SFT-stability and Krull dimension in power series rings over an almost pseudo-valuation domain

Abstract

Let $R$ be an APVD with maximal ideal $M$. We show that the power series ring $R[[x_1,\ldots ,x_n]]$ is an SFT-ring if and only if the integral closure of $R$ is an SFT-ring if and only if ($R$ is an SFT-ring and $M$ is a Noether strongly primary ideal of $(M:M)$). We deduce that if $R$ is an $m$-dimensional APVD that is a residually *-domain, then dim $R[[x_1,\ldots ,x_n]]\,=\,nm+1$ or $nm+n$.

1 Introduction

In this paper, all rings are commutative with identity and the dimension of a ring means its Krull dimension. Throughout this paper, if D denotes an integral domain with quotient field K, then $D^{\prime }$ denotes its integral closure, $\text{ Min }(D)$ denotes the set of height-one prime ideals of D and if $X_n=\{x_1,\ldots ,x_n\}$ is a set of indeterminates over K, then we write $D[[X_n]]$ rather than $D[[x_1,\ldots ,x_n]]$. An ideal $I$ is called an SFT-ideal if there exist a natural number $k$ and a finitely generated ideal $J\subseteq I$ such that $a^k\in J$ for each $a\in I$. An SFT-ring is a ring in which every ideal is an SFT-ideal. SFT-rings are similar to Noetherian rings, and they have many nice properties. For properties about SFT-rings, the readers are referred to [1, 2, 4].

In [2], Arnold studied the Krull dimension of power series ring and showed that if a ring $R$ fails to have the SFT-property, then $R[[X]]$ has infinite dimension. Arnold’s result forces us to consider only SFT-rings when we study finite-dimensional power series extensions. Also, it forces us to raise the question: does $R[[X]]$ is an SFT-ring when $R$ is an SFT-ring?. In fact, this is an old question which was raised by R. Gilmer. In Coykendall [8], answered this question in the negative by constructing a one-dimensional SFT-domain R such that $R[[X_n]]$ is not an SFT-ring. At the same time (2002), in Badawi and Houston [5], introduced the concept of an almost pseudo-valuation domain. An integral domain $R$ is said to be an almost pseudo-valuation domain (or, for short, APVD) if $R$ is a quasi-local domain with maximal ideal $M$, and there is a valuation overring in which $M$ is a primary ideal. One very remarkable thing in [8] is that Coykendall’s example is a one-dimensional APVD (see [13, Example 2.4]). So it is natural to study the SFT-instability via power series extension over an APVD regardless of its dimension and ask which things cause the SFT-instability in such extensions, which is the first purpose of this paper.

In this paper, we give a necessary and sufficient condition to win the SFT-stability via power series extension over an APVD. Recall from [7] that a P-primary ideal $Q$ of an integral domain $R$ is called a Noether strongly primary ideal of $R$ if $P^k\subseteq Q$ for some positive integer $k$. We prove that if $R$ is an APVD with maximal ideal $M$, then $R[[x_1]]$ is an SFT-ring if and only if $R[[x_1,x_2]]$ is an SFT-ring if and only if the integral closure of $R$ is an SFT-ring if and only if $R$ is an SFT-ring and $M$ is a Noether strongly primary ideal of $(M:M)$ (Theorem 2.3).

We show first that if $R$ is a one-dimensional APVD with maximal ideal $M$, then $R[[X]]$ is an SFT-ring if and only if $R$ is an accp-ring (i.e., satisfies the ascending chain condition on principal ideals) and $M$ is a Noether strongly primary ideal of $(M:M)$.

We use rings of the form $D+M$ to construct several examples which prove the SFT-instability via power series extension over an APVD regardless of its dimension (Proposition 2.7 and Example 2.8).

Recall from [13], an integral domain $R$ is called a $*$-domain if the set $\text{ Min }(R)$ is nonempty and if for each $P\in \text{ Min }(R),\,R$ satisfies the ascending chain condition on $P$-principal ideals (ideals of the form $aP,\,a\in R$). Also from [13], $R$ is said to be a residually $*$-domain if for each nonmaximal prime ideal $P$ of $R$, the quotient ring $R/P$ is a $*$-domain. In [13], we proved that if $R$ is an APVD with nonzero finite dimension, then $\dim R[[X]]<\infty$ if and only if $\dim R[[X_n]]<\infty$ if and only if $R$ is a residually $*$-domain if and only if the ring $R[[X]]$ is catenary ([13, Theorem 3.5]). And, in this case, we have $\dim R[[X]]=1+\dim R,\,1+n\dim R\le \dim R[[X_n]]\le n+(n+1)\dim R$ and ht$P[[X_n]]=n$ ht$P$ for each nonmaximal prime ideal $P$ of $R$ ([13, Theorem 2.9 and Corollary 2.17]). As an application of Theorem 2.3 (instability theorem on an APVD), we show that $\dim R[[X_n]]=n\dim R+1$ or $n\dim R+n$ and ht$M[[X_n]]=n($ht$M-1)+1$ or $n$ ht$M$ and for each value we give sufficient and necessary conditions on $R$ (Theorem 3.1, Corollary 3.4 and Corollary 3.5). Also we give examples for each value (Example 3.8 and Example 3.9). Finally, we show that the ring $R[[X_n]]$ is not catenary if $n>1$ and $\dim R>1$.

2 When is ${\varvec{R}}{\varvec{[[}}{\varvec{X}}_{\varvec{n}}{\varvec{]]}}$ an SFT-ring?

Throughout this section, $R$ will denote an APVD with quotient filed $K$ and maximal ideal $M$ and $V$ will denote the overring $(M:M)=\{x\in K; xM\subseteq M\}$ of $R$. By [5, Theorem 3.4], $V$ is a valuation domain and $M$ is primary to the maximal ideal $N$ of $V$.

Lemma 2.1

• 1. $V$ and $R$ have the same set of nonmaximal prime ideals. In particular, $\dim V=\dim R$ and every prime ideal of $R$ is a proper ideal of $V$.

• 2. Let $P$ be a nonmaximal prime ideal of $R$. Then $V/P=(\widetilde{M}:\widetilde{M})$ where $\widetilde{M}=M/P$.

Proof

• (1) By [13, Lemma 2.13], every nonmaximal prime ideal $P$ of $R$ is also a nonmaximal (because $P\subset M\subseteq N$) prime ideal of $V$. Since $R\subseteq V$, it is easy to show that every nonmaximal prime ideal $P$ of $V$ (thus $P\subset M$ because $\sqrt{MV}=N$) is also a nonmaximal prime ideal of $R$.

• (2) Since $\widetilde{M}$ is an ideal of $V/P,\,(\widetilde{M}:\widetilde{M})$ is an overring of the valuation domain $V/P$. Thus $(\widetilde{M}:\widetilde{M})=(V/P)_{Q/P}$ for some prime ideal $Q$ of $V$ which contains $P$. Then $(\widetilde{M}:\widetilde{M})=V_Q/PV_Q=V_Q/P$. So $Q/P$ is the maximal ideal of $(\widetilde{M}:\widetilde{M})$. Since $\widetilde{M}$ is a proper ideal of $(\widetilde{M}:\widetilde{M}),\,\widetilde{M}\subseteq Q/P$ and thus $M\subseteq Q$. So $N=\sqrt{MV}\subseteq Q$, and then $Q=N$. Hence $(\widetilde{M}:\widetilde{M})=V/P$. $\square$

The following lemma gives new characterizations of a one-dimensional APVD $R$ so that $R[[X]]$ is an SFT-ring.

Lemma 2.2

If $\dim R=1$, then the following assertions are equivalent:

• 1. $R[[X]]$ is an SFT-ring.

• 2. $R$ is a $*$-domain.

• 3. $R$ is an SFT-ring and $M$ is a Noether strongly primary ideal of $(M:M)$.

• 4. $R$ is an accp-ring and $M$ is a Noether strongly primary ideal of $(M:M)$.

• 5. $(M:M)$ is an SFT-ring

Proof

For $(1)\Leftrightarrow (2)\Leftrightarrow (5)$, see [13, Theorem 2.2 and Theorem 2.9].

$(5)\Rightarrow (4)$ Note that a rank-one valuation domain is SFT if and only if it is Noetherian if and only if it is discrete. If $(a_nR)_{n\ge 0}$ is an ascending chain of nonzero proper principal ideals of $R$, then $(a_nV)_{n\ge 0}$ is an ascending chain of principal ideals of $V$. Thus, there exists a positive integer $k$ such that $a_nV=a_kV$ for every integer $n\ge k$. Let $c\in V$ and $d\in R$ such that $a_n=ca_k$ and $a_k=da_n$. We have $dc=1$, and thus, $d$ is invertible in $V$. So $d\notin M$, and then $d$ is invertible in $R$ because $R$ is quasi-local. Hence $R$ is an accp-ring. Suppose that $N^n\nsubseteq M$ for each integer $n>0$. Then $M\subseteq \bigcap _{n=1}^{\infty }N^n$ which is a prime ideal of $V$ by [9, Theorem 17.1]. Thus $N=\sqrt{M}\subseteq N^2$, a contradiction.

$(4)\Rightarrow (3)$ Note that for a nonzero proper ideal $I$ of a valuation domain, $I$ is an SFT-ideal if and only if $I^2\ne I$. Let $s$ be a positive integer such that $N^s\subseteq M$ and suppose that $M^2=M$. Then $V$ is not SFT and so not Noetherian. So there exists a strictly increasing chain $(a_nV)_{n\ge 0}$ of principal ideals of $V$. Thus $\frac{a_n}{a_{n+1}}\in N$ and so $a_{n}^s\in a_{n+1}^sM$. Then $(a_{n}^sR)_{n\ge 0}$ is a strictly increasing chain of principal ideals of $R$, a contradiction. Necessarily $M^2\ne M$, and hence $R$ is SFT by [1, Proposition 2.2] and [13, Proposition 1.4].

$(3)\Rightarrow (5)$ Let $s$ be a positive integer such that $N^s\subseteq M$. By [13, Proposition 1.4], there exists an $a\in M$ such that $M^6\subseteq aR$. So $N^{6s}\subseteq aV$. Thus, $N$ is an SFT-ideal of $V$. Hence, $V$ is an SFT-ring. $\square$

In the following result, we give a necessary and sufficient condition on an APVD $R$ so that $R[[X_n]]$ is an SFT-ring. Also many characterizations of nonzero finite-dimensional APVDs which are residually $*$-domains are given.

Theorem 2.3

Let R be an APVD with maximal ideal M. The following statements are equivalent:

• 1. $R[[X_n]]$ is an SFT-ring.

• 2. $R[[X]]$ is an SFT-ring.

• 3. The integral closure of $R$ is an SFT-ring.

• 4. $R$ is an SFT-ring and $M$ is a Noether strongly primary ideal of $(M:M)$.

• 5. $(M:M)$ is an SFT-ring. Moreover, if $0<\dim R<\infty$, then each of the previous statements is equivalent to the following:

• 6. $R$ is a residually $*$-domain.

Proof

$(1)\Rightarrow (2)$ Note that the homomorphic image of an SFT-ring is also an SFT-ring [1, Proposition 2.3]. The result follows from the fact that $R[[X]]\cong R[[X_n]]/<X_{n-1}>$.

$(2)\Rightarrow (4)$ Since $R\cong R[[X]]/<X>,\,R$ is an SFT-ring. We can assume that $M\ne 0$ (because in this case $N=0$). Denote $Q$ to be the union of the nonmaximal prime ideals of $R$. Since $R$ is a quasi-local domain with linearly ordered prime ideals, $Q$ is a prime ideal of $R$ by [12, Theorem 9]. Also $Q$ is a prime ideal of $V$ by Lemma 2.1. If $Q=M$, then $N=M$. So, we can assume that $Q\ne M$. Then $Q$ is the prime ideal just below $M$ in $R$, and so $\dim \widetilde{R}=1$ where $\widetilde{R}=R/Q$. Note that the homomorphic image (not be field) of a finite-dimensional APVD with the property residually $*$ is also a finite-dimensional APVD with the property residually $*$ [13, Remark 1.2-(4) and Proposition 1.9]. Since $\widetilde{R}[[X]]\cong R[[X]]/Q[[X]]$ is an SFT-domain, $\widetilde{M}$ is a Noether strongly primary ideal of $(\widetilde{M}:\widetilde{M})$ by Lemma 2.2, where $\widetilde{M}=M/Q$. By Lemma 2.1, $(\widetilde{M}:\widetilde{M})=V/Q$. Thus $(N/Q)^s\subseteq M/Q$ for some integer $s>0$. Hence $N^s\subseteq M+Q=M$.

$(4)\Rightarrow (1)$ Suppose that $R[[X_n]]$ is not an SFT-domain. Then there exists an infinite chain $Q_1\subset Q_2\subset \cdots$ of prime ideals in $R[[X_n]][[x_{n+1}]]$ by [2, Proof of Theorem 1]. Denote $P=(\bigcup _{i=1}^{\infty }Q_i)\cap R$. Since $P$ is an SFT-ideal of $R,\,Q_k\supseteq P$ for some integer $k>0$. Then $Q_k\supseteq P[[X_{n+1}]]$ by [4, Proposition 2.1]. We have an infinite chain of prime ideals $\widetilde{Q_{t}}\subset \widetilde{Q_{t+1}}\subset \cdots$ of $\widetilde{R}[[X_{n+1}]]$ where $\widetilde{R}=R/P,\,\widetilde{Q_i}=Q_i/P[[X_{n+1}]]$. Since $\widetilde{Q_i}\cap \widetilde{R}=(0)$ for all integer $i\ge t,\,\dim \widetilde{R}[[X_{n+1}]]_{\widetilde{R}\backslash (0)}=\infty$ (in particular $P\ne M$ and so $\dim \widetilde{R}\ge 1$), and so $\text{ Min }(\widetilde{R})$ is nonempty [13, Remark 2.1]. Let $s>0$ be an integer such that $N^s\subseteq M$. Thus $(N/P)^s\subseteq \widetilde{M}$, and then $\widetilde{M}$ is a Noether strongly primary ideal of $(\widetilde{M}:\widetilde{M})$ by Lemma 2.1. Hence $\widetilde{R}$ is a $*$-domain by Lemma 2.2 and [13, Lemma 1.11], a contradiction by [13, Theorem 2.2].

$(4)\Rightarrow (5)$ Let $P$ be a nonzero prime ideal of $V$. If $P$ is not maximal in $V$, then $P$ is a prime ideal of $R$ by Lemma 2.1. Thus $P^2\ne P$. So $P$ is an SFT-ideal of $V$. Now, we suppose that $P=N$. Let $s>0$ be an integer such that $N^s\subseteq M$, and let $a\in M\backslash M^2$. Then $M^6\subseteq aR$ by [13, Proposition 1.4]. Thus $N^{6s}\subseteq aV\subseteq N$. Then $N$ is an SFT-ideal of $V$.

$(5)\Rightarrow (4)$ Every nonzero prime ideal of $R$ is a proper ideal of $V$, and so it is not idempotent. Hence $R$ is an SFT-ring. Note that a nonidempotent maximal ideal of a valuation domain is a principal ideal. Let $a\in N$ and $s>0$ an integer such that $N=aV$ and $a^s\in M$ for some integer $s>0$. Hence $N^s=a^sV\subseteq MV=M$.

$(5)\Leftrightarrow (3)$ By [5, Proposition 3.7], $R^{\prime }$ is a PVD with maximal ideal $Q:=\sqrt{MR^{\prime }}$ (Recall that a domain $R$ with quotient field $K$ is called a pseudo-valuation domain (for short, PVD) if each prime ideal $P$ of $R$ is strongly prime, in the sense that $x,y\in K$ and $xy\in P$ implies that $x\in P$ or $y\in P$). Let $W=(Q:Q)$ be the valuation domain associated with $R^{\prime }$. Since $R^{\prime }\subseteq V$ [9, Theorem 19.8], $Q\subseteq N$ and so $V\subseteq W$. Let $P$ be a prime ideal of $V$ such that $W=V_P$. Thus, $Q=PV_P=P$ is a prime ideal of $V$, and then $W=V_Q$. Since $R\subseteq R^{\prime }$ is an integral extension and $Q$ is a maximal ideal of $R^{\prime },\,Q\cap R=M$. Thus $Q=N$ by Lemma 2.1. Hence $V=W$. The result follows from the fact that a PVD is SFT if and only its associated valuation overring is SFT [6, Chapitre 11-Proposition 8.10].

In case $0<\dim R<\infty$, the equivalence $(6)\Leftrightarrow (2)$ follows from [13, Theorem 2.9]. $\square$

The proof of the following result is straightforward because the maximal ideal $M$ of a PVD is always a Noether strongly primary ideal of $(M:M)$ (in fact, M is the maximal ideal of $(M:M)$, see [10, Corollary 6]).

Corollary 2.4

Let $R$ be a PVD with maximal ideal M. The following statements are equivalent:

• 1. $R[[X_n]]$ is an SFT-ring.

• 2. $R[[X]]$ is an SFT-ring.

• 3. The integral closure of $R$ is an SFT-ring.

• 4. $R$ is an SFT-ring.

• 5. $(M:M)$ is an SFT-ring. Moreover, if $0<\dim R<\infty$, then each of the previous statements is equivalent to the following:

• 6. $R$ is a residually $*$-domain.

In the following example, we show that for each $m\in \mathbb N ^*\cup \{\infty \}$, there exists an $m$-dimensional APVD $R$ (not PVD) such that $R$ and $R[[X_n]]$ are SFT-rings.

Example 2.5

Let $F$ be a field. It is known that there exists an SFT-valuation domain $V$ of the form $F+N$ with maximal ideal $N$ and Krull dimension $m$. By [4, Lemma 3.1], there exists $a\in V$ such that $N=aV$. Denote $R=F+N^2$. The ideal $N^2$ is maximal in $R$ because $R/N^2\cong F$. Since the conductor $(R:V)=N^2\ne (0),\,V$ is a valuation overring of $R$. Since $a^2\in N^2$ and $a\notin N^2,\,N^2$ is not strongly prime and then $R$ is not a PVD. Since $N^2$ is a primary ideal of $V,\,N^2$ is a strongly primary ideal of $R$ by [5, Theorem 2.11] and then $R$ is an APVD by [5, Theorem 3.4]. Since $(N^2:N^2)=(a^2V:a^2V)=V,\,\dim R=m$ (by Lemma 2.1) and $R[[X_n]]$ is an SFT-ring by Theorem 2.3.

Example 2.6

[8, Theorem 4.1](Coykendall’s example). We follow [13, Example 2.4]. Take $R=V_1=\mathbb F _{2}+XV$ where $V=\mathbb F _{2}[X^\alpha ]_{M}$, with $\mathbb F _{2}[X^\alpha ]=\{\sum _{i=0}^{n}\epsilon _iX^{\alpha _i}|\epsilon _i\in \mathbb F _{2},\alpha _i\in \mathbb Q ^{+}\}$ and $M$ is the maximal ideal of $\mathbb F _{2}[X^\alpha ]$ generated by the monomials. The ring R is a one-dimensional SFT–APVD which is not a residually $*$-domain. Hence, by Theorem 2.3, $R[[X_n]]$ is not an SFT-domain and so $(MV)^n\nsubseteq XV$ for each integer $n>0$. Thus, $XV$ is not a Noether strongly primary ideal of $V$. This last fact is the cause of the SFT-instability in Coykendall’s example.

Coykendall [8] constructed a one-dimensional APVD $R$ which is SFT but $R[[X_n]]$ is not an SFT-ring. We will show that for each $m\in \mathbb N ^*\cup \{\infty \}$, there exists an $m$-dimensional APVD $R$ which is SFT but $R[[X_n]]$ is not an SFT-ring. The following result will help us to construct the desired examples.

Proposition 2.7

Let $m\in \mathbb N ^*\cup \{\infty \}$ and $V$ be an $m$-dimensional valuation domain of the form $F+N$, where $F$ is a field and $N$ is the maximal ideal of $V$. Let $D$ be a domain with quotient field $F$ and set $R=D+N$. If $V$ is an SFT-ring and $D$ is a one-dimensional SFT–APVD with maximal ideal $M$, then $R$ is an $(m+1)$-dimensional SFT–APVD with maximal ideal $M+N$.

Proof

Since $R/(M+N)\cong D/M,\,M+N$ is a maximal ideal of $R$. The domains $R$ and $V$ have the same quotient field $K$ because the conductor $(R:V)=N$ is nonzero. Let $x\in K$ such that $x^n\notin M+N$ for each integer $n>0$. Since $V$ is a valuation overring of $R$ and $x\notin R$, either $x^{-1}\in N$ or $x$ is a unit of $V$. If $x^{-1}\in N$, then $x^{-1}(M+N)\subseteq N\subseteq (M+N)$. If $x=f+z$ is a unit of $V$, where $f$ is a nonzero element of $F$ and $z\in N$. Thus $f^n\notin M$ for each integer $n>0$ and there exists $z^{\prime }\in N$ such that $x^{-1}=f^{-1}+z^{\prime }$. So $f^{-1}M\subseteq M$ by [5, Lemma 2.3]. Thus $x^{-1}(M+N)\subseteq (M+N)$. Then $R$ is an APVD by [5, Lemma 2.3 and Theorem 3.4]. Easily one can show that $N$ is a strongly prime ideal of $R$ and $\text{ ht }_RN=m$. Since $N$ is a divided prime ideal of $R,\,\dim R=\text{ ht }N+\dim (R/N)=m+\dim D=m+1$. Let $P$ be a nonzero nonmaximal prime ideal of $R$. So $P$ is a prime ideal of $V$. Thus, $P^2\ne P$ and then $P$ is an SFT-ideal of $R$. By [13, Proposition 1.4], $M^2\ne M$ and $M^6\subseteq aD$ for each $a\in M\backslash M^2$. Since $N$ is a divided prime ideal of $R,\,N\subset aR$. So $(M+N)^6\subseteq M^6+N\subseteq aR$. Hence $(M+N)$ is an SFT-ideal of $R$. $\square$

In the following example, we show that for every integer $m\in \mathbb N ^*\cup \{\infty \}$, there exists an $m$-dimensional APVD $R$ (not PVD) such that $R$ is SFT but $R[[X_n]]$ is not SFT.

Example 2.8

Let $W$ be a rank-one nondiscrete valuation domain (so $Q^2=Q$ and $W$ is not SFT) of the form $H+Q$, where $H$ is a field and $Q$ is the maximal ideal of $W$. Let $0\ne a\in Q$ and set $D=H+a^2W$. By an argument similar to the one just given in Example 2.5, one can show that $D$ is an APVD (not PVD) with maximal ideal $a^2W$, that $\dim D=1$ and that $D[[X_n]]$ is not an SFT-ring. The ideal $a^2W$ is the unique nonzero prime ideal of $D$ and $(a^2W)^2=a^4W\subseteq a^2D$. Hence $D$ is an SFT-ring. Let $m\in (\mathbb N \cup \{\infty \})-\{0,1\}$ and $F$ be the quotient field of $D$. Let $V$ be an $(m-1)$-dimensional SFT-valuation domain of the form $F+N$, where $N$ is the maximal ideal of $V$. Set $R=D+N$. By Proposition 2.7, $R$ is an $m$-dimensional SFT–APVD. Since $R[[X_n]]/N[[X_n]]\cong D[[X_n]],\,R[[X_n]]$ is not an SFT-ring.

3 Application: Krull dimension of ${\varvec{R}}{\varvec{[[}}{\varvec{X}}_{\varvec{n}}{\varvec{]]}}$

Let $F\subset L$ be fields, let $K_0$ be the maximal separable extension of $F$ in $L$, and let $p$ denote the characteristic of $F$ if $F$ has nonzero characteristic but set $p:=1$ if $F$ has characteristic zero. We say that $L$ has finite exponent over $K_0$ if $L^{p^n}\subseteq K_0$ for some positive integer $n$.

Park [14] proved that if $R$ is a finite-dimensional SFT-PVD with maximal ideal $M$, then $\dim R[[X_n]]=\left\{ \begin{array}{clrcll} n\dim R+1 &{} \text{ if } L \text{ has } \text{ finite } \text{ exponent } \text{ over } K_0 \text{ and } [K_0:F]<\infty \\ n\dim R+n &{} \text{ otherwise }\\ \end{array}\right.$

where $F=R/M,\,L$ is the residue field of $(M:M)$ and $K_0$ is the maximal separable extension of $F$ in $L$.

We will show that Park’s result also remains true if $R$ is a nonzero finite-dimensional APVD such that it is a residually $*$-domain. In fact, Park proved that if $R$ is a finite-dimensional SFT GPVD with maximal ideal $M$, then $\dim R[[X_n]]=n\dim R+1$ or $n\dim R+n$, where GPVDs is a generalization of the concept of PVD in the nonquasi-local case (Recall that an integral domain $R$ is called a globalized pseudo-valuation domain (or, for short, GPVD) if there exists a Prüfer overring $T$ satisfying the following two conditions: $(1)\,R\subseteq T$ is a unibranched extension; $(2)$ there exists a nonzero radical ideal $I$ common to $T$ and $R$ such that each prime ideal of $T$ (respectively, $R$) that contains $I$ is a maximal ideal of $T$ (respectively, $R$).). It is easy to see that an APVD is a GPVD if and only if it is a PVD. As our goal was proved by Park in the case of a PVD, so we can suppose that our APVD is not a PVD and consequently necessarily is not a GPVD. However, Park’s proof in [14, Theorem 2.4] is also valid for an APVD which is not PVD.

An immediate consequence of Theorem 2.3 and Park’s technique is that we can calculate $\dim R[[X_n]]$ as the following result shows.

Theorem 3.1

Let R be an APVD with maximal ideal $M$, residue field $F$ and nonzero finite Krull dimension such that $R$ is a residually $*$-domain. Let $L$ be the residue field of $(M:M)$ and $K_0$ be the maximal separable extension of $F$ in $L$. Then $\dim R[[X_n]]=\left\{ \begin{array}{clrcll} n\dim R+1 &{} \text{ if } L \text{ has } \text{ finite } \text{ exponent } \text{ over } K_0 \text{ and } [K_0:F]<\infty \\ n\dim R+n &{} \text{ otherwise } \end{array}\right.$

Proof

Assume that $L$ has finite exponent over $K_0$ and $[K_0:F]<\infty$. By [14, Lemma 2.1] (also [3, (3.8) p. 107 and Theorem 3.9]), $L[[X_n]]$ is integral over $F[[X_n]]$. By Theorem 2.3, $N^s\subseteq M$ for some integer $s>0$ and so $N[[X_n]]^s\subseteq M[[X_n]]$. It follows that $V[[X_n]]$ is integral over $R[[X_n]]$ and hence $\dim R[[X_n]]=\dim V[[X_n]]=n\dim R+1$ by [4, Theorem 3.6] and Lemma 2.1.

For the case “$L$ has infinite exponent over $K_0$ or $[K_0:F]=\infty$”, we follow the same way in [14, Theorem 2.4] given by Park (the proof needs [14, Lemma 2.3] which is also true if we replace an SFT-PVD by an APVD which is a residually $*$-domain, the proof is similar to Park’s proof). $\square$

Let $R$ be an APVD with maximal ideal $M$, residue field $F$ and nonzero finite Krull dimension such that $R$ is a residually $*$-domain. Let $L$ be the residue field of $(M:M)$ and $K_0$ be the maximal separable extension of $F$ in $L$. In [13], we proved that $\text{ ht }P[[X_n]]=n \text{ ht }P$ for each prime ideal $P\ne M$ of $R$ [13, Corollary 2.17]. We will show that $\text{ ht }M[[X_n]]=n \text{ ht }M$ or $n(\text{ ht }M-1)+1$, but first we need to recall some background material.

Let $D$ and $J$ be two integral domains such that $D\subseteq J$. Let $Z=\{z_i\}_{i=1}^{\infty }$ be a countable set of indeterminates over $J$. Let $\mathcal F =\{f_i\}_{i=2}^{\infty }$ be a subset of $z_1J[[z_1]]$. By Arnold ([4, p. 899]), $\mathcal F$ is said to be a suitable subset of $z_1J[[[z_1]]$ if $\{i; f_i\notin z_{1}^{k}J[[z_1]]\}$ is finite for each positive integer $k$. If $\mathcal F$ is a suitable subset of $z_1J[[z_1]]$, then we can define a unique $D[[z_1]]$-homomorphism $\varPhi _\mathcal{F }: D[[Z]]\longrightarrow J[[z_1]]$ by $\varPhi (z_i)=f_i,\,i\ge 2$ (where $D[[Z]]$ denotes (in the notation of [9, p. 10]) the full power series ring $D[[\{z_i\}_{i=1}^{\infty }]]_3$). By Arnold, $J$ is called a special algebraic extension of $D$ if for each suitable subset $\mathcal E =\{g_i\}_{i=2}^{\infty }$ of $z_1J[[z_1]]$, the $D[[z_1]]$-homomorphism $\varPhi _\mathcal{E }$ is not an isomorphism.

In other words, J is a special algebraic extension of D if and only if for each integral domain T such that $D[[X]]\subseteq T\subseteq J[[X]],\,T\ncong D[[X]][[\{Y_i\}_{i=1}^{\infty }]]$ via a $D[[X]]$-isomorphism.

Lemma 3.2

Suppose that $\dim R=1$. If $L$ has infinite exponent over $K_0$ or $[K_0:F]=\infty$, then $\text{ ht } M[[X_n]]=n$.

Proof

$R\subseteq V$ is a special algebraic extension because the conductor $(R:V)=M$ is nonzero. By [14, Lemma 2.1], $F\subseteq L$ is not a special algebraic extension and hence $\text{ ht }M[[X_n]]\ge n$ by [4, Proposition 2.3]. Since $\text{ ht }M[[X_n]]+n\le \text{ ht }(M+<X_n>),\,\text{ ht }M[[X_n]]\le n$ by Theorem 3.1. $\square$

Remark 3.3

In fact, another proof for the last lemma is similar to the one given by Park in [14, Theorem 2.4], but we choose to give a new proof.

Corollary 3.4

If $L$ has infinite exponent over $K_0$ or $[K_0:F]=\infty$, then ht$M[[X_n]]=n\,\text{ ht }M$.

Proof

Since $\text{ ht }M[[X_n]]+n\le \text{ ht }(M+(X_n)),\,\text{ ht }M[[X_n]]\le n\,\text{ ht }M$ by Theorem 3.1. Let $P$ be the prime ideal just below $M$ in $R$. Denote $\widetilde{R}=R/P$ and $\widetilde{M}=M/P$. Note that $\widetilde{R}$ is a one-dimensional APVD with maximal ideal $\widetilde{M}$ and is a residually $*$-domain [13, Remark 1.2-(4) and Proposition 1.10]. Also note that the valuation domain associated with $\widetilde{R}$ is $V/P$ by Lemma 2.1. Since residue field $(\widetilde{R})=F$ and residue field $(V/P)=L,\,\text{ ht }\widetilde{M}[[X_n]]=n$ by Lemma 3.2. Hence $\text{ ht }M[[X_n]]\ge \,\text{ ht }\widetilde{M}[[X_n]]+\text{ ht }Q[[X_n]]=n+n\,\text{ ht }Q=n\,\text{ ht }M$ by [13, Corollary 2.17]. $\square$

Corollary 3.5

If $L$ has finite exponent over $K_0$ and $[K_0:F]<\infty$, then $\text{ ht }M[[X_n]]=n(\text{ ht }M-1)+1$.

Proof

It is easy to see that ht$M[[X_n]]\le \,\text{ ht }(M+(X_n))-n=\dim R[[X_n]]-n$. Thus $\text{ ht }M[[X_n]]\le n(\text{ ht }M-1)+1$ by Theorem 3.1. Since $N[[X_n]]\cap R[[X_n]]=M[[X_n]]$ and $R[[X_n]]\subseteq V[[X_n]]$ is an integral extension (so it satisfies incomparability), $\text{ ht }M[[X_n]]\ge \,\text{ ht }N[[X_n]]=n(\text{ ht }M-1)+1$ by [11, Theorem 13]. $\square$

It is well known that if $Q$ is a prime ideal of $D[X_n]$ with $Q\cap D=P$, then $\text{ ht }Q=\,\text{ ht }P[X_n]+\text{ ht }(Q/P[X_n])$ [9, Theorem 30.18]. Kang and Park [11] asked if the power series analogue of this last result is true. They were especially interested in Prüfer domains and they answered their question in some cases of Prüfer domains [11, Corollary 15]. The following result gives an answer for their question in some cases of an APVD.

Corollary 3.6

If $L$ has infinite exponent over $K_0$ or $[K_0:F]=\infty$, then $\text{ ht }Q=\text{ ht }P[[X_n]]+\text{ ht }(Q/P[[X_n]])$ for all prime ideal $Q$ of $R[[X_n]]$ with $Q\cap R=P$.

Proof

“$\ge$” is obvious. “$\le$”: Use induction on $m:=\mathrm{ht}P$. If $m=0$, then it is obvious. Suppose $m>0$ and the statement holds when ht$P<m$. Let $Q$ be a prime ideal of $R[[X_n]]$ such that ht$P=m$ where $P=Q\cap R$. Let $(0)=Q_0\subset Q_1\subset \ldots \subset Q_s=Q$ be a chain of prime ideals of $R[[X_n]]$ arriving at $Q$. Let $k$ be the smallest integer $i\in \{1,\ldots ,s\}$ such that $Q_i\cap R\ne (0)$. Thus $Q_0\cap R=\ldots =Q_{k-1}\cap R=(0)$. So $k-1\le \dim R[[X_n]]_{R\backslash (0)}=n$ by [13, Theorem 2.2]. Denote $P_0=Q_k\cap R$. Hence $P_0[[X_n]]\subseteq Q_k$.

• Case 1 $P_0=M$ (and so $P=M$). If $M[[X_n]]=Q_k$, then $s=k+(s-k)\le$ ht$M[[X_n]]+$ht$(Q/M[[X_n]])$. If $M[[X_n]]\subset Q_k$, then $s=(k-1)+(s-k+1)\le n+$ht$(Q/M[[X_n]])\le$ ht$M[[X_n]]+$ht$(Q/M[[X_n]])$ by Corollary 3.4.

• Case 2 $P_0\ne M$. Denote $\widetilde{R}=R/P_0$ and $\widetilde{P}=P/P_0$. So, by considering $R[[X_n]]/P_0[[X_n]]\cong \widetilde{R}[[X_n]]$, since $ht\widetilde{P}<m$ and $(Q/P_0[[X_n]])\cap \widetilde{R}=\widetilde{P}$, then ht$(Q/P_0[[X_n]])=$ ht$\widetilde{P}[[X_n]]+$ ht$(Q/P[[X_n]])$ by induction hypothesis.

• Case 2.1 $Q_k=P_0[[X_n]]$. Then $s=k+(s-k)\le$ ht$P_0[[X_n]]+$ht$(Q/P_0[[X_n]]=$ ht$P_0[[X_n]]+$ht$\widetilde{P}[[X_n]]+$ht$(Q/P[[X_n]])\le$ ht$P[[X_n]]+$ht$(Q/P[[X_n]])$.

• Case 2.2 $P_0[[X_n]]\subset Q_k$. Then $s=(k-1)+(s-k+1)\le n+$ht$(Q/P_0[[X_n]])=n+$ht$\widetilde{P}[[X_n]]+$ht$(Q/P[[X_n]])\le$ ht$P_0[[X_n]]+$ht$\widetilde{P}[[X_n]]+$ht$(Q/P[[X_n]])$ because $n\le n$ ht$P_0=$ht$P_0[[X_n]]$. Thus $s\le$ ht$P[[X_n]]+$ht$(Q/P[[X_n]])$. $\square$

Of course, the last result forces us to ask if the ring $R[[X_n]]$ is catenarian (Recall that a ring $D$ is said to be catenary if for each pair $P\subset Q$ of prime ideals of $D$, all saturated chains of prime ideals of $D$ between $P$ and $Q$ have a common finite length). In [13], we proved that if $D$ is an APVD with nonzero finite dimension, then $D[[X]]$ is catenarian if and only if $D$ is a residually $*$-domain. The question is still open for $R[[X_n]]$ with $n>1$.

Corollary 3.7

If $n>1$ and $\dim R>1$, then $R[[X_n]]$ is not catenarian.

Proof

Let $n>1$ be an integer and suppose that $R[[X_n]]$ is catenarian. Then $\mathrm{ht}(x_n)+\mathrm{ht}(M+(X_{n}))/(x_n)=\mathrm{ht}(M+(X_n))$. Thus $1+\mathrm{ht}(M+(X_{n-1}))=\mathrm{ht}(M+(X_n))$. If $L$ has finite exponent over $K_0$ and $[K_0:F]<\infty$, then $1+(n-1)\dim R+1=n\dim R+1$ by Theorem 3.1. Thus $\dim R=1$, a contradiction. If $L$ has infinite exponent over $K_0$ or $[K_0:F]=\infty$, then $1+(n-1)(\dim R+1)=n(\dim R+1)$ again by Theorem 3.1. Thus $\dim R=0$, a contradiction. $\square$

In the following example, we show that for every integer $m\ge 1$, there exists an $m$-dimensional APVD $R$ such that $R$ is a residually $*$-domain and ($L$ has infinite exponent over $K_0$ or $[K_0:F]=\infty$).

Example 3.8

Let $F\subset L$ be an extension of fields such that $L$ is not algebraic over $F$. Let $K_0$ be the maximal separable extension of $F$ in $L$. It is known that there exists an SFT-valuation domain $V$ of the form $L+N$ with maximal ideal $N$ and Krull dimension $m$. By [4, Lemma 3.1], there exists $a\in V$ such that $N=aV$. Denote $R=F+N^2$. By an argument similar to the one just given in Example 2.5, we show that $R$ is an APVD with maximal ideal $N^2,\,\dim R=m$ and $R$ is a residually $*$-domain. It is clear that $F$ and $L$ are the residue fields, respectively, of $R$ and $(N^2:N^2)=V$. Since $F\subset L$ is not an algebraic extension, $F\subset L$ is not a special algebraic extension by [3, Theorem 2.1]. Then $L$ has infinite exponent over $K_0$ or $[K_0:F]=\infty$ by [14, Lemma 2.1].

In the following example, we show that for every integer $m\ge 1$, there exists an $m$-dimensional APVD $R$ such that $R$ is a residually $*$-domain and ($L$ has finite exponent over $K_0$ and $[K_0:F]<\infty$).

Example 3.9

Let $F, m, V, N$ and $R$ be as in Example 2.5. The result follows from the fact that $R/N^2\cong F\cong V/N$.

SFT-stability and Krull dimension questions in power series rings over an APVD are solved, but catenarity question is not yet. It is still open if $R[[X_n]]$ is catenarian where $n>1$ and $R$ is a one-dimensional APVD.