Divisorial Ideals in the Power Series Ring $A+XB[\![X]\!]$

Abstract

Let $A\subseteq B$ be an extension of integral domains, $B[\![X]\!]$ be the power series ring over B,  and $R=A + XB[\![X]\!]$ be a subring of $B[\![X]\!].$ In this paper, we give a complete description of v-invertible v-ideals (with nonzero trace in A) of R. We show that if B is a completely integrally closed domain and I is a fractional divisorial v-invertible ideal of R with nonzero trace over A,  then $I = u(J_1 + XJ_2[\![X]\!])$ for some $u\in qf(R),$ $J_2$ an integral divisorial v-invertible ideal of B and $J_1\subseteq J_2$ a nonzero ideal of A.

1 Introduction

Let A be an integral domain with quotient field K. Let $\mathcal {F}(A)$ be the set of nonzero fractional ideals of D. For an $I\in \mathcal {F}(A),$ set $I^{-1}=\{x\in K \; \mid \; xI\subseteq A\}.$ The mapping on $\mathcal {F}(A)$ defined by $I\mapsto I_v=(I^{-1})^{-1}$ is called the v-operation on A. A nonzero fractional ideal I is said to be a v-ideal or divisorial if $I=I_v,$ and I is said to be v-invertible if $(II^{-1})_v=A.$ For properties of the v-operation the reader is referred to [8, Section 34]. However, we will be mostly interested in the t-operation defined on $\mathcal {F}(A)$ by $I\mapsto I_t=\cup \{J_v,\; J$ is a nonzero finitely generated fractional subideal of $I\}.$ (For properties of the t-operation the reader may consult [1]). A fractional ideal I is called a t-ideal if $I=I_t.$ A t-ideal (respectively, v-ideal) I has t- (respectively, v-) finite type if $I=J_t$ (respectively, $I=J_v)$ for some finitely generated fractional ideal J of A. The set of v-ideals may be a proper subset of the set of t-ideals. A fractional ideal I is said to be t-invertible if $(II^{-1})_t=A.$ The set T(A) of t-invertible fractional t-ideals of A is a group under the t-multiplication $I\star J:=(IJ)_t,$ and the set P(A) of nonzero principal fractional ideals of A is a subgroup of T(A). Following [7], we define the t-class group of A, denoted Cl$_{t}(A),$ to be the t-group of t-invertible fractional t-ideals of A under t-multiplication modulo its subgroup of principal fractional ideals that is, Cl$_{t}(A)= T(A)/ P(A).$The t-class group of an integral domain was studied by many authors ([7,8,9,10,11]).

Let $A\subseteq B$ be an extension of integral domains. In [2], the authors study when the natural mapping $\varphi : Cl_{t}(A) \rightarrow Cl_{t}(A+XB[X]);$ $[I] \mapsto [I(A+XB[X])]$ is an isomorphism. They showed that if B is an integrally closed domain and qf$(A)\subseteq B,$ then $Cl_{t}(A) \cong Cl_{t}(A+XB[X])$ ([2, Theorem 4.7]). Also, the authors study the form of v-invertible (respectively, t-invertible) ideals of the polynomial ring of the form $A+XB[X].$ Let $A\subseteq B$ be an extension of integral domains such that B is an integrally closed domain and $A+XB[X].$ The authors proved that if I is a fractional divisorial v-invertible ideal of R,  then $I = u(J_1 + XJ_2[X])$ for some $u\in qf(A+XB[X]),$ $J_2$ an integral divisorial v-invertible ideal of B and $J_1\subseteq J_2$ a nonzero ideal of A ([2, Theorem 2.3]). In this paper, we extend these results to the ring of formal power series of the form $A+XB[\![X]\!].$ In particular, we give a relationship between v-invertible v-ideals of an integral domain and those of its power series ring of the form $A + XB[\![X]\!].$

Let $A\subseteq B$ be an extension of integral domains, $B[\![X]\!]$ be the power series ring over B,  and $R=A + XB[\![X]\!].$ In the first part of this paper, we study when the natural mapping

$$\begin{array}{ccccc} \varphi &{} : &{} Cl_{t}(A) &{} \rightarrow &{} Cl_{t}(R) \\ &{} &{} [I] &{} \mapsto &{} [(IR)_t] \\ \end{array}$$

is an injective homomorphism. We show that if B is a flat A-module, then the mapping $\varphi$ is an injective homomorphism. Also, we prove that the mapping $\varphi$ is not surjective in general (Remark 2.7). In the second part of this paper, we give a complete description of v-invertible v-ideals (with nonzero trace in A) of $A+XB[\![X]\!].$ First, we show that if $A\subseteq B$ is an extension of integral domains such that B is completely integrally closed, then for each divisorial ideal I of $R = A+XB[\![X]\!]$ such that $I\cap A\ne (0),$ there exist a divisorial ideal J of B and a nonzero ideal $H\subseteq J$ of A such that $I= H+XJ[\![X]\!]$ (Proposition 3.2). Based on the above result, we prove that if I is a fractional divisorial v-invertible ideal of R such that $I\cap A \ne (0),$ then $I = u(J_1 + XJ_2[\![X]\!])$ for some $u\in qf(R),$ $J_2$ an integral divisorial v-invertible ideal of B and $J_1\subseteq J_2$ a nonzero ideal of A,  where B satisfies $\circledast .$

2 The t-Class Group of A+XB[[X]]

Let A be an integral domain. A fractional ideal I of A is said to be v-invertible (respectively, t-invertible, invertible) if $(II^{-1})_v=A$ (respectively, $(II^{-1})_t=A$, $II^{-1}=A)$. Following [7], we define the t-class group of A, denoted by $Cl_t(A),$ to be the group T(A) of t-invertible fractional t-ideals of A under t-multiplication (i.e., $I\star J:=(IJ)_t$) modulo its subgroup P(A) of principal fractional ideals, that is, $Cl_t(A)= T(A)/ P(A).$ When A is a Krull domain, then the t-class group and the divisor class group coincide. We denote by [I] the equivalence class of a t-invertible t-ideal I of A. Let $A\subseteq B$ be an extension of integral domains and $R = A+XB[\![X]\!].$ In this section we show that the natural mapping $\varphi : Cl_{t}(A) \rightarrow Cl_{t}(R);$ $[I] \mapsto [(IR)_t]$ is an injective homomorphism. To prove it, we need the following lemmas.

Lemma 2.1

Let $A\subseteq B$ be an extension of integral domains and $R = A+XB[\![X]\!].$ Let $F_1$ (respectively, $F_2$) be a fractional ideal of A (respectively, B) such that $F_1\subseteq F_2.$ Then $F_1 + XF_2[\![X]\!]$ is a fractional ideal of R and

$$(F_1 + XF_2[\![X]\!])^{-1} = F_1^{-1}\cap F_2^{-1} +XF_2^{-1}[\![X]\!].$$

Proof

Let $I = F_1 + XF_2[\![X]\!].$ Since $F_1 \subseteq I,$ we obtain $I^{-1} =R:I \subseteq R:F_1,$ where $R:F_1=$ $\{g\in qf(R), gF_1 \subseteq R\}.$ This implies that $I^{-1}\subseteq$ $K[\![X]\!],$ where $K=qf(B).$ Indeed, let $u\in I^{-1}$ and $\alpha \in F_1\backslash (0).$ Since $uF_1 \subseteq R,$ $\displaystyle {u =\frac{\alpha u}{\alpha } \in \frac{1}{\alpha }R}$ $\subseteq K[\![X]\!].$

Now we show that $u \in I^{-1}$ if and only if $u(0)F_1 \subseteq A$ and $uF_2[\![X]\!]\subseteq B[\![X]\!].$

$(\Rightarrow )$ Let $u \in I^{-1}.$ Since $uI \subseteq R,$ we get $uF_1 + uXF_2[\![X]\!]$ $\subseteq$ $A + XB[\![X]\!].$ Chose $X = 0,$ we obtain $u(0)F_1 \subseteq A.$ Moreover, $uF_2[\![X]\!]\subseteq B[\![X]\!].$

$(\Leftarrow )$ Assume that $u(0)F_1 \subseteq A$ and $uF_2[\![X]\!]\subseteq B[\![X]\!].$ We prove that $u\in I^{-1}.$

As $u\in K[\![X]\!],$ we can write $u =\displaystyle \sum _{i=0}^\infty a_iX^i,$ where $a_i \in K.$ It is clearly that

$$uI = uF_1 + XuF_2[\![X]\!]\subseteq u(0)F_1 + \displaystyle (\sum _{i=1}^\infty a_iX^i)F_1 + XuF_2[\![X]\!].$$

Moreover, $\displaystyle (\sum _{i=1}^\infty a_iX^i)F_1$ $=$ $(u - u(0))F_1$ $\subseteq$ $uF_1 + u(0)F_1.$ Then $uF_1 + u(0)F_1$ $\subseteq B[\![X]\!],$ because $uF_1$ $\subseteq$ $uF_2 \subseteq B[\![X]\!]$ and $u(0)F_1$ $\subseteq$ $A \subseteq B[\![X]\!].$ This implies that $(\displaystyle \sum _{i=1}^\infty a_iX^i)F_1$ $\subseteq$ $B[\![X]\!].$ Now let P be an element of $(\displaystyle \sum _{i=1}^\infty a_iX^i)F_1.$ Then there exists an element $\alpha$ of $F_1$ such that $P = X(\displaystyle \sum _{i=1}^\infty a_iX^{i-1})\alpha .$ Since $(\displaystyle \sum _{i=1}^\infty a_iX^i)F_1$ $\subseteq$ $B[\![X]\!],$ $(\displaystyle \sum _{i=1}^\infty a_iX^{i-1})\alpha$ $\in$ $B[\![X]\!].$ Thus $P\in XB[\![X]\!],$ and so $(\displaystyle \sum _{i=1}^\infty a_iX^i)F_1$ $\subseteq$ $XB[\![X]\!].$ This shows that

$$\begin{array}{ccl} uI &{}\subseteq &{} u(0)F_1 + XuF_2[\![X]\!]+ (\displaystyle \sum _{i=1}^\infty a_iX^i)F_1 \\ &{} \subseteq &{} A + XB[\![X]\!]\\ &{} = &{} R. \\ \end{array}$$

Hence $u\in I^{-1}.$

Now $u \in I^{-1}$ if and only if $u(0)F_1 \subseteq A$ and $uF_2[\![X]\!]$ $\subseteq B[\![X]\!]$ which equivalent to $u(0)\in F_1^{-1}$ and $u \in (F_2[\![X]\!])^{-1}.$ But $(F_2[\![X]\!])^{-1}$ $=$ $F_2^{-1}[\![X]\!].$ Hence $u\in I^{-1}$ if and only if $u \in F_1^{-1}\cap F_2^{-1} +XF_2^{-1}[\![X]\!].$ $\square$

Example 2.2

Let A $= \mathbb {Z},$ B $= \mathbb {Z}[i]$ and $R=\mathbb {Z}+X\mathbb {Z}[i][\![X]\!].$ Let $I=2\mathbb {Z}+(1+i)X\mathbb {Z}[i][\![X]\!].$ We show that I is a divisorial ideal of R,  i.e., $I_{\upsilon }=I.$

It is clear that I is an ideal of R. Now by Lemma 2.1,

$$\begin{array}{ccl} I^{-1} &{}=&{} \frac{1}{2}\mathbb {Z} \bigcap ((1+i)\mathbb {Z}[i])^{-1} + X((1+i)\mathbb {Z}[i])^{-1}[\![X]\!]\\ {} &{}=&{} \frac{1}{2} \mathbb {Z} \bigcap (1+i)^{-1}\mathbb {Z}[i] + (1+i)^{-1}X\mathbb {Z}[i][\![X]\!].\end{array}.$$

But if $x\in$ $\frac{1}{2}\mathbb {Z}\bigcap \frac{1}{1+i}\mathbb {Z}[i],$ then x $= \frac{1}{2}r$ $= \frac{1}{1+i}u,$ with $r\in$ $\mathbb {Z}$ and $u\in$ $\mathbb {Z}[i].$ This implies that $(1+i)r$ $= 2u.$ Write u $= \alpha + i\beta .$ Then $2\alpha$ $= r$ and $2\beta$ $=r$ thus 2 divided r,  and so x $= \alpha$ $\in$ $\mathbb {Z}.$ Hence $I^{-1}$ $=\mathbb {Z}$ $+ X\frac{1-i}{2}\mathbb {Z}[i][[\![X]\!].$ Again by Lemma 2.1,

$$\begin{array}{ccl} I_{\upsilon } &{}=&{} (I^{-1})^{-1} \\ &{}=&{} \mathbb {Z}\bigcap ((1+i)^{-1}\mathbb {Z}[i])^{-1} + X((1+i)^{-1}\mathbb {Z}[i])^{-1}[\![X]\!]\\ &{}=&{} \mathbb {Z}\bigcap (1+i)\mathbb {Z}[i]+ (1+i)X\mathbb {Z}[i][\![X]\!]\\ &{}=&{} 2\mathbb {Z} + (1+i)X\mathbb {Z}[i][\![X]\!]\\ &{}=&{} I.\end{array}.$$

This shows that I is a divisorial ideal of R.

Let $A\subseteq B$ be an extension of integral domains. Following [3], we say that B is t-linked over A,  if for each finitely generated fractional ideal I of A with $I^{-1}=A,$ we have $(IB)^{-1}=B.$

Lemma 2.3

Let $A\subseteq B$ be an extension of integral domains and $R = A+XB[\![X]\!].$ If B is t-linked over A,  then the extension $A\subseteq R$ is t-linked.

Proof

Let I be a finitely generated fractional ideal of A such that $I^{-1}=A.$ Since $IR\subseteq$ $I+(IB)[\![X]\!],$ then by Lemma 2.1,

$$I^{-1}\cap (IB)^{-1}+X(IB)^{-1}[\![X]\!]=(I+(IB)[\![X]\!])^{-1}\subseteq (IR)^{-1}.$$

But B is t-linked over A,  then $R=A+XB[\![X]\!]$ $=I^{-1}\cap (IB)^{-1}+X(IB)^{-1}[\![X]\!]$ $\subseteq$ $(IR)^{-1},$ and hence R $\subseteq$ $(IR)^{-1}.$

Now we will show that $(IR)^{-1}\subseteq R.$ Let u be an element of $(IR)^{-1}.$ It is easy to prove that $u\in$ $L+XK[\![X]\!],$ where $L=qf(A)$ and $K=qf(B).$ Put $u=\displaystyle \sum _{i=0}^\infty a_iX^i\in L+XK[\![X]\!],$ and let $\alpha \in I.$ Since $\alpha u= \sum _{i=0}^\infty (\alpha a_i)X^i\in R,$ $\alpha a_0\in A,$ and hence $a_0\in I^{-1}.$ Moreover, if $r\in IB,$ then $urX\in$ $u(IR)\subseteq R.$ This implies that for each $i\ge 1,$ $ra_i\in B.$ Therefore for each $i\ge 1,$ $a_i\in (IB)^{-1}.$ Hence $u\in I^{-1}+ X(IB)^{-1}[\![X]\!]=A+ XB[\![X]\!]=R$ since B is t-linked over A. Hence $(IR)^{-1}= R.$ $\square$

Proposition 2.4

Let $A\subseteq B$ be an extension of integral domains such that B is t-linked over A. Then the mapping

$$\begin{array}{ccccc} \varphi &{} : &{} Cl_{t}(A) &{} \rightarrow &{} Cl_{t}(R) \\ &{} &{} [I] &{} \mapsto &{} [(IR)_t] \\ \end{array}$$

is an homomorphism.

Proof

Follows from Lemma 2.3 and [3, Theorem 2.2]. $\square$

Let $A\subseteq B$ be an extension of integral domains and I a finitely generated ideal of A. It well known that $I.A[\![X]\!]=(IA)[\![X]\!]=I[\![X]\!].$ Using the same proof we can prove that $I.B[\![X]\!]=(IB)[\![X]\!].$

Lemma 2.5

Let $A\subseteq B$ be an extension of integral domains such that B is a flat A-module, I an ideal of A and $R=A+XB[\![X]\!].$ We assume that I and $I^{-1}$ are v-ideals of finite type. Then $(IR)_v=I+X(IB)[\![X]\!].$

Proof

Since I and $I^{-1}$ are v-ideals of finite type, $I=J_v$ and $I^{-1}=L_v$ for some finitely generated ideals J and L of A. Since $JR=J+X(JB)[\![X]\!],$ by Lemma 2.1,

$$\begin{array}{ccl} (JR)^{-1} &{}=&{} (J+X(JB)[\![X]\!])^{-1} \\ &{}=&{} J^{-1}\cap (JB)^{-1}+X(JB)^{-1}[\![X]\!]\\ &{}=&{} J^{-1}\cap J^{-1}B +X(J^{-1}B)[\![X]\!]\\ &{}=&{} J^{-1},\end{array}$$

where the third equality follows from the fact that B is a flat A-module.

Again apply Lemma 2.1, $(JR)_v=$ $J_v\cap (J^{-1}B)^{-1}$ $+X(J^{-1}B)^{-1}[\![X]\!].$

Since $L_v=I^{-1}=J^{-1},$

$$(J^{-1}B)^{-1}=(L_vB)^{-1}=(LB)^{-1}=L^{-1}B=J_vB,$$

where the second equality follow from the proof of [5, Proposition 2.2]. So

$$\begin{array}{ccl} (JR)_v &{}=&{} J_v\cap (J_vB)+X(J_vB)[\![X]\!]\\ &{}=&{} J_v+X(J_vB)[\![X]\!]\\ &{}=&{}I+X(IB)[\![X]\!].\end{array}$$

This implies that $I+X(IB)[\![X]\!]\subseteq (IR)_v.$ Now, using Lemma 2.1, we can prove that

$$(I+X(IB)[\![X]\!])_v=I+X(IB)[\![X]\!].$$

This shows that $(IR)_v\subseteq$ $I+X(IB)[\![X]\!],$ and hence $(IR)_v=$ $I+X(IB)[\![X]\!].$ $\square$

We are now ready to prove the main result of this section.

Theorem 2.6

Let $A\subseteq B$ be an extension of integral domains such that B is a flat A-module. Then the mapping

$$\begin{array}{ccccc} \varphi &{} : &{} Cl_{t}(A) &{} \rightarrow &{} Cl_{t}(R) \\ &{} &{} [I] &{} \mapsto &{} [(IR)_t] \\ \end{array}$$

is an injective homomorphism.

Proof

Since B is a flat A-module, B is t-linked over A. So by Proposition 2.4, the mapping $\varphi$ is an homomorphism. We show that $\varphi$ is injective. Let I be a t-invertible t-ideal of A such that $(IR)_t$ is a principal ideal of R. We will prove that I is principal. Since $(IR)_t$ is principal, $(IR)_t=fR$ for some $f\in (IR)_t.$

Case 1: I is an integral ideal of A.

As $(IR)_t=fR,$ then $(IR)_v=fR.$ By Lemma 2.5, $(IR)_v=$ $I+X(IB)[\![X]\!];$ so $I=f(0)A$ is a principal ideal of A.

Case 2: I is a fractional ideal of A.

Let $d\in A\backslash (0)$ such that $dI\subseteq A.$ Put $I^\prime =dI.$ Then $I^\prime$ is an integral t-invertible t-ideal of A. Moreover, $(I^\prime R)_t=dfR$ is a principal ideal of R. By case 1, $I^\prime$ is a principal ideal of A. So I is a principal ideal of A,  and hence $\varphi$ is injective. $\square$

Remark 2.7

Let $A\subseteq B$ be an extension of integral domains and let $\varphi : Cl_{t}(A) \rightarrow Cl_{t}(R)$ be the natural mapping. Note that $\varphi$ is not surjective in general. Indeed, let A $= \mathbb {Z},$ B $= \mathbb {Z}[i]$ and $R=\mathbb {Z}+X\mathbb {Z}[i][\![X]\!].$ Assume that $\varphi$ is surjective.

By [6, Chapter 1, Proposition 2], $\mathbb {Z}[i]=\mathbb {Z}\oplus i\mathbb {Z}$ is a flat $\mathbb {Z}$-module; so by Theorem 2.6, $\varphi$ is an injective homomorphism, and hence $\varphi$ is an isomorphism. This implies that

$$Cl_{t}(\mathbb {Z}) \cong Cl_{t}(\mathbb {Z}+X\mathbb {Z}[i][\![X]\!]).$$

Since $\mathbb {Z}$ is a PID (principal ideal domain), $Cl_{t}(\mathbb {Z})=0$ which implies that $Cl_{t}(\mathbb {Z}+X\mathbb {Z}[i][\![X]\!])=0.$ Now we prove that $Cl_{t}(\mathbb {Z}+X\mathbb {Z}[i][\![X]\!])\ne 0,$ and hence we obtain a contradiction. Let $I=2\mathbb {Z}+(1+i)X\mathbb {Z}[i][\![X]\!].$

Claim 1: I and $I^{-1}$ are ideals of R of v-finite type.

It is clear that $(2,(1+i)X)$ $\subseteq$ I. Conversely, let f $\in$ I. Then f $= 2r + X(1+i)Q,$ for some $r\in \mathbb {Z}$ and $Q\in \mathbb {Z}[i][\![X]\!]$ $= \mathbb {Z} + i\mathbb {Z} +X\mathbb {Z}[i][\![X]\!].$ So there exist $s,t\in \mathbb {Z}$ and $h\in \mathbb {Z}[i][\![X]\!]$ such that f $= 2r + X(1+i)(s + it + Xh)$ $= 2(r - tX) + (1+i)X(s + t + Xh)$ $\in$ $(2,(1+i)X).$ Hence I $= (2,(1+i)X).$ Now, by Example 2.2, $I^{-1}$ $=\mathbb {Z}$ $+ X\frac{1-i}{2}\mathbb {Z}[i][\![X]\!].$ In the same way, we can show that $I^{-1}$ $= (1, \frac{1-i}{2}X).$

Claim 2: I is a v-invertible ideal of R.

Note that

$$II^{-1}=(1, \frac{1-i}{2}X)(2,(1+i)X)=(2, (1+i)X, (1-i)X, X^2).$$

Let $u\in$ qf(R) such that $(2, (1+i)X,$ $(1-i)X, X^2)$ $\subseteq$ uR. Since $2\in$ $(2, (1+i)X,$ $(1-i)X, X^2)$ $\subseteq$ uR,  then u $=\frac{2}{f},$ with $f\in R$ and $X^2\in$ $(2, (1+i)X,$ $(1-i)X, X^2)$ $\subseteq$ uR $= \frac{2}{f}R.$ Thus $X^2f= 2\,g,$ for some $g = a_0 + a_1X + \cdots +a_nX^n$ $\in R.$ This implies that $a_0 = a_1 = 0,$ and so g $= X^2\,h,$ where h $= (a_2 + \cdots +a_nX^{n-2})$ $\in \mathbb {Z}[i][\![X]\!].$ Then f(0) $= 2\,h(0)\in$ $\mathbb {Z}.$ But $\mathbb {Z}[i]$ $= \mathbb {Z} + i\mathbb {Z},$ then h(0) $= s + it$ $\in$ $\mathbb {Z} + i\mathbb {Z}.$ Since $2\,h(0)\in$ $\mathbb {Z},$ then h(0) $\in \mathbb {Z},$ and so 1 $= uh$ $\in$ uR. Thus

$$(II^{-1})_v= (2, (1+i)X, (1-i)X, X^2)_{v}= R.$$

Using claim 1 and 2, it is easy to prove that I is a t-invertible t-ideal of R. This implies that $[I]\in$ $Cl_{t}(R).$ Now we show that $[I]\ne 0$ which equivalent to I is not a principal ideal of R. Assume the contrary that I is principal. Then $I=PR$ for some $P\in R.$ Since $2\in I,$ $P(0)\ne 0.$ In fact $P(0) \in \{\pm 1, \pm 2\}.$ Moreover, as $(1+i)X \in I,$ we obtain $P(0)\in \{\pm 1\}$ which implies that P(0) is a unit in $\mathbb {Z}.$ A routine calculation (by induction) shows that P is a unit in R. This implies that $I=PR=R,$ a contradiction. Then $[I]\ne 0,$ and hence $Cl_{t}(\mathbb {Z}+X\mathbb {Z}[i][\![X]\!])\ne 0.$

3 v-Invertible v-Ideals of A+XB[[X]]

In this section, we investigate a relationship between v-invertible v-ideals of an integral domain and those of its power series ring of the form $A + XB[\![X]\!],$ where $A\subseteq B$ is an extension of integral domain. We begin this section by the following proposition.

Proposition 3.1

Let $A\subseteq B$ be an extension of integral domain, J an ideal of A and $R= A + XB[\![X]\!]$.

• (1) If $(JR)_{v}$ $= R,$ then $J_{v}$ $= A.$

• (2) If $(JR)_t$ $= R,$ then $J_t$ $= A.$

Proof

• (1). Assume that $(JR)_{v}$ $= R$ and let u $\in$ qf(A) such that J $\subseteq$ uA. Then JR $\subseteq$ uAR $\subseteq$ uR which implies that R $=(JR)_{v}$ $\subseteq$ $(uR)_{v}$ $= uR.$ Thus

  $$A \subseteq \bigcap _{u\in \hbox {qf}(A), J \subseteq uA}Au=J_v.$$

  This shows that $A\subseteq J_{v}\subseteq A,$ and hence $J_{v}$ $= A.$

• (2). Suppose that $(JR)_t$ $= R.$ Then

  $$\begin{aligned} R =\bigcup \{(FR)_{v}, F\subseteq J \text { of } \text { finite type of } A\}. \end{aligned}$$

  Thus there exists a finitely generated ideal $F_0$ of A such that $F_0\subseteq$ J and 1 $\in$ $(F_0R)_{v}.$ This implies that R $= (F_0R)_{v}.$ Now, by (1), $(F_0)_{v}$ $= A;$ so

  $$\begin{aligned} A \subseteq \bigcup \{F_{v}, F\subseteq J \text { of } \text { finite type of } A\} =J_t \subseteq A. \end{aligned}$$

  Hence A $= J_t.$

$\square$

Let A be an integral domain. According to [12, Theorem 2.11], A is completely integrally closed if and only if for each $f,g\in A[\![X]\!],$ $(A_fA_g)_{v}=(A_{fg})_{v}.$ Using this result we prove a complete description of v-invertible v-ideals (with nonzero trace in A) of R. First we need to prove the following proposition.

Proposition 3.2

Let $A\subseteq B$ be an extension of integral domains such that B completely integrally closed and $R = A+XB[\![X]\!].$ Then for each divisorial ideal I of R such that $I\cap A\ne (0),$ there exist a divisorial ideal J of B and a nonzero ideal $H\subseteq J$ of A such that $I= H+XJ[\![X]\!].$

Proof

Let $H=I\bigcap A$ and J the ideal of B generated the coefficients of all elements of I.

It is clear that $H\subseteq J$ and $H\subseteq I.$ We show that $XJ_v[\![X]\!]\subseteq I.$ Let $f,g\in R,\;g\ne 0$ such that $I\subseteq \frac{f}{g}R.$ Let $0 \ne a\in H.$ Since $a\in H\subseteq$ $I\subseteq$ $\frac{f}{g}R,$ then there exists an $r\in R{\setminus }(0)$ such that $\frac{a}{r}=\frac{f}{g}.$ Let $0 \ne h\in I$ $\subseteq \frac{f}{g}R$ $=\frac{a}{r}R.$ Then $rh\in aR$ which implies that $rh\in aB[\![X]\!].$ So $(A_{rh})_v\subseteq aB.$ By hypothesis B is a completely integrally closed domain, then $A_{r}A_{h}\subseteq$ $(A_{r}A_{h})_v$ $\subseteq$ aB. This implies that $rA_h[\![X]\!]\subseteq aB[\![X]\!].$ Now we show that $rJ[\![X]\!]$ $\subseteq$ $aB[\![X]\!].$ Indeed, if $f\in rJ[\![X]\!],$ then $f=rf_1$ for some $f_1=\displaystyle \sum _{i=0}^{\infty }a_iX^i\in J[\![X]\!].$ Put $r=\displaystyle \sum _{i=0}^{\infty }\beta _iX^i.$ Then $f=\displaystyle \sum _{n=0}^{\infty }(\displaystyle \sum _{i=0}^na_i\beta _{n-i})X^n.$ But $a_i=\displaystyle \sum _{k=0}^{m_i}\alpha _{i,k}t_{i,k}$ with $t_{i,k}$ $\in B,$ $\alpha _{i,k}$ $\in A_{f_{i,k}},$ then

$$a_i\beta _{n-i}= \displaystyle \sum _{k=0}^{n_i}\alpha _{i,k}t_{i,k}\beta _{n-i} \in A_{r}A_{f_{i,k}} \subseteq aB.$$

Which implies that $rJ[\![X]\!]\subseteq aB[\![X]\!].$ So

$$r(J[\![X]\!])_v=(rJ[\![X]\!])_v\subseteq (aB[\![X]\!])_v = a(B[\![X]\!])_v = aB[\![X]\!].$$

Since $(J[\![X]\!])_v$ = $J_v[\![X]\!],$ $rJ_v[\![X]\!]$ $\subseteq$ $aB[\![X]\!].$ This implies that $\frac{aX}{r}B[\![X]\!]$ $\subseteq$ $\frac{a}{r}R;$ so $\frac{rXJ_v[\![X]\!]}{r}$ $\subseteq$ $\frac{aX}{r}B[\![X]\!]$ $\subseteq$ $\frac{a}{r}R$ =$\frac{f}{g}R$ which implies that $XJ_v[\![X]\!]$ $\subseteq$ $\frac{f}{g}R.$ Thus

$$XJ_v[\![X]\!]\subseteq \displaystyle \cap _{f,g\in R, I\subseteq \frac{f}{g}}\frac{f}{g}R =I_v=I,$$

and hence $H+XJ[\![X]\!]$ $\subseteq$ $H+XJ_v[\![X]\!]$ $\subseteq I.$ Now we will show that $I\subseteq H+XJ[\![X]\!].$ Let $f\in I.$ Then $f=a_0+\displaystyle \sum _{i=1}^\infty a_iX^i,$ where $a_0\in A$ and $\;a_i\in B$ for each $i\ge 1.$

As $J=<A_f,\;f\in I>,$ then for each $i\ge 1,$ $a_i\in J;$ so

$$\displaystyle \sum _{i=1}^\infty a_iX^i =X\displaystyle \sum _{i=1}^\infty a_iX^{i-1}\in XJ[\![X]\!]\subseteq XJ_v[\![X]\!].$$

Since $XJ_v[\![X]\!]$ $\subseteq I,$ $\displaystyle \sum _{i=1}^\infty a_iX^i$ $\in I.$ This implies that $a_0=f-\displaystyle \sum _{i=1}^\infty a_iX^i\in I.$ Thus $a_0\in A\cap I=H,$ and hence $f\in H+XJ[\![X]\!].$ Now we have

$$H+XJ[\![X]\!]\subseteq H+XJ_v[\![X]\!]\subseteq I \subseteq H+XJ[\![X]\!].$$

Hence $I=H+XJ[\![X]\!]$ and $J_v=J.$ $\square$

Our next result give a complete description of v-invertible v-ideals of $A+XB[\![X]\!]$ with nonzero trace in A.

Theorem 3.3

Let $A\subseteq B$ be an extension of integral domains such that B is completely integrally closed and $R = A+XB[\![X]\!].$ Let I be a fractional divisorial v-invertible ideal of R such that $I\cap A \ne (0).$ Then $I = u(J_1 + XJ_2[\![X]\!])$ for some $u\in qf(R),$ $J_2$ an integral divisorial v-invertible ideal of B and $J_1\subseteq J_2$ a nonzero ideal of A.

Proof

Since I is a divisorial ideal of R and B is completely integrally closed, by Proposition 3.2, $I=H+XJ[\![X]\!]$ for some divisorial ideal J of B and a nonzero ideal $H\subseteq J$ of A. We show that there exists nonzero $c\in K$ such that $cH\subseteq A$ and $cJ\subseteq B.$

Let $a\in H$ be a nonzero element. We have $aI^{-1}$ is a divisorial ideal of R. Using Lemma 2.1, it is easy to prove that $aI^{-1}\cap A\ne (0).$ Then by Proposition 3.2, $aI^{-1}=H^\prime +XJ^\prime [\![X]\!]$ for some divisorial ideal $J^\prime$ of B and a nonzero ideal $H^\prime \subseteq J^\prime$ of A.

$$\begin{array}{ccl} aR &{}=&{} a(II^{-1})_v\\ {} &{}=&{} (a(II^{-1}))_v\\ &{}=&{} (I(aI^{-1}))_v\\ &{}=&{} ((H + XJ[\![X]\!])(H^{\prime } + XJ^{\prime }[\![X]\!]))_v. \end{array}.$$

So $(H + XJ[\![X]\!])(H^{\prime } + XJ^{\prime }[\![X]\!])$ $\subseteq$ $aR=aA+aXB[\![X]\!].$ Then $HH^{\prime }$ $\subseteq$ aA and $JJ^{\prime }$ $\subseteq$ aB. This implies that $\frac{1}{a}HH^{\prime }$ $\subseteq$ A and $\frac{1}{a}JJ^{\prime }$ $\subseteq$ B.

Let $c\in$ $\frac{1}{a}H^{\prime }$ be a nonzero element. Then $J_1$ $= cH$ $\subseteq$ $\frac{1}{a}HH^{\prime }$ $\subseteq A$ and $J_2$ $= cJ\subseteq B.$ We have $J_1 \ne (0)$ and $J_2$ is a divisorial ideal of B.

Since $I=H + XJ[\![X]\!],$ then

$$I= \frac{1}{c}(cH + XcJ[\![X]\!])= \frac{1}{c}(J_1 +XJ_2[\![X]\!])= u(J_1 + XJ_2[\![X]\!]),$$

where $u = \frac{1}{c}$ $\in qf(R).$ Now we will show that $J_2$ is v-invertible. By Lemma 2.1, we have

$$\begin{aligned} I^{-1} = \frac{1}{u}(J_1^{-1} \cap J_2^{-1} + XJ_2^{-1}[\![X]\!]). \end{aligned}$$

Thus

$$\begin{array}{ccl} II^{-1} &{}\subseteq &{} J_1(J_1^{-1} \cap J_2^{-1})+XJ_2(J_2^{-1}[\![X]\!])\\ &{}\subseteq &{} J_1J_1^{-1} +X(J_2J_2^{-1})[\![X]\!]\\ &{}\subseteq &{} A + XB[\![X]\!]\\ &{}=&{} R.\end{array}$$

Since I is v-invertible, we get

$$\begin{aligned} R =(J_1(J_1^{-1}\cap J_2^{-1}) + X(J_2J_2^{-1})[\![X]\!])^{-1}.\end{aligned}$$

Again by Lemma 2.1, R $= (J_1(J_1^{-1}\cap J_2^{-1}))^{-1} \cap (J_2J_2^{-1})^{-1}$ $+$ $X(J_2J_2^{-1})^{-1}[\![X]\!].$ Then $B[\![X]\!]$ $= (J_2J_2^{-1})^{-1}[\![X]\!],$ and this implies that B $= (J_2J_2^{-1})^{-1}.$ Hence $J_2$ is v-invertible. $\square$

Clearly that every Krull domain is completely integrally closed. Using Theorem 3.3, we obtain a new characterization of divisorial v-invertible ideals of the power series ring of the form $A+XB[\![X]\!].$

Corollary 3.4

Let I be a fractional divisorial v-invertible ideal of $R=A+XB[\![X]\!]$ such that $I\cap A \ne (0).$ Assume that B is a Krull domain. Then $I = u(J_1 + XJ_2[\![X]\!])$ for some $u\in qf(R),$ $J_2$ an integral divisorial v-invertible ideal of B and $J_1\subseteq J_2$ a nonzero ideal of A.

Recall from [4] that an integral domain A is called formally integrally closed if $(A_{fg})_{t}=(A_fA_g)_{t}$ for all $f,g\in A[\![X]\!]\backslash (0).$ It was shown in [4] that if A is formally integrally closed, then A is completely integrally closed, but the converse is false in general ([4, Example 3.2]).

Proposition 3.5

[4, Proposition 3.6] Let A be a formally integrally closed domain. If I is a finite type v-ideal of $A[\![X]\!]$ with $J\cap A\ne 0,$ then $I=J[\![X]\!]$ for some v-ideal J of A.

Note that in [4] Anderson and Kang characterized the v-ideals of finite type of the power series ring $A[\![X]\!]$ with nonzero trace in A in the case when A is a formally integrally closed domain. Now, using Proposition 3.2, in the particular case when $A=B,$ we obtain a new approach to characterize the divisorial ideals of the ring $A[\![X]\!]$ with nonzero trace in A.

Proposition 3.6

Let A be a completely integrally closed domain and I a fractional divisorial ideal of $A[\![X]\!]$ such that $I\cap A \ne (0).$ Then $I = J_1 + XJ_2[\![X]\!]$ for some nonzero ideal $J_1$ of A and some divisorial ideal $J_2$ of A such that $J_1\subseteq J_2.$