $\mathbb {Z}$-graded ring such that $\mathcal {R}_n= \overline{Q^n}t^n$ for $n>0$ and $\mathcal {R}^\prime _n=R^\prime t^{n}$ for $n\le 0$. Consider the map

$$\begin{aligned} \mathcal {R}^\prime /(xt)\mathcal {R}^\prime \rightarrow R^\prime [t, t^{-1}]/(xt)R^\prime [t, t^{-1}]. \end{aligned}$$

If we localize at any prime ideal $\mathcal {P}$ of $R^\prime [Qt, t^{-1}]$ that does not contain $(Qt, t^{-1})$, then we note that $(\mathcal {R}^\prime /(xt)\mathcal {R}^\prime )_{\mathcal {P}}$ is integrally closed inside $(R^\prime [t, t^{-1}]/(xt)R^\prime [t, t^{-1}])_{\mathcal {P}}$. To see this, one can “unlocalize" the ring $R^\prime$, and consider the integral closure of $R[t_1,\dots , t_d][Qt, t^{-1}]$ inside $R[t_1,\dots ,t_d][t, t^{-1}]$, call this ring $\mathcal {R}$. If one localizes the map

$$\mathcal {R}/(xt)\mathcal {R}\rightarrow R[t_1,\dots ,t_d][t, t^{-1}]/(xt)R[t_1,\dots ,t_d][t, t^{-1}]$$

at any prime ideal that does not contain $(Qt, t^{-1})$ (say it does not contain $x_1t$), then the resulting map is a localization of $R[t_2,\dots ,t_d][\overline{Q^n}t^n, t^{-1}][\frac{1}{x_1t}] \rightarrow R[t_2,\dots , t_d][t, t^{-1}][\frac{1}{x_1t}]$, and the former is already integrally closed in the latter.

Since the radical of $(Qt, t^{-1})$ is the unique homogeneous maximal ideal of $R[Qt, t^{-1}]$, it follows that $\mathcal {R}^\prime /(xt)\mathcal {R}^\prime$ and the integral closure of $R^\prime [Qt, t^{-1}]/(xt)R^\prime [Qt, t^{-1}]$ inside$R^\prime [t, t^{-1}]/(xt)R^\prime [t, t^{-1}]$ agree in large degree. But note that for $n>0$,

$$[\mathcal {R}^\prime /(xt)\mathcal {R}^\prime ]_n\cong \frac{\overline{Q^n}}{x\overline{Q^{n-1}}}\cdot t^n\cong \frac{\overline{Q^n}}{x(\overline{Q^n}: x)}\cdot t^n\cong \frac{\overline{Q^n}}{(xR^\prime )\cap \overline{Q^n}}\cdot t^n\cong \overline{Q^n}(R^\prime /xR^\prime )\cdot t^n,$$

where we have used [21, Corollary 6.8.13] again, while the degree n part of the integral closure of $R'[Qt, t^{-1}]/(xt)R^\prime [Qt, t^{-1}]$ inside $R^\prime [t, t^{-1}]/(xt)R^\prime [t, t^{-1}]$ is $\overline{Q^n(R^\prime /xR^\prime )}\cdot t^n$. Thus the fact that they agree in degree $n\gg 0$ is precisely saying that $\overline{Q^n}(R^\prime /xR^\prime )= \overline{Q^n(R^\prime /xR^\prime )}$ for $n\gg 0$.$\square$

Now we come back to the proof of the theorem. Let S be the $S_2$-ification of $R^{\prime \prime }$. We have a short exact sequence

$$0\rightarrow R^{\prime \prime } \rightarrow S \rightarrow S/R^{\prime \prime } \rightarrow 0$$

such that $S/R^{\prime \prime }$ has finite length (since $R^{\prime \prime }$ is $S_2$ on the punctured spectrum). Also note that $(S,\mathfrak {n})$ is (complete) local by [17, Proposition (3.9)] and that S is reduced (since S is a subring of the total quotient ring of $R^{\prime \prime }$). Since $R^{\prime \prime }\rightarrow S$ is an integral extension, we have $\overline{IS}\cap R^{\prime \prime } =\overline{I}$ for every ideal $I\subseteq R^{\prime \prime }$ by [21, Proposition 1.6.1]. It follows that $\ell _{R^{\prime \prime }}(S/\overline{Q^nS})\ge \ell _{R^{\prime \prime }}(R^{\prime \prime }/\overline{Q^nR^{\prime \prime }})$ for all $n\ge 0$. Thus for $n\gg 0$ we have

$$\begin{aligned} & \overline{e}_0(QR^{\prime \prime })\left( {\begin{array}{c}n+d\\ d\end{array}}\right) -\overline{e}_1(QR^{\prime \prime })\left( {\begin{array}{c}n+d-1\\ d-1\end{array}}\right) + o(n^{d-1})\\= & \ell _{R^{\prime \prime }}(R^{\prime \prime }/\overline{Q^{n+1}R^{\prime \prime }}) \\\le & \ell _{R^{\prime \prime }}(S/\overline{Q^{n+1}S}) \\= & [S/\mathfrak {n}: R/\mathfrak {m}]\cdot \ell _S(S/\overline{Q^{n+1}S})\\= & [S/\mathfrak {n}: R/\mathfrak {m}]\cdot \left( \overline{e}_0(QS)\left( {\begin{array}{c}n+d\\ d\end{array}}\right) -\overline{e}_1(QS)\left( {\begin{array}{c}n+d-1\\ d-1\end{array}}\right) +o(n^{d-1})\right) . \end{aligned}$$

Since S is a rank one module over $R^{\prime \prime }$, we also know that

$$\overline{e}_0(QR^{\prime \prime })=e(QR^{\prime \prime }, R^{\prime \prime }) = [S/\mathfrak {n}: R/\mathfrak {m}] \cdot e(QS, S)=[S/\mathfrak {n}: R/\mathfrak {m}] \cdot \overline{e}_0(QS),$$

where the second equality is the projection formula for the Hilbert-Samuel multiplicity (which can be seen by combining [21, Theorems 11.2.4 and 11.2.7]). Putting these together we have

$$ [S/\mathfrak {n}: R/\mathfrak {m}] \cdot \overline{e}_1(QS)\le \overline{e}_1(QR^{\prime \prime })=0.$$

But since $\overline{e}_1(QS)\ge 0$ by [10, Theorem 1.1] (see Theorem 3.1), we must have $\overline{e}_1(QS)=0$. Now $(S,\mathfrak {n})$ is a reduced complete local ring that is $S_2$ and $\dim (S)=d-1$, such that $\overline{e}_1(QS)=0$. By our inductive hypothesis, we know that S is regular. But since $S/R^{\prime \prime }$ has finite length, by the long exact sequence of local cohomology induced by $0\rightarrow R^{\prime \prime } \rightarrow S \rightarrow S/R^{\prime \prime } \rightarrow 0$, we obtain that

$$\begin{aligned}H_{\mathfrak {m}}^i(R^{\prime \prime })=0\, \text {for all}\, i<\dim (R^{\prime \prime })\, \text {and}\, i\ne 1, \text {and} \, H_{\mathfrak {m}}^1(R^{\prime \prime })\cong S/R^{\prime \prime }.\end{aligned}$$

At this point, we consider the long exact sequence of local cohomology induced by $0\rightarrow \widehat{R^\prime } \xrightarrow {\cdot x} \widehat{R^\prime } \rightarrow R^{\prime \prime }\rightarrow 0$, we get

$$0=H_{\mathfrak {m}}^1(R^\prime )\rightarrow H_{\mathfrak {m}}^1(R^{\prime \prime }) \rightarrow H_{\mathfrak {m}}^2(R^\prime )\xrightarrow {\cdot x} H_{\mathfrak {m}}^2(R^\prime ) \rightarrow H_{\mathfrak {m}}^2(R^{\prime \prime }) \rightarrow \cdots .$$

If $d\ge 4$, then $\dim (R^{\prime \prime })\ge 3$ and thus $H_{\mathfrak {m}}^2(R^{\prime \prime })=0$. Since $\widehat{R^\prime }$ is $S_2$, $H_{\mathfrak {m}}^2(R^\prime )$ has finite length and the above exact sequence tells us that $H_{\mathfrak {m}}^2(R^\prime )=0$ by Nakayama’s lemma. But then by the above exact sequence again, we have $H_{\mathfrak {m}}^1(R^{\prime \prime })=0$ and hence $S/R^{\prime \prime }=0$. Thus $R^{\prime \prime }\cong S$ is regular. But then $\widehat{R^\prime }$ and hence R is regular as wanted.

Finally, suppose $d=3$. Let B be a balanced big Cohen-Macaulay algebra of $\widehat{R^\prime }$ that is $\mathfrak {m}$-adic complete, then B/xB is a balanced big Cohen-Macaulay algebra of $R^{\prime \prime }$. It follows that the canonical map $R^{\prime \prime }\rightarrow B/xB$ factors through S.

Claim 3.9

B/xB is a balanced big Cohen-Macaulay algebra over S.

Proof of the Claim

It is clear that some system of parameters of S (namely those coming from $R^{\prime \prime }$) are regular sequences on B/xB. To see that every system of parameters of S is a regular sequence on B/xB, we first note that B/xB is $\mathfrak {m}$-adically complete: since B is $\mathfrak {m}$-adic complete, B/xB is derived $\mathfrak {m}$-complete by [29, Tag 091U], take (y, z) that is a system of parameters of $R^{\prime \prime }$, then as y, z is a regular sequence on B/xB, the derived completion with respect to (y, z), which is B/xB itself, agrees with the usual completion with respect to (y, z) by [29, Tag 0920] (equivalently, with respect to $\mathfrak {m}$ as $\sqrt{(y,z)}=\mathfrak {m}$). Hence by [4, Corollary 8.5.3], every system of parameters of S is a regular sequence on $\widehat{B/xB}\cong B/xB$.$\square$

Note that $\dim (R^{\prime \prime })=\dim (S)=2$ and S is regular, thus the long exact sequence of local cohomology induced by $0\rightarrow R^{\prime \prime } \rightarrow S \rightarrow S/R^{\prime \prime } \rightarrow 0$ implies that $H_{\mathfrak {m}}^2(R^{\prime \prime })\cong H_{\mathfrak {m}}^2(S)$. Hence we have the following commutative algebra:

where the injectivity of $\phi$ follows from the fact that B/xB is a balanced big Cohen-Macaulay algebra over S and thus faithfully flat over S (as S is regular). Chasing this diagram we find that the map $H_{\mathfrak {m}}^2(R^\prime )\xrightarrow {\cdot x} H_{\mathfrak {m}}^2(R^\prime )$ is surjective. But since $\widehat{R^\prime }$ is $S_2$, $H_{\mathfrak {m}}^2(R^\prime )$ has finite length, thus $H_{\mathfrak {m}}^2(R^\prime )=0$ by Nakayama’s lemma. Hence $\widehat{R^\prime }$ is Cohen-Macaulay and thus $R^{\prime \prime }$ is also Cohen-Macaulay. But then $R^{\prime \prime }\cong S$ and so $R^{\prime \prime }$ is regular and thus $\widehat{R^\prime }$ is regular. Thus R is regular as wanted.

Now we have established that R is regular, we can repeat the argument in the first paragraph of the proof to show that $\nu (\mathfrak {m}/Q)\le 1$ (essentially, this follows from the main result of [8]). $\square$

As a consequence, we answer the problem raised in [10, Section 3] for excellent rings.

Corollary 3.10

Let R be an excellent local ring such that $\widehat{R}$ is reduced and equidimensional. Suppose that $I\subseteq R$ is an $\mathfrak {m}$-primary ideal such that $\overline{e}_1(I)=0$. Then $R^{N }$, the normalization of R, is regular and $IR^N$ is normal (i.e., all powers of $IR^N$ are integrally closed in $R^N$).

Proof

Replacing R by $R[t]_{\mathfrak {m} R[t]}$, we may assume that the residue field of R is infinite (we leave it to the readers to check that the hypotheses and conclusions are stable under such a base change). Let S be the $S_2$-ification of R. We will show that the $\mathfrak {m}$-adic completion of $\widehat{S}$ is regular. Since R is excellent, $\widehat{S}$ agrees with the $S_2$-ification of $\widehat{R}$ by [17, Proposition 3.8]. Thus $\widehat{S}$ is semilocal, reduced, and $S_2$. Since $\overline{J\widehat{S}}\cap \widehat{R}=\overline{J}$ for every $\mathfrak {m}$-primary ideal $J\subseteq \widehat{R}$ by [21, Proposition 1.6.1], we have $\ell _{\widehat{R}}(\widehat{R}/\overline{J})\le \ell _{\widehat{R}}(\widehat{S}/\overline{J\widehat{S}})$.

Let $\mathfrak {n}_1,\dots , \mathfrak {n}_s$ be the maximal ideals of $\widehat{S}$ and let $S_i:= (\widehat{S})_{\mathfrak {n}_i}$ (in fact, since $\widehat{S}$ is complete, we have $\widehat{S}\cong \prod _{i=1}^{s}S_i$, and each $S_i$ is complete local, reduced, and $S_2$). Then we have

$$\begin{aligned} & \overline{e}_0(I)\left( {\begin{array}{c}n+d\\ d\end{array}}\right) -\overline{e}_1(I)\left( {\begin{array}{c}n+d-1\\ d-1\end{array}}\right) + o(n^{d-1})\\= & \ell _R(R/\overline{I^{n+1}})=\ell _{\widehat{R}}(\widehat{R}/\overline{I^{n+1}\widehat{R}}) \\\le & \ell _{\widehat{R}}(\widehat{S}/\overline{I^{n+1}\widehat{S}}) \\= & \sum _{i=1}^{s}[S_i/\mathfrak {n}_i: R/\mathfrak {m}]\cdot \ell _{S_i}(S_i/\overline{I^{n+1}S_i})\\= & \sum _{i=1}^{s}[S_i/\mathfrak {n}_i: R/\mathfrak {m}]\cdot \left( \overline{e}_0(IS_i)\left( {\begin{array}{c}n+d\\ d\end{array}}\right) -\overline{e}_1(IS_i)\left( {\begin{array}{c}n+d-1\\ d-1\end{array}}\right) +o(n^{d-1})\right) , \end{aligned}$$

where we have used [21, Lemma 9.1.1] for the equality in the second line. Since $\widehat{S}$ is a rank one module over $\widehat{R}$, we also know that

$$\overline{e}_0(I)=e(I\widehat{R}, \widehat{R}) = \sum _{i=1}^{s}[S_i/\mathfrak {n}_i: R/\mathfrak {m}] \cdot e(IS_i, S_i)=\sum _{i=1}^{s}[S_i/\mathfrak {n}_i: R/\mathfrak {m}] \cdot \overline{e}_0(IS_i),$$

where we have used the projection formula for the Hilbert-Samuel multiplicity (see [21, Theorems 11.2.4 and 11.2.7]). The above inequality implies that

$$\sum _{i=1}^{s}[S_i/\mathfrak {n}_i: R/\mathfrak {m}] \cdot \overline{e}_1(IS_i) \le \overline{e}_1(I)=0.$$

But since $\overline{e}_1(IS_i)\ge 0$ by [10, Theorem 1.1], we must have $\overline{e}_1(IS_i)=0$ for all i. Let Q be a minimal reduction of I (note that Q is a parameter ideal of R, since we have reduced to the case that R has an infinite residue field). It follows that $\overline{e}_1(QS_i)=0$ and thus by Theorem 3.7, $S_i$ is regular and $\nu (\mathfrak {n}_i/Q)\le 1$. But then $QS_i$ is normal in $S_i$. It follows that $\widehat{S}\cong \prod _{i=1}^{s}S_i$ is regular, $Q\widehat{S}$ is normal in $\widehat{S}$ and in particular, $Q\widehat{S}=I\widehat{S}$.

Since $S\rightarrow \widehat{S}\cong \widehat{R}\otimes _RS$ is faithfully flat with geometrically regular fibers (as R is excellent). We have S is regular and $QS=IS$ is normal in S by [21, Theorem 19.2.1]. Finally, since S is regular, S agrees with the normalization $R^\text {N}$ of R.

Remark 3.11

The condition $\widehat{R}$ is $S_2$ cannot be dropped in Theorem 3.7. This was already observed in [10, Section 3]. We give a different example that is a complete local domain. Let $R=k[[x,xy,y^2, y^3]]$, where k is a field. Then the $S_2$-ification of R is $S=k[[x,y]]$ and we have $0\rightarrow R\rightarrow S\rightarrow S/R\cong k\cdot \overline{y}\rightarrow 0$. Let $Q=(x, y^2)\subseteq R$ and we claim that $\overline{e}_1(Q)=0$. To see this, note that $QS=(x,y^2)\subseteq S$ is normal and $\ell (S/Q^{n+1}S)=2\cdot \left( {\begin{array}{c}n+2\\ 2\end{array}}\right)$. It follows from the short exact sequence

$$0\rightarrow R/\overline{Q^{n+1}}\rightarrow S/Q^{n+1}S \rightarrow k\rightarrow 0$$

that $\ell (R/\overline{Q^{n+1}})=2\cdot \left( {\begin{array}{c}n+2\\ 2\end{array}}\right) -1$. In particular, $\overline{e}_1(Q)=0$.

Recall that a Noetherian local ring $(R,\mathfrak {m})$ of prime characteristic $p>0$ is called F-rational if every ideal generated by a system of parameters is tightly closed. It was mentioned in [5] that Huneke asked that when $\widehat{R}$ is reduced and equidimensional of prime characteristic $p>0$, whether $e_1^*(Q)=0$ for some system of parameters $Q\subseteq R$ implies R is F-rational. In general, counter-examples to the question were constructed in [5, Examples 5.4 and 5.5] (in fact, the example in Remark 3.11 is a counter-example that is a complete local domain). However, all these examples do not satisfy Serre’s $S_2$ condition.

Let $(R,\mathfrak {m})$ be a Noetherian local ring and let B be a big Cohen-Macaulay R-algebra. Recall that R is called $\mathrm{{BCM}}_B$-rational if R is Cohen-Macaulay and the natural map $H_{\mathfrak {m}}^d(R)\rightarrow H_{\mathfrak {m}}^d(B)$ is injective, where $d=\dim (R)$. If R is an excellent local ring of prime characteristic $p>0$, then R is F-rational if and only if R is $\mathrm{{BCM}}_B$-rational for all big Cohen-Macaulay algebra B, see [24, Proposition 3.5].

We propose the following conjecture relating the vanishing of $e_1^B(Q)$ and $\mathrm{{BCM}}_B$-rational singularities, which modifies Huneke’s question and makes sense in all characteristics.

Conjecture 3.12

Let $(R,\mathfrak {m})$ be a Noetherian local ring such that $\widehat{R}$ is reduced and $S_2$. Let B be a balanced big Cohen-Macaulay R-algebra that satisfies ($\dagger$). If $e_1^B(Q) = 0$ for some parameter ideal $Q\subseteq R$, then R is $\mathrm{{BCM}}_B$-rational.

In particular, if R is excellent and has characteristic $p>0$ (such that $\widehat{R}$ is reduced and $S_2$), and $e_1^*(Q) = 0$ for some parameter ideal $Q\subseteq R$, then R is F-rational.

We have the following partial result towards Conjecture 3.12, which is an analog of the main result of [27].

Proposition 3.13

Let $(R, \mathfrak {m})$ be a Noetherian local ring such that $\widehat{R}$ is reduced and equidimensional. Let B be a balanced big Cohen-Macaulay R-algebra that satisfies ($\dagger$). If $e_1^B(Q) = e_1(Q)$ for some parameter ideal $Q\subseteq R$, then R is $\mathrm{{BCM}}_B$-rational.

In particular, if R is excellent and has characteristic $p>0$, and $e_1^*(Q) = e_1(Q)$ for some parameter ideal $Q\subseteq R$, then R is F-rational.

Proof

By Theorem 3.1, we know that $e_1^B(Q)=e_1(Q)=0$. By the main result of [6], $e_1(Q)=0$ implies that R is Cohen-Macaulay. By [19, Corollary 4.9], we have $Q^B = Q$. Now we consider the commutative diagram:

where the injectivity of the top row follows from $Q^B=Q$, the injectivity of the left column is because R is Cohen-Macaulay, and the injectivity of the right column is because B is balanced big Cohen-Macaulay. Since R is Cohen-Macaulay, we know that $\textrm{Soc}(R/Q) \cong \textrm{Soc}(H^d_{\mathfrak {m}}(R))$. Chasing the commutative diagram we find that the map $H^d_{\mathfrak {m}}(R) \rightarrow H^d_{\mathfrak {m}}(B)$ is injective. Therefore, R is $\textrm{BCM}_B$-rational.$\square$

Remark 3.14

It is clear from the proof of Proposition 3.13 that Conjecture 3.12 holds when R is Cohen-Macaulay, and this essentially follows from [19, Corollary 4.9].