# Characterizations of graded Prüfer $\star$-multiplication domains

Parviz Sahandi  Department of Mathematics, University of Tabriz, Tabriz, Iran,
and School of Mathematics, Institute for Research in Fundamental Sciences
(IPM), P.O. Box: 19395-5746, Tehran Iran<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain
graded by an arbitrary grading torsionless monoid $\Gamma$, and $\star$ be a
semistar operation on $R$. In this paper we define and study the graded
integral domain analogue of $\star$-Nagata and Kronecker function rings of $R$
with respect to $\star$. We say that $R$ is a graded Prüfer
$\star$-multiplication domain if each nonzero finitely generated homogeneous
ideal of $R$ is $\star_{f}$-invertible. Using $\star$-Nagata and Kronecker
function rings, we give several different equivalent conditions for $R$ to be
a graded Prüfer $\star$-multiplication domain. In particular we give new
characterizations for a graded integral domain, to be a P$v$MD.

###### Key words and phrases:

semistar operation, Nagata ring, Kronecker function ring, Prüfer domain,
graded domain, P$v$MD

###### 2000 Mathematics Subject Classification:

Primary 13A15, 13G05, 13A02

Parviz Sahandi was in part supported by a grant from IPM (No. 91130030)

##  1\. Introduction

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded (commutative)
integral domain graded by an arbitrary grading torsionless monoid $\Gamma$,
that is $\Gamma$ is a commutative cancellative monoid (written additively).
Let $\langle\Gamma\rangle=\\{a-b|a,b\in\Gamma\\},$ be the quotient group of
$\Gamma$, which is a torsionfree abelian group.

Let $H$ be the saturated multiplicative set of nonzero homogeneous elements of
$R$. Then $R_{H}=\bigoplus_{\alpha\in\langle\Gamma\rangle}(R_{H})_{\alpha}$,
called the _homogeneous quotient field of $R$_, is a graded integral domain
whose nonzero homogeneous elements are units. For a fractional ideal $I$ of
$R$ let $I_{h}$ denote the fractional ideal generated by the set of
homogeneous elements of $R$ in $I$. It is known that if $I$ is a prime ideal,
then $I_{h}$ is also a prime ideal (cf. [29, Page 124]). An integral ideal $I$
of $R$ is said to be homogeneous if $I=\bigoplus_{\alpha\in\Gamma}(I\cap
R_{\alpha})$; equivalently, if $I=I_{h}$. A fractional ideal $I$ of $R$ is
_homogeneous_ if $sI$ is an integral homogeneous ideal of $R$ for some $s\in
H$ (thus $I\subseteq R_{H}$). For $f\in R_{H}$, let $C_{R}(f)$ (or simply
$C(f)$) denote the fractional ideal of $R$ generated by the homogeneous
components of $f$. For a fractional ideal $I$ of $R$ with $I\subseteq R_{H}$,
let $C(I)=\sum_{f\in I}C(f)$. For more on graded integral domains and their
divisibility properties, see [3, 29].

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ and $N_{v}(H)=\\{f\in
R|C(f)^{v}=R\\}$. (Definitions related to the $v$-operation will be reviewed
in the sequel.) Then $N_{v}(H)$ is a saturated multiplicative subset of $R$ by
[4, Lemma 1.1(2)]. The graded integral domain analogue of the well known
Nagata ring is the ring $R_{N_{v}(H)}$. In [4], Anderson and Chang, studied
relationships between the ideal-theoretic properties of $R_{N_{v}(H)}$ and the
homogeneous ideal-theoretic properties of $R$. For example it is shown that if
$R$ has a unit of nonzero degree, $Pic(R_{N_{v}(H)})=0$ and that $R$ is a
P$v$MD if and only if each ideal of $R_{N_{v}(H)}$ is extended from a
homogeneous ideal of $R$, if and only if $R_{N_{v}(H)}$ is a Prüfer (or
Bézout) domain [4, Theorems 3.3 and 3.4]. Also, they generalized the notion of
Kronecker function ring, (for e.a.b. star operations on $R$) and then showed
that this ring is a Bézout domain [4, Theorem 3.5]. For the definition and
properties of semistar-Nagata and Kronecker function rings of an integral
domain see the interesting survey article [21]. Recall that the _Picard group
(or the ideal class group)_ of an integral domain $D$, is
$Pic(D)=Inv(D)/Prin(D)$, where $Inv(D)$ is the multiplicative group of
invertible fractional ideals of $D$, and $Prin(D)$ is the subgroup of
principal fractional ideal of $D$.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be an integral domain, and
$\star$ be a semistar operation on $R$. In Section 2 of this paper we study
the homogeneous elements of $\operatorname{QSpec}^{\star}(R)$ denoted by
$h$-$\operatorname{QSpec}^{\star}(R)$. We show that if $\star$ is a finite
type semistar operation on $R$ which sends homogeneous fractional ideals to
homogeneous ones, and such that $R^{\star}\subsetneq R_{H}$, then each
homogeneous quasi-$\star$-ideal of $R$, is contained in a homogeneous
quasi-$\star$-prime ideal of $R$. One of key results in this paper is
Proposition 2.3, which shows that if $R^{\star}\subsetneq R_{H}$, the
$\widetilde{\star}$ sends homogeneous fractional ideals to homogeneous ones.
We also define and study the Nagata ring of $R$ with respect to $\star$. The
$\star$-Nagata ring is defined by the quotient ring $R_{N_{\star}(H)}$, where
$N_{\star}(H)=\\{f\in R|C(f)^{\star}=R^{\star}\\}$. Among other things, it is
shown that $Pic(R_{N_{\star}(H)})=0$. In Section 3 we define and study the
Kronecker function ring of $R$ with respect to $\star$. The Kronecker function
ring, inspired by [20, Theorem 5.1], is defined by
$\operatorname{Kr}(R,\star):=\\{0\\}\cup\\{f/g|0\neq f,g\in R,\text{ and there
is }0\neq h\in R\text{ such that }C(f)C(h)\subseteq(C(g)C(h))^{\star}\\}$. It
is shown that if $\star$ sends homogeneous fractional ideals to fractional
ones, then $\operatorname{Kr}(R,\star)$ is a Bézout domain. In Section 3 we
define the notion of graded Prüfer $\star$-multiplication domains and give
several different equivalent conditions to be a graded P$\star$MD. A graded
integral domain $R$, is called a _graded Prüfer $\star$-multiplication domain
(graded P$\star$MD)_ if every finitely generated homogeneous ideal of $R$ is a
$\star_{f}$-invertible, i.e., $(II^{-1})^{\star_{f}}=R^{\star}$ for each
finitely generated homogeneous ideal $I$ of $R$. Among other results we show
that $R$ is a graded P$\star$MD if and only if $R_{N_{\star}(H)}$ is a Prüfer
domain if and only if $R_{N_{\star}(H)}$ is a Bézout domain if and only if
$R_{N_{\star}(H)}=\operatorname{Kr}(R,\widetilde{\star})$ if and only if
$\operatorname{Kr}(R,\widetilde{\star})$ is a flat $R$-module.

To facilitate the reading of the paper, we review some basic facts on semistar
operations. Let $D$ be an integral domain with quotient field $K$. Let
$\overline{\mathcal{F}}(D)$ denote the set of all nonzero $D$-submodules of
$K$. Let $\mathcal{F}(D)$ be the set of all nonzero _fractional_ ideals of
$D$; i.e., $E\in\mathcal{F}(D)$ if $E\in\overline{\mathcal{F}}(D)$ and there
exists a nonzero element $r\in D$ with $rE\subseteq D$. Let $f(D)$ be the set
of all nonzero finitely generated fractional ideals of $D$. Obviously,
$f(D)\subseteq\mathcal{F}(D)\subseteq\overline{\mathcal{F}}(D)$. As in [30], a
semistar operation on $D$ is a map
$\star:\overline{\mathcal{F}}(D)\rightarrow\overline{\mathcal{F}}(D)$,
$E\mapsto E^{\star}$, such that, for all $x\in K$, $x\neq 0$, and for all
$E,F\in\overline{\mathcal{F}}(D)$, the following three properties hold:

  * $\star_{1}$

: $(xE)^{\star}=xE^{\star}$;

  * $\star_{2}$

: $E\subseteq F$ implies that $E^{\star}\subseteq F^{\star}$;

  * $\star_{3}$

: $E\subseteq E^{\star}$ and $E^{\star\star}:=(E^{\star})^{\star}=E^{\star}$.

Let $\star$ be a semistar operation on the domain $D$. For every
$E\in\overline{\mathcal{F}}(D)$, put $E^{\star_{f}}:=\cup F^{\star}$, where
the union is taken over all finitely generated $F\in f(D)$ with $F\subseteq
E$. It is easy to see that $\star_{f}$ is a semistar operation on $D$, and
${\star_{f}}$ is called _the semistar operation of finite type associated to_
$\star$. Note that $(\star_{f})_{f}=\star_{f}$. A semistar operation $\star$
is said to be of _finite type_ if $\star=\star_{f}$; in particular
${\star_{f}}$ is of finite type. We say that a nonzero ideal $I$ of $D$ is a
_quasi- $\star$-ideal_ of $D$, if $I^{\star}\cap D=I$; a _quasi-
$\star$-prime_ (ideal of $D$), if $I$ is a prime quasi-$\star$-ideal of $D$;
and a _quasi- $\star$-maximal_ (ideal of $D$), if $I$ is maximal in the set of
all proper quasi-$\star$-ideals of $D$. Each quasi-$\star$-maximal ideal is a
prime ideal. It was shown in [16, Lemma 4.20] that if $D^{\star}\neq K$, then
each proper quasi-$\star_{f}$-ideal of $D$ is contained in a
quasi-$\star_{f}$-maximal ideal of $D$. We denote by
$\operatorname{QMax}^{\star}(D)$ (resp., $\operatorname{QSpec}^{\star}(D)$)
the set of all quasi-$\star$-maximal ideals (resp., quasi-$\star$-prime
ideals) of $D$.

If $\star_{1}$ and $\star_{2}$ are semistar operations on $D$, one says that
$\star_{1}\leq\star_{2}$ if $E^{\star_{1}}\subseteq E^{\star_{2}}$ for each
$E\in\overline{\mathcal{F}}(D)$ (cf. [30, page 6]). This is equivalent to
saying that
$(E^{\star_{1}})^{\star_{2}}=E^{\star_{2}}=(E^{\star_{2}})^{\star_{1}}$ for
each $E\in\overline{\mathcal{F}}(D)$ (cf. [30, Lemma 16]). Obviously, for each
semistar operation $\star$ defined on $D$, we have $\star_{f}\leq\star$. Let
$d_{D}$ (or, simply, $d$) denote the identity (semi)star operation on $D$.
Clearly, $d_{D}\leq\star$ for all semistar operations $\star$ on $D$.

It has become standard to say that a semistar operation $\star$ is stable if
$(E\cap F)^{\star}=E^{\star}\cap F^{\star}$ for all $E$,
$F\in\overline{\mathcal{F}}(D)$. (“Stable” has replaced the earlier usage,
“quotient”, in [30, Definition 21].) Given a semistar operation $\star$ on
$D$, it is possible to construct a semistar operation $\widetilde{\star}$,
which is stable and of finite type defined as follows: for each
$E\in\overline{\mathcal{F}}(D)$,

$E^{\widetilde{\star}}:=\\{x\in K|xJ\subseteq E,\text{ for some }J\subseteq
R,J\in f(R),J^{\star}=D^{\star}\\}.$

It is well known that [16, Corollary 2.7]

$E^{\widetilde{\star}}:=\cap\\{ED_{P}|P\in\operatorname{QMax}^{\star_{f}}(D)\\}\text{,
for each }E\in\overline{\mathcal{F}}(D).$

The most widely studied (semi)star operations on $D$ have been the identity
$d$, $v$, $t:=v_{f}$, and $w:=\widetilde{v}$ operations, where
$A^{v}:=(A^{-1})^{-1}$, with $A^{-1}:=(R:A):=\\{x\in K|xA\subseteq D\\}$.

Let $\star$ be a semistar operation on an integral domain $D$. We say that
$\star$ is an _e.a.b. (endlich arithmetisch brauchbar) semistar operation_ of
$D$ if, for all $E,F,G\in f(D)$, $(EF)^{\star}\subseteq(EG)^{\star}$ implies
that $F^{\star}\subseteq G^{\star}$ ([20, Definition 2.3 and Lemma 2.7]). We
can associate to any semistar operation $\star$ on $D$, an e.a.b. semistar
operation of finite type $\star_{a}$ on $D$, called the _e.a.b. semistar
operation associated to $\star$_, defined as follows for each $F\in f(D)$ and
for each $E\in\overline{F}(D)$:

$\displaystyle F^{\star_{a}}:=$
$\displaystyle\bigcup\\{((FH)^{\star}:H^{\star})|H\in f(R)\\},$ $\displaystyle
E^{\star_{a}}:=$ $\displaystyle\bigcup\\{F^{\star_{a}}|F\subseteq E,F\in
f(R)\\}$

[20, Definition 4.4 and Proposition 4.5] (note that
$((FH)^{\star}:H^{\star})=((FH)^{\star}:H)$). It is known that
$\star_{f}\leq\star_{a}$ [20, Proposition 4.5(3)]. Obviously
$(\star_{f})_{a}=\star_{a}$. Moreover, when $\star=\star_{f}$, then $\star$ is
e.a.b. if and only if $\star=\star_{a}$ [20, Proposition 4.5(5)].

Let $\star$ be a semistar operation on a domain $D$. Recall from [17] that,
$D$ is called a _Prüfer $\star$-multiplication domain_ (for short, a
P$\star$MD) if each finitely generated ideal of $D$ is _$\star_{f}$
-invertible_; i.e., if $(II^{-1})^{\star_{f}}=D^{\star}$ for all $I\in f(D)$.
When $\star=v$, we recover the classical notion of P$v$MD; when $\star=d_{D}$,
the identity (semi)star operation, we recover the notion of Prüfer domain.

##  2\. Nagata ring

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain,
$\star$ be a semistar operation on $R$, $H$ be the set of nonzero homogeneous
elements of $R$. An overring $T$ of $R$, with $R\subseteq T\subseteq R_{H}$
will be called a _homogeneous overring_ if
$T=\bigoplus_{\alpha\in\langle\Gamma\rangle}(T\cap(R_{H})_{\alpha})$. Thus $T$
is a graded integral domain with $T_{\alpha}=T\cap(R_{H})_{\alpha}$.

In this section we study the homogeneous elements of
$\operatorname{QSpec}^{\star}(R)$, denoted by
$h$-$\operatorname{QSpec}^{\star}(R)$, and the graded integral domain analogue
of $\star$-Nagata ring. Let $h$-$\operatorname{QMax}^{\star}(R)$ denote the
set of ideals of $R$ which are maximal in the set of all proper homogeneous
quasi-$\star$-ideals of $R$. The following lemma shows that, if
$R^{\star}\subsetneq R_{H}$ and $\star=\star_{f}$ sends homogeneous fractional
ideals to homogeneous ones, then $h$-$\operatorname{QMax}^{\star_{f}}(R)$ is
nonempty and each proper homogeneous quasi-$\star_{f}$-ideal is contained in a
maximal homogeneous quasi-$\star_{f}$-ideal.

######  Lemma 2.1.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain,
$\star$ a finite type semistar operation on $R$ which sends homogeneous
fractional ideals to homogeneous ones, and such that $R^{\star}\subsetneq
R_{H}$. If $I$ is a proper homogeneous quasi-$\star$-ideal of $R$, then $I$ is
contained in a proper homogeneous quasi-$\star$-prime ideal.

###### Proof.

Let $X:=\\{I|I$ is a homogeneous quasi-$\star$-ideal of $R\\}$. Then it is
easy to see that $X$ is nonempty. Indeed, in this case $R^{\star}$ is a
homogeneous overring of $R$, and if $u\in H$ is a nonunit in $R^{\star}$, then
$uR^{\star}\cap R$ is a proper homogeneous quasi-$\star$-ideal of $R$. Also
$X$ is inductive (see proof of [16, Lemma 4.20]). From Zorn’s Lemma, we see
that every proper homogeneous quasi-$\star$-ideal of $R$ is contained in some
maximal element $Q$ of $X$.

Now we show that $Q$ is actually prime. Take $f,g\in H\backslash Q$ and
suppose that $fg\in Q$. By the maximality of $Q$ we have
$(Q,f)^{\star}=R^{\star}$ (note that $(Q,f)^{\star}\cap R$ is a homogeneous
quasi-$\star$-ideal of $R$ and properly contains $Q$). Since $\star$ is of
finite type, we can find a finitely generated ideal $J\subseteq Q$ such that
$(J,f)^{\star}=R^{\star}$. Then $g\in gR^{\star}\cap R=g(J,f)^{\star}\cap
R\subseteq Q^{\star}\cap R=Q$ a contradiction. Thus $Q$ is a prime ideal. ∎

The following example shows that we can not drop the condition that, $\star$
sends homogeneous fractional ideals to homogeneous ones, in the above lemma.

######  Example 2.2.

Let $k$ be a field and $X,Y$ be indeterminates over $k$. Let $R=k[X,Y]$, which
is a ($\mathbb{N}_{0}$-)graded Noetherian integral domain with $\deg X=\deg
Y=1$. Set $M:=(X,Y+1)$ which is a maximal non-homogeneous ideal of $R$. Let
$T$ be a DVR [11], with maximal ideal $N$, dominating the local ring $R_{M}$.
If $R_{H}\subseteq T$, then there exists a prime ideal $P$ of $R$ such that,
$P\cap H=\emptyset$ and $N\cap R_{H}=PR_{H}$. Thus $M=N\cap R=N\cap R_{H}\cap
R=PR_{H}\cap R=P$. Hence $M\cap H=\emptyset$, which is a contradiction, since
$X\in M\cap H$. So that, $R_{H}\nsubseteq T$. Let $\star$ be a semistar
operation on $R$ defined by $E^{\star}=ET\cap ER_{H}$ for each
$E\in\overline{\mathcal{F}}(R)$. Then clearly $\star=\star_{f}$ and
$R^{\star}\subsetneq R_{H}$. If $P$ is a nonzero prime ideal of $R$, such that
$P\cap H=\emptyset$, then $P^{\star_{f}}\cap R=PT\cap PR_{H}\cap R=PT\cap
P=P$. Thus $P$ is a quasi-$\star_{f}$-prime ideal. On the other hand if $P$ is
any nonzero prime ideal of $R$ such that $P\cap H\neq\emptyset$, then
$PT=N^{k}$, for some integer $k\geq 1$. Therefore, if we assume that $P$ is a
quasi-$\star_{f}$-ideal of $R$, then we would have $P=PT\cap PR_{H}\cap
R=PT\cap R=N^{k}\cap R\supseteq M^{k}$, which implies that $P=M$. Thus
$\operatorname{QSpec}^{\star_{f}}(R)=\\{M\\}\cup\\{P\in\operatorname{Spec}(R)|P\neq
0$ and $P\cap H=\emptyset\\}$. Therefore by [16, Lemma 4.1, Remark 4.5], we
have
$\operatorname{QSpec}^{\widetilde{\star}}(R)=\\{Q\in\operatorname{Spec}(R)|0\neq
Q\subseteq M\\}\cup\\{P\in\operatorname{Spec}(R)|P\neq 0$ and $P\cap
H=\emptyset\\}$. Hence in the present example we have
$h$-$\operatorname{QSpec}^{\star_{f}}(R)=h$-$\operatorname{QMax}^{\star_{f}}(R)=\emptyset$,
and
$h$-$\operatorname{QSpec}^{\widetilde{\star}}(R)=h$-$\operatorname{QMax}^{\widetilde{\star}}(R)=\\{(X)\\}$.
Note that in this example
$h$-$\operatorname{QMax}^{\widetilde{\star}}(R)\nsubseteq\operatorname{QMax}^{\widetilde{\star}}(R)=\operatorname{QMax}^{\star_{f}}(R)$.

From now on in this paper, we are interested and consider, the semistar
operations $\star$ on $R$, such that $R^{\star}\subsetneq R_{H}$ and sends
homogeneous fractional ideals to homogeneous ones. For any such semistar
operation, if $I$ is a homogeneous ideal of $R$, we have
$I^{\star_{f}}=R^{\star}$ if and only if $I\nsubseteq Q$ for each $Q\in
h$-$\operatorname{QMax}^{\star_{f}}(R)$. Also if $P$ is a quasi-$\star$-prime
ideal of $R$, then either $P_{h}=0$ or $P_{h}$ is a quasi-$\star$-prime ideal
of $R$. Indeed, if $P_{h}\neq 0$, then $P_{h}\subseteq(P_{h})^{\star}\cap
R\subseteq P^{\star}\cap R=P$, which implies that $P_{h}=(P_{h})^{\star}\cap
R$, since $(P_{h})^{\star}\cap R$ is a homogeneous ideal.

The following proposition is the key result in this paper.

######  Proposition 2.3.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain, and
$\star$ be a semistar operation on $R$ such that $R^{\star}\subsetneq R_{H}$.
Then, $\widetilde{\star}$ sends homogeneous fractional ideals to homogeneous
ones. In particular
$h$-$\operatorname{QMax}^{\widetilde{\star}}(R)\neq\emptyset$, and
$R^{\widetilde{\star}}$ is a homogeneous overring of $R$.

###### Proof.

Let $E$ be a homogenous fractional ideal of $R$. To show that
$E^{\widetilde{\star}}$ is homogeneous let $f\in E^{\widetilde{\star}}$. Then
$fJ\subseteq E$ for some finitely generated ideal $J$ of $R$ such that
$J^{\star}=R^{\star}$. Suppose that $J=(g_{1},\cdots,g_{n})$. Using [4, Lemma
1.1(1)], there is an integer $m\geq 1$ such that
$C(g_{i})^{m+1}C(f)=C(g_{i})^{m}C(fg_{i})$ for all $i=1,\cdots,n$. Since $E$
is a homogeneous fractional ideal and $fg_{i}\in E$, we have
$C(fg_{i})\subseteq E$. Thus we have $C(g_{i})^{m+1}C(f)\subseteq E$. Let
$J_{0}:=C(g_{1})^{m+1}+\cdots+C(g_{n})^{m+1}$. Thus $J_{0}$ is a finitely
generated homogeneous ideal of $R$ such that $J^{\star}_{0}=R^{\star}$. Since
$C(f)J_{0}\subseteq E$, $C(f)\subseteq E^{\widetilde{\star}}$. Therefore
$E^{\widetilde{\star}}$ is a homogeneous ideal. ∎

######  Lemma 2.4.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain,
$\star$ a semistar operation on $R$ which sends homogeneous fractional ideals
to homogeneous ones. Then $\star_{f}$ sends homogeneous fractional ideals to
homogeneous ones.

###### Proof.

Let $E$ be a homogenous fractional ideal of $R$. Let $0\neq x\in
E^{\star_{f}}$. Then, there exists an $F\in f(R)$ such that $F\subseteq E$ and
$x\in F^{\star}$. Suppose that $F$ is generated by $y_{1},\cdots,y_{n}\in
R_{H}$. Let $G$ be a homogeneous fractional ideal of $R$, generated by
homogeneous components of $y_{1},\cdots,y_{n}$. Note that $F\subseteq
G\subseteq E$ and $x\in G^{\star}$. Thus homogeneous components of $x$ belong
to $G^{\star}\subseteq E^{\star_{f}}$. This shows that $E^{\star_{f}}$ is
homogeneous. ∎

Note that the $v$-operation sends homogeneous fractional ideals to homogeneous
ones by [3, Proposition 2.5]. Using the above two results, the $t$ and
$w$-operations also, send homogeneous fractional ideals to homogeneous ones.

It it well-known that
$\operatorname{QMax}^{\star_{f}}(R)=\operatorname{QMax}^{\widetilde{\star}}(R)$,
see [5, Theorem 2.16], for star operation case, and [18, Corollary 3.5(2)], in
general semistar operations. Although Example 2.2, shows that it may happen
that $h$-$\operatorname{QMax}^{\star_{f}}(R)\neq
h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$, we have the following
proposition whose proof is almost the same as [4, Theorem 2.16].

######  Proposition 2.5.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain,
$\star$ a semistar operation on $R$ such that $R^{\star}\subsetneq R_{H}$,
which sends homogeneous fractional ideals to homogeneous ones. Then
$h$-$\operatorname{QMax}^{\star_{f}}(R)=h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$.

###### Proof.

Assume that $Q\in h$-$\operatorname{QMax}^{\star_{f}}(R)$. Then since
$\widetilde{\star}\leq\star_{f}$ by [18, Lemma 2.7(1)], we have $Q\subseteq
Q^{\widetilde{\star}}\cap R\subseteq Q^{\star_{f}}\cap R=Q$, that is $Q$ is a
quasi-$\widetilde{\star}$-ideal. Suppose that $Q\notin
h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$. Then $Q$ is properly
contained in some $P\in h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$. So
since $Q\in h$-$\operatorname{QMax}^{\star_{f}}(R)$, using Lemma 2.1, we must
have $P^{\star_{f}}=R^{\star}$. Thus there is some finitely generated ideal
$F\subseteq P$ such that $F^{\star}=R^{\star}$. So for any $r\in R$,
$rF\subseteq F\subseteq P$. But then, $r\in P^{\widetilde{\star}}$, so
$R\subseteq P^{\widetilde{\star}}$, which implies that
$P^{\widetilde{\star}}=R^{\widetilde{\star}}$, a contradiction. Therefore, we
must have $Q\in h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$.

If $Q\in h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$, then
$Q=Q^{\widetilde{\star}}\cap R\subseteq Q^{\star_{f}}\cap R\subseteq R$.
Suppose that $Q^{\star_{f}}\cap R=R$, which implies that
$Q^{\star_{f}}=R^{\star}$. Then there is a finitely generated ideal
$F\subseteq Q$ such that $F^{\star}=R^{\star}$. Now for any $r\in R$,
$rF\subseteq F\subseteq Q$. Therefore $R\subseteq Q^{\widetilde{\star}}$, and
so $R=Q^{\widetilde{\star}}\cap R=Q$, which is a contradiction. So
$Q^{\star_{f}}\cap R\subsetneq R$. Now, since $Q^{\star_{f}}\cap R$ is a
homogeneous quasi-$\star_{f}$-ideal, there is a $P\in
h$-$\operatorname{QMax}^{\star_{f}}(R)$ such that $Q\subseteq
Q^{\star_{f}}\cap R\subseteq P$. From the first half of the proof, we know
that $P\in h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$. So we must have
$P=Q$. Therefore $Q\in h$-$\operatorname{QMax}^{\star_{f}}(R)$. ∎

Park in [31, Lemma 3.4], proved that $I^{w}=\bigcap_{P\in
h\text{-}QMax^{w}(R)}IR_{H\backslash P}$ for each homogeneous ideal $I$ of
$R$.

######  Proposition 2.6.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain,
$\star$ a semistar operation on $R$ such that $R^{\star}\subsetneq R_{H}$.
Then $I^{\widetilde{\star}}=\bigcap_{P\in
h\text{-}\operatorname{QMax}^{\widetilde{\star}}(R)}IR_{H\backslash P}$ for
each homogeneous ideal $I$ of $R$. Moreover
$I^{\widetilde{\star}}R_{H\backslash P}=IR_{H\backslash P}$ for all
homogeneous ideal $I$ of $R$ and all $P\in
h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$.

###### Proof.

By Proposition 2.3, $I^{\widetilde{\star}}$ is a homogeneous ideal. Also note
that $\bigcap_{P\in
h\text{-}\operatorname{QMax}^{\widetilde{\star}}(R)}IR_{H\backslash P}$ is a
homogeneous ideal of $R$. Let $f\in I^{\widetilde{\star}}$ be homogeneous.
Then $fJ\subseteq I$ for some homogeneous finitely generated ideal $J$ of $R$
such that $J^{\star}=R^{\star}$. It is easy to see that
$J^{\widetilde{\star}}=R^{\widetilde{\star}}$. Hence we have $J\nsubseteq P$
for all $P\in h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$. Thus $f\in
IR_{H\backslash P}$ for all $P\in
h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$. Conversely, let
$f\in\bigcap_{P\in
h\text{-}\operatorname{QMax}^{\widetilde{\star}}(R)}IR_{H\backslash P}$ be
homogeneous. Then $(I:f)$ is a homogeneous ideal which is not contained in any
$P\in h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$. Therefore
$(I:f)^{\widetilde{\star}}=R^{\widetilde{\star}}$. So that there exist a
finitely generated ideal $J\subseteq(I:f)$ such that $J^{\star}=R^{\star}$.
Thus $fJ\subseteq I$, i.e., $f\in I^{\widetilde{\star}}$. The second assertion
follows from the first one. ∎

Let $D$ be a domain with quotient field $K$, and let $X$ be an indeterminate
over $K$. For each $f\in K[X]$, we let $c_{D}(f)$ denote the content of the
polynomial $f$, i.e., the (fractional) ideal of $D$ generated by the
coefficients of $f$. Let $\star$ be a semistar operation on $D$. If
$N_{\star}:=\\{g\in D[X]|g\neq 0\text{ and }c_{D}(g)^{\star}=D^{\star}\\}$,
then
$N_{\star}=D[X]\backslash\bigcup\\{P[X]|P\in\operatorname{QMax}^{\star_{f}}(D)\\}$
is a saturated multiplicative subset of $D[X]$. The ring of fractions

$\operatorname{Na}(D,\star):=D[X]_{N_{\star}}$

is called the $\star$-Nagata domain (of $D$ with respect to the semistar
operation $\star$). When $\star=d$, the identity (semi)star operation on $D$,
then $\operatorname{Na}(D,d)$ coincides with the classical Nagata domain
$D(X)$ (as in, for instance [28, page 18], [23, Section 33] and [18]).

Let $N_{\star}(H)=\\{f\in R|C(f)^{\star}=R^{\star}\\}$. It is easy to see that
$N_{\star}(H)$ is a saturated multiplicative subset of $R$. Indeed assume
$f,g\in N_{\star}(H)$. Then $C(f)^{n+1}C(g)=C(f)^{n}C(fg)$ for some integer
$n\geq 1$ by [4, Lemma 1.1(2)], and $C(fg)\subseteq C(f)C(g)$. Thus $fg\in
N_{\star}(H)\Leftrightarrow C(fg)^{\star}=R^{\star}\Leftrightarrow
C(f)^{\star}=C(g)^{\star}=R^{\star}\Leftrightarrow f,g\in N_{\star}(H)$. Also
it is easy to show that
$N_{\star}(H)=N_{\star_{f}}(H)=N_{\widetilde{\star}}(H)$. We define the graded
integral domain analogue of $\star$-Nagata ring, by the quotient ring
$R_{N_{\star}(H)}$. When $\star=v$, $R_{N_{\star}(H)}$ was studied in [4],
denoted by $R_{N(H)}$.

######  Lemma 2.7.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain, and
$\star$ be a semistar operation on $R$ such that $R^{\star}\subsetneq R_{H}$,
which sends homogeneous fractional ideals to homogeneous ones.

  * (1)

$N_{\star}(H)=R\backslash\bigcup_{Q\in
h\text{-}\operatorname{QMax}^{\star_{f}}(R)}Q$.

  * (2)

$\operatorname{Max}(R_{N_{\star}(H)})=\\{QR_{N_{\star}(H)}|Q\in
h\text{-}\operatorname{QMax}^{\star_{f}}(R)\\}$ if and only if $R$ has the
property that if $I$ is a nonzero ideal of $R$ with $C(I)^{\star}=R^{\star}$,
then $I\cap N_{\star}(H)\neq\emptyset$.

###### Proof.

(1) Let $x\in R$. Then $x\in N_{\star}(H)\Leftrightarrow
C(x)^{\star}=R^{\star}\Leftrightarrow C(x)\nsubseteq Q$ for all $Q\in
h\text{-}\operatorname{QMax}^{\star_{f}}(R)\Leftrightarrow x\notin Q$ for all
$Q\in h\text{-}\operatorname{QMax}^{\star_{f}}(R)\Leftrightarrow$ $x\in
R\backslash\bigcup_{Q\in h\text{-}\operatorname{QMax}^{\star_{f}}(R)}Q$.

(2) $(\Rightarrow)$ Let $I$ is a nonzero ideal of $R$ with
$C(I)^{\star}=R^{\star}$. Then $I\nsubseteq Q$ for all $Q\in
h\text{-}\operatorname{QMax}^{\star_{f}}(R)$, and hence
$IR_{N_{\star}(H)}=R_{N_{\star}(H)}$. Thus $I\cap N_{\star}(H)\neq\emptyset$.

$(\Leftarrow)$ Let $I$ be a nonzero ideal of $R$ such that
$I\subseteq\bigcup_{Q\in h\text{-}\operatorname{QMax}^{\star_{f}}(R)}Q$. If
$C(I)^{\star_{f}}=R^{\star}$, then, by assumption, there exists an $f\in I$
with $C(f)^{\star}=R^{\star}$. But, since $I\subseteq\bigcup_{Q\in
h\text{-}\operatorname{QMax}^{\star_{f}}(R)}Q$, we have $f\in Q$ for some
$Q\in h\text{-}\operatorname{QMax}^{\star_{f}}(R)$, a contradiction. Thus
$C(I)^{\star}\subsetneq R^{\star}$, and hence $I\subseteq Q$ for some $Q\in
h\text{-}\operatorname{QMax}^{\star_{f}}(R)$. Thus $\\{QR_{N_{\star}(H)}|Q\in
h\text{-}\operatorname{QMax}^{\star_{f}}(R)\\}$ is the set of maximal ideals
of $R_{N_{\star}(H)}$ by [23, Proposition 4.8]. ∎

We will say that $R$ satisfies property $(\\#_{\star})$ if, for any nonzero
ideal $I$ of $R$, $C(I)^{\star}=R^{\star}$ implies that there exists an $f\in
I$ such that $C(f)^{\star}=R^{\star}$.

######  Example 2.8.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain, and
let $\star$ be a semistar operation on $R$. If $R$ contains a unit of nonzero
degree, then $R$ satisfies property $(\\#_{\star})$ (see [4, Example 1.6] for
the case $\star=t$).

The next result is a generalization of the fact that
$I^{\widetilde{\star}}=I\operatorname{Na}(R,\star)\cap K$, where $K$ is the
quotient field of $R$ [18, Proposition 3.4(3)].

######  Lemma 2.9.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain, and
$\star$ be a semistar operation on $R$ such that $R^{\star}\subsetneq R_{H}$,
with property $(\\#_{\star})$. Then
$I^{\widetilde{\star}}=IR_{N_{\star}(H)}\cap R_{H}$ and
$I^{\widetilde{\star}}R_{N_{\star}(H)}=IR_{N_{\star}(H)}$ for each homogeneous
ideal $I$ of $R$. In particular $R^{\widetilde{\star}}$ is integrally closed
if and only if $R_{N_{\star}(H)}$ is integrally closed.

###### Proof.

If $I^{\widetilde{\star}}=IR_{N_{\star}(H)}\cap R_{H}$, then it is easy to see
that $I^{\widetilde{\star}}R_{N_{\star}(H)}=IR_{N_{\star}(H)}$. Hence it
suffices to show that $I^{\widetilde{\star}}=IR_{N_{\star}(H)}\cap R_{H}$.

$(\subseteq)$ Let $f\in I^{\widetilde{\star}}(\subseteq R_{H})$, and let $J$
be a finitely generated ideal of $R$ such that $J^{\star}=R^{\star}$ and
$fJ\subseteq I$. Then $C(J)^{\star}=R^{\star}$, and since $R$ satisfies
property $(\\#_{\star})$, there exists an $h\in J$ with
$C(h)^{\star}=R^{\star}$. Hence $h\in N_{\star}(H)$ and $fh\in I$. Thus $f\in
IR_{N_{\star}(H)}\cap R_{H}$.

$(\supseteq)$ Let $f=\frac{g}{h}\in IR_{N_{\star}(H)}\cap R_{H}$, where $g\in
I$ and $h\in N_{\star}(H)$. Then $fh=g\in I$, and since
$C(h)^{m+1}C(f)=C(h)^{m}C(fh)$ for some integer $m\geq 1$ by [4, Lemma
1.1(1)], we have $fC(h)^{m+1}\subseteq
C(f)C(h)^{m+1}=C(h)^{m}C(fh)=C(h)^{m}C(g)\subseteq I$. Also note that
$(C(h)^{m+1})^{\star}=R^{\star}$, since $C(h)^{\star}=R^{\star}$. Thus $f\in
I^{\widetilde{\star}}$.

For the in particular case, assume that $R_{N_{\star}(H)}$ is integrally
closed. Using [3, Proposition 2.1], $R_{H}$ is a GCD-domain, hence is
integrally closed. Therefore $R^{\widetilde{\star}}=R_{N_{\star}(H)}\cap
R_{H}$ is integrally closed. Conversely, assume that $R^{\widetilde{\star}}$
is integrally closed. Then $R_{Q}$ is integrally closed by [14, Proposition
3.8] for all $Q\in\operatorname{QSpec}^{\widetilde{\star}}(R)$. Let
$QR_{N_{\star}(H)}$ be a maximal ideal of $R_{N_{\star}(H)}$ for some $Q\in
h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$. Then
$(R_{N_{\star}(H)})_{QR_{N_{\star}(H)}}=R_{Q}$ is integrally closed. Thus
$R_{N_{\star}(H)}$ is integrally closed. ∎

######  Lemma 2.10.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain, and
$\star$ be a semistar operation on $R$ such that $R^{\star}\subsetneq R_{H}$,
with property $(\\#_{\star})$. Then for each nonzero finitely generated
homogeneous ideal $I$ of $R$, $I$ is $\star_{f}$-invertible if and only if,
$IR_{N_{\star}(H)}$ is invertible.

###### Proof.

Let $I$ be nonzero finitely generated homogeneous ideal of $R$, such that $I$
is $\star_{f}$-invertible. Let
$QR_{N_{\star}(H)}\in\operatorname{Max}(R_{N_{\star}(H)})$, where $Q\in
h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$ by Lemma 2.7(2). Thus by [22,
Theorem 2.23], $(IR_{N_{\star}(H)})_{QR_{N_{\star}(H)}}=IR_{Q}$ is invertible
(is principal) in $R_{Q}$. Hence $IR_{N_{\star}(H)}$ is invertible by [23,
Theorem 7.3]. Conversely, assume that $I$ is finitely generated, and
$IR_{N_{\star}(H)}$ is invertible. By flatness we have
$I^{-1}R_{N_{\star}(H)}=(R:I)R_{N_{\star}(H)}=(R_{N_{\star}(H)}:IR_{N_{\star}(H)})=(IR_{N_{\star}(H)})^{-1}$.
Therefore,
$(II^{-1})R_{N_{\star}(H)}=(IR_{N_{\star}(H)})(I^{-1}R_{N_{\star}(H)})=(IR_{N_{\star}(H)})(IR_{N_{\star}(H)})^{-1}=R_{N_{\star}(H)}$.
Hence $II^{-1}\cap N_{\star}(H)\neq\emptyset$. Let $f\in II^{-1}\cap
N_{\star}(H)$. So that
$R^{\star}=C(f)^{\star}\subseteq(II^{-1})^{\star_{f}}\subseteq R^{\star}$.
Thus $I$ is $\star_{f}$-invertible. ∎

######  Corollary 2.11.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain, and
$\star$ be a semistar operation on $R$ such that $R^{\star}\subsetneq R_{H}$,
with property $(\\#_{\star})$ and $0\neq f\in R$. Then the following
conditions are equivalent:

  * (1)

$C(f)$ is $\star_{f}$-invertible.

  * (2)

$C(f)R_{N_{\star}(H)}$ is invertible.

  * (3)

$C(f)R_{N_{\star}(H)}=fR_{N_{\star}(H)}$.

###### Proof.

Exactly is the same as [4, Corollary 1.9]. ∎

Let $\mathbb{Z}$ be the additive group of integers. Clearly, the direct sum
$\Gamma\oplus\mathbb{Z}$ of $\Gamma$ with $\mathbb{Z}$ is a torsionless
grading monoid. So if $y$ is an indeterminate over
$R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$, then $R[y,y^{-1}]$ is a graded
integral domain graded by $\Gamma\oplus\mathbb{Z}$. In the following
proposition we use a technique for defining semistar operations on integral
domains, due to Chang and Fontana [9, Theorem 2.3].

######  Proposition 2.12.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain with
quotient field $K$, let $y$, $X$ be two indeterminates over $R$ and let
$\star$ be a semistar operation on $R$ such that $R^{\star}\subsetneq R_{H}$.
Set $T:=R[y,y^{-1}]$, $K_{1}:=K(y)$ and take the following subset of
$\operatorname{Spec}(T)$:

$\triangle^{\star}:=\\{Q\in\operatorname{Spec}(T)|\text{ }Q\cap R=(0)\text{ or
}Q=(Q\cap R)R[y,y^{-1}]\text{ and }(Q\cap R)^{\star_{f}}\subsetneq
R^{\star}\\}.$

Set $S^{\star}:=T[X]\backslash(\bigcup\\{Q[X]|Q\in\triangle^{\star}\\})$ and:

$E^{\star\prime}:=E[X]_{S^{\star}}\cap K_{1},\text{ for all
}E\in\overline{\mathcal{F}}(T).$

  * (a)

The mapping
$\star\prime:\overline{\mathcal{F}}(T)\to\overline{\mathcal{F}}(T)$, $E\mapsto
E^{\star\prime}$ is a stable semistar operation of finite type on $T$, i.e.,
$\widetilde{\star\prime}=\star\prime$.

  * (b)

$(\widetilde{\star})\prime=(\star_{f})\prime=\star\prime$.

  * (c)

$(ER[y,y^{-1}])^{\star\prime}\cap K=E^{\widetilde{\star}}$ for all
$E\in\overline{\mathcal{F}}(R)$.

  * (d)

$(ER[y,y^{-1}])^{\star\prime}=E^{\widetilde{\star}}R[y,y^{-1}]$ for all
$E\in\overline{\mathcal{F}}(R)$.

  * (e)

$T^{\star\prime}\subsetneq T_{H^{\prime}}$, where $H^{\prime}$ is the set of
nonzero homogeneous elements of $T$, and $\star\prime$ sends homogeneous
fractional ideals to homogeneous ones.

  * (f)

$\operatorname{QMax}^{\star\prime}(T)=\\{Q|Q\in\operatorname{Spec}(T)\text{
such that }Q\cap R=(0)\text{ and
}c_{R}(Q)^{\star_{f}}=R^{\star}\\}\cup\\{PR[y,y^{-1}]|P\in\operatorname{QMax}^{\star_{f}}(R)\\}$.

  * (g)

$h$-$\operatorname{QMax}^{\star\prime}(T)=\\{PR[y,y^{-1}]|P\in
h\text{-}\operatorname{QMax}^{\widetilde{\star}}(R)\\}$.

  * (h)

$(w_{R})\prime=(t_{R})\prime=(v_{R})\prime=w_{T}$.

###### Proof.

Set $\nabla^{\star}:=\\{Q\in\operatorname{Spec}(T)|\text{ }Q\cap R=(0)$ and
$c_{D}(Q)^{\star_{f}}=R^{\star}$ or $Q=PR[y,y^{-1}]$ and
$P\in\operatorname{QMax}^{\star_{f}}(D)\\}.$ Then it is easy to see that the
elements of $\nabla^{\star}$ are the maximal elements of $\triangle^{\star}$
(see proof of [9, Theorem 2.3]). Thus

$S^{\star}:=T[X]\backslash(\bigcup\\{Q[X]|Q\in\triangle^{\star}\\})=T[X]\backslash(\bigcup\\{Q[X]|Q\in\nabla^{\star}\\}).$

(a) It follows from [9, Theorem 2.1 (a) and (b)], that $\star\prime$ is a
stable semistar operation of finite type on $T$.

(b) Since
$\operatorname{QMax}^{\star_{f}}(D)=\operatorname{QMax}^{\widetilde{\star}}(D)$,
the conclusion follows easily from the fact that
$S^{\widetilde{\star}}=S^{\star_{f}}=S^{\star}$.

(c) and (d) Exactly are the same as proof of [9, Theorem 2.3(c) and (d)].

(e) From part (d) we have
$T^{\star\prime}=R^{\widetilde{\star}}R[y,y^{-1}]\subsetneq
R_{H}R[y,y^{-1}]=T_{H^{\prime}}$. The second assertion follows from
Proposition 2.3, since $\widetilde{\star\prime}=\star\prime$ by (a).

(f) Follows from [9, Theorem 2.1(e)] and the remark in the first paragraph in
the proof.

(g) Let $M\in h$-$\operatorname{QMax}^{\star\prime}(T)$. Since $y,y^{-1}\in
T$, clearly we have $M\cap R\neq(0)$. Then by (f), there is
$P\in\operatorname{QMax}^{\star_{f}}(R)$ such that $M\subseteq PR[y,y^{-1}]$.
If $P\in h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$, then
$M=PR[y,y^{-1}]$ and we are done. So suppose that $P\notin
h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$. Then note that $P_{h}\in
h$-$\operatorname{QSpec}^{\widetilde{\star}}(R)$ and $M\subseteq
P_{h}R[y,y^{-1}]=(PR[y,y^{-1}])_{h}$; hence $M=P_{h}R[y,y^{-1}]$, because $M$
is a homogeneous maximal quasi-$\star\prime$-ideal. Note that in this case
$P_{h}\in h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$ by [16, Lemma 4.1,
Remark 4.5]. So that $M\in\\{PR[y,y^{-1}]|P\in
h\text{-}\operatorname{QMax}^{\widetilde{\star}}(R)\\}$. The other inclusion
is trivial.

(h) Suppose that $\star_{f}=t$. Note that if
$M\in\operatorname{QMax}^{\star\prime}(T)$, and $M\cap R\neq(0)$, then,
$M=(M\cap R)[y,y^{-1}]$ and $M\cap R\in\operatorname{QMax}^{t}(R)$ (cf. [24,
Proposition 1.1]). Moreover, if $Q\in\operatorname{Spec}(T)$ is such that
$Q\cap R=(0)$, then $Q$ is a quasi-$t$-maximal ideal of $T$ if and only if
$c_{R}(Q)^{t}=R$. Indeed, if $Q$ is a quasi-$t$-maximal ideal of $T$, and
$c_{R}(Q)^{t}\subsetneq R$, then there exists a quasi-$t$-maximal ideal $P$ of
$R$ such that $c_{R}(Q)^{t}\subseteq P$. Hence $Q\subseteq P[y,y^{-1}]$, and
therefore $Q=P[y,y^{-1}]$. Consequently $(0)=Q\cap R=P[y,y^{-1}]\cap R=P$
which is a contradiction. Conversely assume that $c_{R}(Q)^{t}=R$. Suppose $Q$
is not a quasi-$t$-maximal ideal of $T$, and let $M$ be a quasi-$t$-maximal
ideal of $T$ which contains $Q$. Since the containment is proper, we have
$M\cap R\neq(0)$. Thus $M=(M\cap R)[y,y^{-1}]$ and $M\cap
R\in\operatorname{QMax}^{t}(R)$ (cf. [24, Proposition 1.1]). Since $Q\subseteq
M$, $c_{R}(Q)$ is contained in the quasi-$t$-ideal $M\cap R$, so that
$c_{R}(Q)^{t}\neq R$ which is a contradiction. Thus we showed that
$\operatorname{QMax}^{t}(T)=\\{Q|Q\in\operatorname{Spec}(T)\text{ such that
}Q\cap R=(0)\text{ and
}c_{R}(Q)^{\star_{f}}=R^{\star}\\}\cup\\{PR[y,y^{-1}]|P\in\operatorname{QMax}^{\star_{f}}(R)\\}=\operatorname{QMax}^{\star\prime}(T)$,
where the second equality is by (f). Thus using (a) and (b), we obtain
$(w_{R})\prime=(t_{R})\prime=(v_{R})\prime=w_{T}$. ∎

It is known that $Pic(D(X))=0$ [1, Theorem 2]. More generally, if $*$ is a
star operation on $D$, then $Pic(\operatorname{Na}(D,*))=0$, [26, Theorem
2.14]. Also in the graded case it is shown in [4, Theorem 3.3], that
$Pic(R_{N_{v}(H)})=0$, where $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ is a
graded integral domain containing a unit of nonzero degree. We next show in
general that $Pic(R_{N_{\star}(H)})=0$.

######  Theorem 2.13.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain with
a unit of nonzero degree, and $\star$ be a semistar operation on $R$ such that
$R^{\star}\subsetneq R_{H}$. Then $Pic(R_{N_{\star}(H)})=0$.

###### Proof.

Let $y$ be an indeterminate over $R$, and $T=R[y,y^{-1}]$. Using Proposition
2.12(e) and (g) and Lemma 2.7, we deduce that
$\operatorname{Max}(T_{N_{\star\prime}(H)})=\\{QT_{N_{\star\prime}(H)}|Q\in
h$-$\operatorname{QMax}^{\star_{f}}(R)\\}$. Next since
$\operatorname{Max}((R_{N_{\star}(H)})(y))=\\{P(y)|P$ is a maximal ideal of
$R_{N_{\star}(H)}\\}$, [23, Proposition 33.1], we have
$\operatorname{Max}((R_{N_{\star}(H)})(y))=\\{(QR_{N_{\star}(H)})(y)|Q\in
h$-$\operatorname{QMax}^{\star_{f}}(R)\\}$. Thus by a computation similar to
the proof of [4, Lemma 3.2], we obtain the equality
$T_{N_{\star\prime}(H)}=(R_{N_{\star}(H)})(y)$. The rest of the proof is
exactly the same as proof of [4, Theorem 3.3], using Proposition 2.12. ∎

Let $D$ be a domain and $T$ an overring of $D$. Let $\star$ and
$\star^{\prime}$ be semistar operations on $D$ and $T$, respectively. One says
that $T$ is _$(\star,\star^{\prime})$ -linked to_ $D$ (or that $T$ is a
$(\star,\star^{\prime})$-linked overring of $D$) if

$F^{\star}=D^{\star}\Rightarrow(FT)^{\star^{\prime}}=T^{\star^{\prime}}$

for each nonzero finitely generated ideal $F$ of $D$. (The preceding
definition generalizes the notion of “$t$-linked overring” which was
introduced in [13].) It is shown in [15, Theorem 3.8], that $T$ is a
$(\star,\star^{\prime})$-linked overring of $D$ if and only if
$\operatorname{Na}(D,\star)\subseteq\operatorname{Na}(T,\star^{\prime})$. We
need a graded analogue of linkedness.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain, and
$T$ be a homogeneous overring of $R$. Let $\star$ and $\star^{\prime}$ be
semistar operations on $R$ and $T$, respectively. We say that _$T$ is
homogeneously $(\star,\star^{\prime})$-linked overring of $R$_ if

$F^{\star}=D^{\star}\Rightarrow(FT)^{\star^{\prime}}=T^{\star^{\prime}}$

for each nonzero homogeneous finitely generated ideal $F$ of $R$. We say that
_$T$ is homogeneously $t$-linked overring of $R$_ if $T$ is homogeneously
$(t,t)$-linked overring of $R$. Also it can be seen that $T$ is homogeneously
$(\star,\star^{\prime})$-linked overring of $R$ if and only if $T$ is
homogeneously $(\widetilde{\star},\widetilde{\star^{\prime}})$-linked overring
of $R$ (cf. [15, Theorem 3.8]).

######  Example 2.14.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain, and
let $\star$ be a semistar operation on $R$ such that $R^{\star}\subsetneq
R_{H}$. Let $P\in h$-$\operatorname{QSpec}^{\widetilde{\star}}(R)$. Then,
$R_{H\backslash P}$ is a homogeneously $(\star,\star^{\prime})$-linked
overring of $R$, for all semistar operation $\star^{\prime}$ on
$R_{H\backslash P}$. Indeed assume that $F$ is a nonzero finitely generated
homogeneous ideal of $R$ such that $F^{\star}=R^{\star}$. Then we have
$F^{\widetilde{\star}}=R^{\widetilde{\star}}$. Thus using Proposition 2.6, we
have $FR_{H\backslash P}=F^{\widetilde{\star}}R_{H\backslash
P}=R^{\widetilde{\star}}R_{H\backslash P}=R_{H\backslash P}$.

######  Lemma 2.15.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain with
a unit of nonzero degree, and let $T$ be a homogeneous overring of $R$. Let
$\star$ (resp. $\star^{\prime}$) be a semistar operation on $R$ (resp. on
$T$). Then, $T$ is a homogeneously $(\star,\star^{\prime})$-linked overring of
$R$ if and only if $R_{N_{\star}(H)}\subseteq T_{N_{\star^{\prime}}(H)}$.

###### Proof.

Let $f\in R$ such that $C_{R}(f)^{\star}=R^{\star}$. Then by assumption
$C_{T}(f)^{\star^{\prime}}=(C_{R}(f)T)^{\star^{\prime}}=R^{\star^{\prime}}$.
Hence $R_{N_{\star}(H)}\subseteq T_{N_{\star^{\prime}}(H)}$. Conversely let
$F$ be a nonzero homogeneous finitely generated ideal of $R$ such that
$F^{\star}=R^{\star}$. Since $R$ has a unit of nonzero degree we can choose an
element $f\in R$ such that $C_{R}(f)=F$. From the fact that
$C_{R}(f)^{\star}=R^{\star}$, we have that $f$ is a unit in $R_{N_{\star}(H)}$
and so by assumption, $f$ is a unit in $T_{N_{\star^{\prime}}(H)}$. This
implies that
$C_{T}(f)^{\star^{\prime}}=(C_{R}(f)T)^{\star^{\prime}}=T^{\star^{\prime}}$,
i.e., $(FT)^{\star^{\prime}}=T^{\star^{\prime}}$. ∎

##  3\. Kronecker function ring

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain, $*$
an e.a.b. star operation on $R$. The graded analogue of the well known
Kronecker function ring (see [23, Theorem 32.7]) of $R$ with respect to $*$ is
defined by

$\operatorname{Kr}(R,*):=\left\\{\frac{f}{g}\bigg{|}\begin{array}[]{l}f,g\in
R,\text{ }g\neq 0,\text{ and }C(f)\subseteq C(g)^{*}\end{array}\right\\}$

in [4]. The following lemma is proved in [4, Theorems 2.9 and 3.5], for an
e.a.b. star operation $*$. We need to state it for e.a.b. semistar operations.
Since the proof is exactly the same as star operation case, we omit the proof.

######  Lemma 3.1.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain,
$\star$ an e.a.b. semistar operation on $R$, and

$\operatorname{Kr}(R,\star):=\left\\{\frac{f}{g}\bigg{|}\begin{array}[]{l}f,g\in
R,\text{ }g\neq 0,\text{ and }C(f)\subseteq C(g)^{\star}\end{array}\right\\}.$

Then

  * (1)

$\operatorname{Kr}(R,\star)$ is an integral domain.

In addition, if $R$ has a unit of nonzero degree, then,

  * (2)

$\operatorname{Kr}(R,\star)$ is a Bézout domain.

  * (3)

$I\operatorname{Kr}(R,\star)\cap R_{H}=I^{\star}$ for every nonzero finitely
generated homogeneous ideal $I$ of $R$.

Inspired by the work of Fontana and Loper in [20], we can generalize this
definition of $\operatorname{Kr}(R,\star)$ to all semistar operations on $R$
which send homogeneous fractional ideals, to homogeneous ones, provided that
$R$ has a unit of nonzero degree. Before doing that we need a lemma.

######  Lemma 3.2.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain,
$\star$ a semistar operation on $R$ which sends homogeneous fractional ideals
to homogeneous ones. Suppose that $a\in R$ is homogeneous and $B,F\in f(R)$,
with $B$ homogeneous and $F\subseteq R_{H}$, such that
$aF\subseteq(BF)^{\star}$. Then there exists a homogeneous $T\in f(R)$ such
that $aT\subseteq(BT)^{\star}$.

###### Proof.

Suppose that $F$ is generated by $y_{1},\cdots,y_{n}\in R_{H}$. Let
$y_{i}=\sum t_{ij}$ be the decomposition of $y_{i}$ to homogeneous elements
for $i=1,\cdots,n$. Then $ay_{i}\in(BF)^{\star}=(\sum
y_{i}B)^{\star}\subseteq(\sum t_{ij}B)^{\star}$. Since $(\sum
t_{ij}B)^{\star}$ is homogeneous we have $at_{ij}\in(\sum t_{ij}B)^{\star}$.
Let $T$ be the fractional ideal of $R$, generated by all homogeneous elements
$t_{ij}$. So that $aT\subseteq(BT)^{\star}$ and $T\in f(R)$ is homogeneous. ∎

######  Theorem 3.3.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain with
a unit of nonzero degree, $\star$ a semistar operation on $R$ which sends
homogeneous fractional ideals to homogeneous ones, and

$\operatorname{Kr}(R,\star):=\left\\{\frac{f}{g}\bigg{|}\begin{array}[]{l}f,g\in
R,g\neq 0,\text{ and there is }0\neq h\in R\\\ \text{ such that
}C(f)C(h)\subseteq(C(g)C(h))^{\star}\end{array}\right\\}.$

Then

  * (1)

$\operatorname{Kr}(R,\star)=\operatorname{Kr}(R,\star_{a})$.

  * (2)

$\operatorname{Kr}(R,\star)$ is a Bézout domain.

  * (3)

$I\operatorname{Kr}(R,\star)\cap R_{H}=I^{\star_{a}}$ for every nonzero
finitely generated homogeneous ideal $I$ of $R$.

  * (4)

If $f,g\in R$ are nonzero such that $C(f+g)^{\star}=(C(f)+C(g))^{\star}$, then
$(f,g)\operatorname{Kr}(R,\star)=(f+g)\operatorname{Kr}(R,\star)$. In
particular, $f\operatorname{Kr}(R,\star)=C(f)\operatorname{Kr}(R,\star)$ for
all $f\in R$.

###### Proof.

It it clear from the definition that
$\operatorname{Kr}(R,\star)=\operatorname{Kr}(R,\star_{f})$. Thus using Lemma
2.4, we can assume, without loss of generality, that $\star$ is a semistar
operation of finite type.

Parts (2) and (3) are direct consequences of (1) using Lemma 3.1. For the
proof of (1) we have two cases:

Case 1: Assume that $\star$ is an e.a.b. semistar operation of finite type. In
this case, for $f,g,h\in R\backslash\\{0\\}$ we have

$C(f)C(h)\subseteq(C(g)C(h))^{\star}\Leftrightarrow C(f)\subseteq
C(g)^{\star}.$

Therefore $\operatorname{Kr}(R,\star)$ -as defined in this theorem- coincides
with $\operatorname{Kr}(R,\star)$ of an e.a.b. semistar operation $\star$, as
defined in Lemma 3.1. Also in this case $\star=\star_{a}$ by [20, Proposition
4.5(5)]. Hence in this case (1) is true.

Case 2: General case. Let $\star$ be a semistar operation of finite type on
$R$. By definition it is easy to see that, given two semistar operations on
$R$ with $\star_{1}\leq\star_{2}$, then
$\operatorname{Kr}(R,\star_{1})\subseteq\operatorname{Kr}(R,\star_{2})$. Using
[20, Proposition 4.5(3)] we have $\star\leq\star_{a}$. Therefore
$\operatorname{Kr}(R,\star)\subseteq\operatorname{Kr}(R,\star_{a})$.
Conversely let $f/g\in\operatorname{Kr}(R,\star_{a})$. Then, by Case 1,
$C(f)\subseteq C(g)^{\star_{a}}$. Set $A:=C(f)$ and $B:=C(g)$. Then
$A\subseteq B^{\star_{a}}=\bigcup\\{((BH)^{\star}:H)|H\in f(R)\\}$. Suppose
that $A$ is generated by homogeneous elements $x_{1},\cdots,x_{n}\in R$. Then
there is $H_{i}\in f(R)$, such that $x_{i}H_{i}\subseteq(BH_{i})^{\star}$ for
$i=1,\cdots,n$. Choose $0\neq r_{i}\in R$ such that $F_{i}=r_{i}H_{i}\subseteq
R$. Thus $x_{i}F_{i}\subseteq(BF_{i})^{\star}$. Therefore Lemma 3.2 gives a
homogeneous $T_{i}\in f(R)$ such that $x_{i}T_{i}\subseteq(BT_{i})^{\star}$.
Now set $T:=T_{1}T_{2}\cdots T_{n}$ which is a finitely generated homogeneous
fractional ideal of $R$ such that $AT\subseteq(BT)^{\star}$. Now since $R$ has
a unit of nonzero degree, we can find an element $h\in R$ such that $C(h)=T$.
Then $C(f)C(h)\subseteq(C(g)C(h))^{\star}$. This means that
$f/g\in\operatorname{Kr}(R,\star)$ to complete the proof of (1).

The proof of (4) is exactly the same as [4, Theorem 2.9(3)]. ∎

##  4\. Graded P$\star$MDs

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain,
$\star$ be a semistar operation on $R$, $H$ be the set of nonzero homogeneous
elements of $R$, and $N_{\star}(H)=\\{f\in R|C(f)^{\star}=R^{\star}\\}$. In
this section we define the notion of graded Prüfer $\star$-multiplication
domain (graded P$\star$MD for short) and give several characterization of it.

We say that a graded integral domain $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$
with a semistar operation $\star$, is a _graded Prüfer $\star$-multiplication
domain (graded P$\star$MD)_ if every nonzero finitely generated homogeneous
ideal of $R$ is a $\star_{f}$-invertible, i.e.,
$(II^{-1})^{\star_{f}}=R^{\star}$ for every nonzero finitely generated
homogeneous ideal $I$ of $R$. It is easy to see that a graded P$\star$MD is
the same as a graded P$\star_{f}$MD by definition, and is the same as a graded
P$\widetilde{\star}$MD by [22, Proposition 2.18]. When $\star=v$ we recover
the classical notion of a _graded Prüfer $v$-multiplication domain (graded
P$v$MD)_ [2]. It is known that $R$ is a graded P$v$MD if and only if $R$ is a
P$v$MD [2, Theorem 6.4].

Also when $\star=d$, a graded P$d$MD is called a _graded Prüfer domain_ [4].
It is clear that every graded Prüfer domain is a graded P$v$MD and hence a
P$v$MD. In particular every graded Prüfer domain is an integrally closed
domain. Although $R$ is a graded P$v$MD if and only if $R$ is a P$v$MD,
Anderson and Chang in [4, Example 3.6] provided an example of a graded Prüfer
domain which is not Prüfer. It is known that if $A,B,C$ are ideals of an
integral domain $D$, then $(A+B)(A+C)(B+C)=(A+B+C)(AB+AC+BC)$. Thus
$R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ is a graded Prüfer domain if and
only if every nonzero ideal of $R$ generated by two homogeneous elements is
invertible. We use this result in this section without comments.

The following proposition is inspired by [23, Theorem 24.3].

######  Proposition 4.1.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain.
Then the following conditions are equivalent:

  * (1)

$R$ is a graded Prüfer domain.

  * (2)

Each finitely generated nonzero homogeneous ideal of $R$ is a cancelation
ideal.

  * (3)

If $A,B,C$ are finitely generated homogeneous ideals of $R$ such that $AB=AC$
and $A$ is nonzero, then $B=C$.

  * (4)

$R$ is integrally closed and there is a positive integer $n>1$ such that
$(a,b)^{n}=(a^{n},b^{n})$ for each $a,b\in H$.

  * (5)

$R$ is integrally closed and there exists an integer $n>1$ such that
$a^{n-1}b\in(a^{n},b^{n})$ for each $a,b\in H$.

###### Proof.

The implications $(1)\Rightarrow(2)\Rightarrow(3)$ and $(4)\Rightarrow(5)$ are
clear.

$(3)\Rightarrow(4)$ By the same argument as in the proof of part
$(2)\Rightarrow(3)$, in [23, Proposition 24.1], we have that $R$ is integrally
closed in $R_{H}$. Therefore by [3, Proposition 5.4], $R$ is integrally
closed. Now if $a,b\in H$ we have $(a,b)^{3}=(a,b)(a^{2},b^{2})$. Thus by (3)
we obtain that $(a,b)^{2}=(a^{2},b^{2})$.

$(5)\Rightarrow(1)$ If (5) holds then [23, Proposition 24.2], implies that
each nonzero homogeneous ideal generated by two homogeneous elements is
invertible. Therefore $R$ is a graded Prüfer domain. ∎

The ungraded version of the following theorem is due to Gilmer (see [23,
Corollary 28.5]).

######  Theorem 4.2.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain with
a unit of nonzero degree. Then $R$ is a graded Prüfer domain if and only if
$C(f)C(g)=C(fg)$ for all $f,g\in R_{H}$.

###### Proof.

$(\Rightarrow)$ Let $f,g\in R_{H}$. Then by [4, Lemma 1.1(1)], there exists
some positive integer $n$ such that $C(f)^{n+1}C(g)=C(f)^{n}C(fg)$. Now since
$R$ is a graded Prüfer domain, the homogeneous fractional ideal $C(f)^{n}$ is
invertible. Thus $C(f)C(g)=C(fg)$ for all $f,g\in R_{H}$.

$(\Leftarrow)$ Let $\alpha\in H$ be a unit of nonzero degree. Assume that
$C(f)C(g)=C(fg)$ for all $f,g\in R_{H}$. Hence $R$ is integrally closed by [2,
Theorem 3.7]. Now let $a,b\in H$ be arbitrary. We can choose a positive
integer $n$ such that $\deg(a)\neq\deg(\alpha^{n}b)$. So that
$C(a+\alpha^{n}b)=(a,b)$. Hence, since
$(a+\alpha^{n}b)(a-\alpha^{n}b)=a^{2}-(\alpha^{n}b)^{2}$, we have
$(a,b)(a,-b)=(a^{2},-b^{2})$. Consequently $(a,b)^{2}=(a^{2},b^{2})$. Thus by
Proposition 4.1, we see that $R$ is a graded Prüfer domain. ∎

######  Lemma 4.3.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain and
$P$ be a homogeneous prime ideal. Then, the following statements are
equivalent:

  * (1)

$R_{H\backslash P}$ is a graded Prüfer domain

  * (2)

$R_{P}$ is a valuation domain.

  * (3)

For each nonzero homogeneous $u\in R_{H}$, $u$ or $u^{-1}$ is in
$R_{H\backslash P}$.

###### Proof.

$(1)\Rightarrow(2)$ Suppose that $R_{H\backslash P}$ is a graded Prüfer
domain. In particular $R_{H\backslash P}$ is a (graded) P$v$MD and each
nonzero homogeneous ideal of $R_{H\backslash P}$ is a $t$-ideal. So that
$h$-$\operatorname{QMax}^{t}(R_{H\backslash P})=\\{PR_{H\backslash P}\\}$.
Thus by [10, Lemma 2.7], we see that $(R_{H\backslash P})_{PR_{H\backslash
P}}=R_{P}$ is a valuation domain.

$(2)\Rightarrow(3)$ Let $0\neq u\in R_{H}$. Thus by the hypothesis $u$ or
$u^{-1}$ is in $R_{P}$. Thus $u$ or $u^{-1}$ is in $R_{H\backslash P}$.

$(3)\Rightarrow(1)$ Let $I,J$ be two nonzero homogeneous ideals of
$R_{H\backslash P}$ and assume that $I\nsubseteq J$. So there is a homogeneous
element $a\in I\backslash J$. For each $b\in J$, we have $\frac{a}{b}\notin
R_{H\backslash P}$, since otherwise we have $a=(\frac{a}{b})b\in J$. Thus by
the hypothesis $\frac{b}{a}\in R_{H\backslash P}$. Hence $b=(\frac{b}{a})a\in
I$. Thus we showed that $J\subseteq I$, and so every two homogeneous ideal are
comparable.

Now Let $(a,b)$ be an ideal generated by two homogeneous elements of
$R_{H\backslash P}$. Now by the first paragraph $(a,b)=(a)$ or $(a,b)=(b)$.
Thus $(a,b)$ is invertible. Hence $R_{H\backslash P}$ is a graded Prüfer
domain. ∎

######  Theorem 4.4.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain, and
$\star$ be a semistar operation on $R$ such that $R^{\star}\subsetneq R_{H}$.
Then, the following statements are equivalent:

  * (1)

$R$ is a graded P$\star$MD.

  * (2)

$R_{H\backslash P}$ is a graded Prüfer domain for each $P\in
h$-$\operatorname{QSpec}^{\widetilde{\star}}(R)$.

  * (3)

$R_{H\backslash P}$ is a graded Prüfer domain for each $P\in
h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$.

  * (4)

$R_{P}$ is a valuation domain for each $P\in
h$-$\operatorname{QSpec}^{\widetilde{\star}}(R)$.

  * (5)

$R_{P}$ is a valuation domain for each $P\in
h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$.

###### Proof.

$(2)\Rightarrow(3)$ is trivial, and, $(2)\Leftrightarrow(4)$ and
$(3)\Leftrightarrow(5)$, follow from Lemma 4.3.

$(1)\Rightarrow(2)$ Let $I$ be a nonzero finitely generated homogeneous ideal
of $R$. Then $I$ is $\widetilde{\star}$-invertible. Therefore, for each $P\in
h$-$\operatorname{QSpec}^{\widetilde{\star}}(R)$, since $II^{-1}\nsubseteq P$,
we have $R_{H\backslash P}=(II^{-1})R_{H\backslash P}=IR_{H\backslash
P}I^{-1}R_{H\backslash P}=(IR_{H\backslash P})(IR_{H\backslash P})^{-1}$. So
that $IR_{H\backslash P}$ is invertible. Thus $R_{H\backslash P}$ is a graded
Prüfer domain for each $P\in h$-$\operatorname{QSpec}^{\widetilde{\star}}(R)$.

$(3)\Rightarrow(1)$ Let $I$ be a nonzero finitely generated homogeneous ideal
of $R$. Suppose that $I$ is not $\widetilde{\star}$-invertible. Hence there
exists $P\in h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$ such that
$II^{-1}\subseteq P$. Thus $R_{H\backslash P}=(IR_{H\backslash
P})(IR_{H\backslash P})^{-1}=II^{-1}R_{H\backslash P}\subseteq PR_{H\backslash
P}$, which is a contradiction. So that $II^{-1}\nsubseteq P$ for each $P\in
h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$. Therefore
$(II^{-1})^{\widetilde{\star}}=R^{\widetilde{\star}}$, that is $I$ is
$\widetilde{\star}$-invertible, and hence $R$ is a graded P$\star$MD. ∎

The ungraded version of the following theorem is due to Chang in the star
operation case [8, Theorem 3.7], and is due to Anderson, Fontana, and
Zafrullah in the case of semistar operations [6, Theorem 1.1].

######  Theorem 4.5.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain with
a unit of nonzero degree, and $\star$ be a semistar operation on $R$ such that
$R^{\star}\subsetneq R_{H}$. Then $R$ is a graded P$\star$MD if and only if
$(C(f)C(g))^{\widetilde{\star}}=C(fg)^{\widetilde{\star}}$ for all $f,g\in
R_{H}$.

###### Proof.

$(\Rightarrow)$ Let $f,g\in R_{H}$. Choose a positive integer $n$ such that
$C(f)^{n+1}C(g)=C(f)^{n}C(fg)$ by [4, Lemma 1.1(1)]. Thus
$(C(f)^{n+1}C(g))^{\widetilde{\star}}=(C(f)^{n}C(fg))^{\widetilde{\star}}$.
Since $R$ is a graded P$\star$MD, the homogeneous fractional ideal $C(f)^{n}$
is ${\widetilde{\star}}$-invertible. Thus
$(C(f)C(g))^{\widetilde{\star}}=C(fg)^{\widetilde{\star}}$ for all $f,g\in
R_{H}$.

$(\Leftarrow)$ Assume that
$(C(f)C(g))^{\widetilde{\star}}=C(fg)^{\widetilde{\star}}$ for all $f,g\in
R_{H}$. Let $P\in h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$. Then using
Proposition 2.6, we have $C(f)R_{H\backslash P}C(g)R_{H\backslash
P}=C(f)C(g)R_{H\backslash P}=(C(f)C(g))^{\widetilde{\star}}R_{H\backslash
P}=C(fg)^{\widetilde{\star}}R_{H\backslash P}=C(fg)R_{H\backslash P}$. Since
$R_{H\backslash P}$ has a unit of nonzero degree, Theorem 4.2 shows that
$R_{H\backslash P}$ is a graded Prüfer domain. Now Theorem 4.4, implies that
$R$ is a graded P$\star$MD. ∎

We now recall the notion of $\star$-valuation overring (a notion due
essentially to P. Jaffard [25, page 46]). For a domain $D$ and a semistar
operation $\star$ on $D$, we say that a valuation overring $V$ of $D$ is a
_$\star$ -valuation overring of $D$_ provided $F^{\star}\subseteq FV$, for
each $F\in f(D)$.

######  Remark 4.6.

(1) Let $\star$ be a semistar operation on a graded integral domain
$R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$. Recall that for each $F\in f(R)$ we
have

$F^{\star_{a}}=\bigcap\\{FV|V\text{ is a }\star\text{-valuation overring of
}R\\},$

by [19, Propositions 3.3 and 3.4 and Theorem 3.5].

(2) We have $N_{\star}(H)=N_{\widetilde{\star}_{a}}(H)$. Indeed, since
$\widetilde{\star}\leq\widetilde{\star}_{a}$ by [20, Proposition 4.5], we have
$N_{\star}(H)=N_{\widetilde{\star}}(H)\subseteq N_{\widetilde{\star}_{a}}(H)$.
Now if $f\in R\backslash N_{\star}(H)$ then,
$C(f)^{\widetilde{\star}}\subsetneq R^{\widetilde{\star}}$. Thus there is a
homogeneous quasi-$\widetilde{\star}$-prime ideal $P$ of $R$ such that
$C(f)\subseteq P$. Let $V$ be a valuation domain dominating $R_{P}$ with
maximal ideal $M$ [23, Corollary 19.7]. Therefore $V$ is a
$\widetilde{\star}$-valuation overring of $R$ by [18, Theorem 3.9], and
$C(f)V\subseteq M$; so $C(f)^{(\widetilde{\star})_{a}}\subsetneq
R^{(\widetilde{\star})_{a}}$ and $f\notin N_{\widetilde{\star}_{a}}(H)$. Thus
we obtain that $N_{\star}(H)=N_{\widetilde{\star}_{a}}(H)$.

In the following theorem we generalize a characterization of P$v$MDs proved by
Arnold and Brewer [7, Theorem 3]. It also generalizes [8, Theorem 3.7], [4,
Theorems 3.4 and 3.5], and [17, Theorem 3.1].

######  Theorem 4.7.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain with
a unit of nonzero degree, and $\star$ be a semistar operation on $R$ such that
$R^{\star}\subsetneq R_{H}$. Then, the following statements are equivalent:

  * (1)

$R$ is a graded P$\star$MD.

  * (2)

Every ideal of $R_{N_{\star}(H)}$ is extended from a homogeneous ideal of $R$.

  * (3)

Every principal ideal of $R_{N_{\star}(H)}$ is extended from a homogeneous
ideal of $R$.

  * (4)

$R_{N_{\star}(H)}$ is a Prüfer domain.

  * (5)

$R_{N_{\star}(H)}$ is a Bézout domain.

  * (6)

$R_{N_{\star}(H)}=\operatorname{Kr}(R,\widetilde{\star})$.

  * (7)

$\operatorname{Kr}(R,\widetilde{\star})$ is a quotient ring of $R$.

  * (8)

$\operatorname{Kr}(R,\widetilde{\star})$ is a flat $R$-module.

  * (9)

$I^{\widetilde{\star}}=I^{\widetilde{\star}_{a}}$ for each nonzero homogeneous
finitely generated ideal of $R$.

In particular if $R$ is a graded P$\star$MD, then $R^{\widetilde{\star}}$ is
integrally closed.

###### Proof.

By Proposition 2.3 and Theorem 3.3, we have
$\operatorname{Kr}(R,\widetilde{\star})$ is well-defined and is a Bézout
domain.

$(1)\Rightarrow(2)$ Let $0\neq f\in R$. Then $C(f)$ is
$\widetilde{\star}$-invertible, because $R$ is a graded P$\star$MD, and thus
$fR_{N_{\star}(H)}=C(f)R_{N_{\star}(H)}$ by Corollary 2.11. Hence if $A$ is an
ideal of $R_{N_{\star}(H)}$, then $A=IR_{N_{\star}(H)}$ for some ideal $I$ of
$R$, and thus $A=(\sum_{f\in I}C(f))R_{N_{\star}(H)}$.

$(2)\Rightarrow(3)$ Clear.

$(3)\Rightarrow(1)$ Is the same as part $(3)\Rightarrow(1)$ in [4, Theorem
3.4].

$(1)\Rightarrow(4)$ Let $A$ be a nonzero finitely generated ideal of
$R_{N_{\star}(H)}$. Then by Corollary 2.11, $A=IR_{N_{\star}(H)}$ for some
nonzero finitely generated homogeneous ideal $I$ of $R$. Since $R$ is a graded
P$\star$MD, $I$ is $\widetilde{\star}$-invertible, and thus
$A=IR_{N_{\star}(H)}$ is invertible by Lemma 2.10.

$(4)\Rightarrow(5)$ Follows from Theorem 2.13.

$(5)\Rightarrow(6)$ Clearly
$R_{N_{\star}(H)}\subseteq\operatorname{Kr}(R,\widetilde{\star})$. Since
$R_{N_{\star}(H)}$ is a Bézout domain, then
$\operatorname{Kr}(R,\widetilde{\star})$ is a quotient ring of
$R_{N_{\star}(H)}$, by [23, Proposition 27.3]. If $Q\in
h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$, then
$Q\operatorname{Kr}(R,\widetilde{\star})\subsetneq\operatorname{Kr}(R,\widetilde{\star})$.
Otherwise
$Q\operatorname{Kr}(R,\widetilde{\star})=\operatorname{Kr}(R,\widetilde{\star})$,
and hence there is an element $f\in Q$, such that
$f\operatorname{Kr}(R,\widetilde{\star})=\operatorname{Kr}(R,\widetilde{\star})$.
Thus $\frac{1}{f}\in\operatorname{Kr}(R,\widetilde{\star})$. Therefore
$R=C(1)\subseteq C(f)^{(\widetilde{\star})_{a}}\subseteq
R^{(\widetilde{\star})_{a}}$, so that
$C(f)^{(\widetilde{\star})_{a}}=R^{(\widetilde{\star})_{a}}$. Hence $f\in
N_{(\widetilde{\star})_{a}}(H)=N_{\star}(H)$ by Remark 4.6(2). This means that
$Q^{\widetilde{\star}}=R^{\widetilde{\star}}$, a contradiction. Thus
$Q\operatorname{Kr}(R,\widetilde{\star})\subsetneq\operatorname{Kr}(R,\widetilde{\star})$,
and so there is a maximal ideal $M$ of
$\operatorname{Kr}(R,\widetilde{\star})$ such that
$Q\operatorname{Kr}(R,\star)\subseteq M$. Hence $M\cap
R_{N_{\star}(H)}=QR_{N_{\star}(H)}$, by Lemma 2.7. Consequently
$R_{Q}\subseteq\operatorname{Kr}(R,\widetilde{\star})_{M}$, and since $R_{Q}$
is a valuation domain, we have
$R_{Q}=\operatorname{Kr}(R,\widetilde{\star})_{M}$. Therefore
$R_{N_{\star}(H)}=\bigcap_{Q\in
h\text{-}\operatorname{QMax}^{\widetilde{\star}}(R)}R_{Q}\supseteq\bigcap_{M\in\operatorname{Max}(\operatorname{Kr}(R,\widetilde{\star}))}\operatorname{Kr}(R,\widetilde{\star})_{M}$.
Hence $R_{N_{\star}(H)}=\operatorname{Kr}(R,\widetilde{\star})$.

$(6)\Rightarrow(7)$ and $(7)\Rightarrow(8)$ are clear.

$(8)\Rightarrow(6)$ Recall that an overring $T$ of an integral domain $S$ is a
flat $S$-module if and only if $T_{M}=S_{M\cap S}$ for all
$M\in\operatorname{Max}(T)$ by [32, Theorem 2].

Let $A$ be an ideal of $R$ such that
$A\operatorname{Kr}(R,\widetilde{\star})=\operatorname{Kr}(R,\widetilde{\star})$.
Then there exists an element $f\in A$ such that
$f\operatorname{Kr}(R,\widetilde{\star})=\operatorname{Kr}(R,\widetilde{\star})$
using Theorem 3.3; so
$\frac{1}{f}\in\operatorname{Kr}(R,\widetilde{\star})=\operatorname{Kr}(R,\widetilde{\star}_{a})$.
Thus $R=C(1)\subseteq C(f)^{\widetilde{\star}_{a}}\subseteq
R^{\widetilde{\star}_{a}}$, and so
$C(f)^{\widetilde{\star}_{a}}=R^{\widetilde{\star}_{a}}$. Hence
$C(f)^{\widetilde{\star}}=R^{\widetilde{\star}}$. Therefore $f\in A\cap
N_{\star}(H)\neq\emptyset$. Hence, if $P_{0}$ is a homogeneous maximal
quasi-$\widetilde{\star}$-ideal of $R$, then
$P_{0}\operatorname{Kr}(R,\widetilde{\star})\subsetneq\operatorname{Kr}(R,\widetilde{\star})$,
and since $P_{0}R_{N_{\star}(H)}$ is a maximal ideal of $R_{N_{\star}(H)}$,
there is a maximal ideal $M_{0}$ of $\operatorname{Kr}(R,\widetilde{\star})$
such that $M_{0}\cap R=(M_{0}\cap R_{N_{\star}(H)})\cap
R=P_{0}R_{N_{\star}(H)}\cap R=P_{0}$. Thus by (8),
$\operatorname{Kr}(R,w)_{M_{0}}=R_{P_{0}}=(R_{N(H)})_{P_{0}R_{N(H)}}$.

Let $M_{1}$ be a maximal ideal of $\operatorname{Kr}(R,\widetilde{\star})$,
and let $P_{1}$ be a homogeneous maximal quasi-$\widetilde{\star}$-ideal of
$R$ such that $M_{1}\cap R_{N_{\star}(H)}\subseteq P_{1}R_{N_{\star}(H)}$. By
the above paragraph, there is a maximal ideal $M_{2}$ of
$\operatorname{Kr}(R,\widetilde{\star})$ such that
$\operatorname{Kr}(R,\widetilde{\star})_{M_{2}}=(R_{N_{\star}(H)})_{P_{1}R_{N_{\star}(H)}}$.
Note that
$\operatorname{Kr}(R,\widetilde{\star})_{M_{2}}\subseteq\operatorname{Kr}(R,\widetilde{\star})_{M_{1}}$
, $M_{1}$ and $M_{2}$ are maximal ideals, and
$\operatorname{Kr}(R,\widetilde{\star})$ is a Prüfer domain; hence
$M_{1}=M_{2}$ (cf. [23, Theorem 17.6(c)]) and
$\operatorname{Kr}(R,\widetilde{\star})_{M_{1}}=(R_{N_{\star}(H)})_{P_{1}R_{N(H)}}$.
Thus

$\operatorname{Kr}(R,\widetilde{\star})=\bigcap_{M\in\operatorname{Max}(\operatorname{Kr}(R,\widetilde{\star}))}\operatorname{Kr}(R,\widetilde{\star})_{M}=\bigcap_{P\in
h\text{-}\operatorname{QMax}^{\widetilde{\star}}(R)}(R_{N_{\star}(H)})_{PR_{N_{\star}(H)}}=R_{N_{\star}(H)}.$

$(6)\Rightarrow(9)$ Assume that
$R_{N_{\star}(H)}=\operatorname{Kr}(R,\widetilde{\star})$. Let $I$ be a
nonzero homogeneous finitely generated ideal of $R$. Then by Lemma 2.9 and
Theorem 3.3(3), we have $I^{\widetilde{\star}}=IR_{N_{\star}(H)}\cap
R_{H}=I\operatorname{Kr}(R,\widetilde{\star})\cap
R_{H}=I^{\widetilde{\star}_{a}}$.

$(9)\Rightarrow(1)$ Let $a$ and $b$ be two nonzero homogeneous elements of
$R$. Then
$((a,b)^{3})^{\widetilde{\star}_{a}}=((a,b)(a^{2},b^{2}))^{\widetilde{\star}_{a}}$
which implies that
$((a,b)^{2})^{\widetilde{\star}_{a}}=(a^{2},b^{2})^{\widetilde{\star}_{a}}$.
Hence $((a,b)^{2})^{\widetilde{\star}}=(a^{2},b^{2})^{\widetilde{\star}}$ and
so $(a,b)^{2}R_{H\backslash P}=(a^{2},b^{2})R_{H\backslash P}$ for each
homogeneous maximal quasi-$\widetilde{\star}$-ideal $P$ of $R$. On the other
hand $R^{\widetilde{\star}}=R^{\widetilde{\star}_{a}}$ by (9). Hence
$R^{\widetilde{\star}}$ is integrally closed. Thus
$R^{\widetilde{\star}}R_{H\backslash P}=R_{H\backslash P}$ is integrally
closed. Therefore by Proposition 4.1, $R_{H\backslash P}$ is a graded Prüfer
domain for each homogeneous maximal quasi-$\star_{f}$-ideal of $R$. Thus $R$
is a graded P$\star$MD by Theorem 4.4. ∎

The following theorem is a graded version of a characterization of Prüfer
domains proved by Davis [12, Theorem 1]. It also generalizes [13, Theorem
2.10], in the $t$-operation, and [15, Theorem 5.3], in the case of semistar
operations.

######  Theorem 4.8.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain with
a unit of nonzero degree, and $\star$ be a semistar operation on $R$ such that
$R^{\star}\subsetneq R_{H}$. Then, the following statements are equivalent:

  * (1)

$R$ is a graded P$\star$MD.

  * (2)

Each homogeneously $(\star,t)$-linked overring of $R$ is a P$v$MD.

  * (3)

Each homogeneously $(\star,d)$-linked overring of $R$ is a graded Prüfer
domain.

  * (4)

Each homogeneously $(\star,t)$-linked overring of $R$, is integrally closed.

  * (5)

Each homogeneously $(\star,d)$-linked overring of $R$, is integrally closed.

###### Proof.

$(1)\Rightarrow(2)$ Let $T$ be a homogeneously $(\star,t)$-linked overring of
$R$. Thus by Lemma 2.15, we have $R_{N_{\star}(H)}\subseteq T_{N_{v}(H)}$.
Since $R$ is a graded P$\star$MD, by Theorem 4.7, we have $R_{N_{\star}(H)}$
is a Prüfer domain. Thus by [23, Theorem 26.1], we have $T_{N_{v}(H)}$ is a
Prüfer domain. Hence, again by Theorem 4.7, we have $T$ is a graded P$v$MD.
Therefore using [2, Theorem 6.4], $T$ is a P$v$MD.

$(2)\Rightarrow(4)\Rightarrow(5)$ and $(3)\Rightarrow(5)$ are clear.

$(5)\Rightarrow(1)$ Let $P\in h$-$\operatorname{QMax}^{\widetilde{\star}}(R)$.
For a nonzero homogeneous $u\in R_{H}$, let $T=R[u^{2},u^{3}]_{H\backslash
P}$. Then $R_{H\backslash P}$ and $T$ are homogeneous $(\star,d)$-linked
overring of $R$ by Example 2.14. So that $R_{H\backslash P}$ and $T$ are
integrally closed. Hence $u\in T$, and since $T=R_{H\backslash
P}[u^{2},u^{3}]$, there exists a polynomial $\gamma\in R_{H\backslash P}[X]$
such that $\gamma(u)=0$ and one of the coefficients of $\gamma$ is a unit in
$R_{H\backslash P}$. So $u$ or $u^{-1}$ is in $R_{H\backslash P}$ by [27,
Theorem 67]. Therefore by Lemma 4.3, $R_{H\backslash P}$ is a graded Prüfer
domain. Thus $R$ is a graded P$\star$MD by Theorem 4.4.

$(1)\Rightarrow(3)$ Is the same argument as in part $(1)\Rightarrow(2)$. ∎

The next result gives new characterizations of P$v$MDs for graded integral
domains, which is the special cases of Theorems 4.4, 4.5, 4.7, and 4.8, for
$\star=v$.

######  Corollary 4.9.

Let $R=\bigoplus_{\alpha\in\Gamma}R_{\alpha}$ be a graded integral domain with
a unit of nonzero degree. Then, the following statements are equivalent:

  * (1)

$R$ is a (graded) P$v$MD.

  * (2)

$R_{H\backslash P}$ is a graded Prüfer domain for each $P\in
h$-$\operatorname{QMax}^{t}(R)$.

  * (3)

$R_{P}$ is a valuation domain for each $P\in h$-$\operatorname{QMax}^{t}(R)$.

  * (4)

Every ideal of $R_{N_{v}(H)}$ is extended from a homogeneous ideal of $R$.

  * (5)

$R_{N_{v}(H)}$ is a Prüfer domain.

  * (6)

$R_{N_{v}(H)}$ is a Bézout domain.

  * (7)

$R_{N_{v}(H)}=\operatorname{Kr}(R,w)$.

  * (8)

$\operatorname{Kr}(R,w)$ is a quotient ring of $R$.

  * (9)

$\operatorname{Kr}(R,w)$ is a flat $R$-module.

  * (10)

Each homogeneously $t$-linked overring of $R$ is a P$v$MD.

  * (11)

Each homogeneously $t$-linked overring of $R$, is integrally closed.

  * (12)

$(C(f)C(g))^{w}=C(fg)^{w}$ for all $f,g\in R_{H}$.

  * (13)

$I^{w}=I^{w_{a}}$ for each nonzero homogeneous finitely generated ideal of
$R$.

ACKNOWLEDGMENT

I am grateful to the referee for carefully reading the first version of this
article.

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