%From mroitman@math2.haifa.ac.il  Sun Dec  3 08:18:28 2000

\documentclass{proc-l}
\usepackage{amsmath,amssymb}%

\renewcommand\theenumi{(\arabic{enumi})}
\renewcommand\labelenumi{\theenumi\ }

\newtheorem{thm}{Theorem}[section]
\newtheorem{prop}[thm]{Proposition}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{cor}[thm]{Corollary}
\newtheorem{rem}[thm]{Remark}
\newtheorem{exmp}[thm]{Example}

\DeclareMathOperator{\Spec}{Spec}
\DeclareMathOperator{\h-Spec}{h-Spec}
\DeclareMathOperator\rad{rad}
\DeclareMathOperator{\rank}{rank}
\DeclareMathOperator{\hgt}{ht}
\commby{Wolmer Vasconcelos}

\title[Homogeneous spectrum ]
{The homogeneous spectrum of a graded  commutative ring}


\author{William Heinzer}
\address{%
Department of Mathematics,
Purdue University, West Lafayette, Indiana 47907-1395}
\email{heinzer@math.purdue.edu}

\author{Moshe Roitman}
\address{%
Department of Mathematics,
University of Haifa, Mount Carmel, Haifa 31905, Israel}
\email{mroitman@math.haifa.ac.il}


\keywords
{ graded ring, homogeneous spectrum, Noetherian spectrum,
torsion-free  cancellative commutative monoid}

\subjclass{ 13A15, 13E99}

\thanks
{This work was prepared while M. Roitman enjoyed the
hospitality of Purdue University.}


\begin{document}

\baselineskip 17pt


\begin{abstract} Suppose $\Gamma$ is a torsion-free
cancellative commutative monoid for which the group of quotients is
finitely generated.  We prove that the spectrum of a $\Gamma $-graded
commutative ring is Noetherian if its homogeneous spectrum is
Noetherian, thus answering a question of David Rush. Suppose $A$ is a
commutative ring having Noetherian spectrum. We determine conditions
in order that the monoid ring $A[\Gamma]$ have Noetherian spectrum. If
$\rank \Gamma \le 2$, we show that $A[\Gamma]$ has Noetherian
spectrum, while for each $n \ge 3$ we establish existence of an
example where the homogeneous spectrum of $A[\Gamma]$ is not
Noetherian.
\end{abstract}

\maketitle



\setcounter{section}{-1}

\section{Introduction.}

All rings we consider are assumed to be nonzero, commutative and with
unity.  All the monoids are assumed to be torsion-free cancellative
commutative monoids.  Let $\Gamma$ be a monoid such that the group of
quotients $G$ of $\Gamma$ is finitely generated, and let $R =
\bigoplus_{\gamma \in \Gamma}R_\gamma$ be a commutative
$\Gamma$-graded ring. A goal of this paper is to answer in the
affirmative a question mentioned to one of us by David Rush as to
whether $\Spec R$ is necessarily Noetherian provided the homogeneous
spectrum, $\h-Spec R$, is Noetherian.


If $I$ is an ideal of a ring $R$, we let $\rad(I)$ denote the radical
of $I$, that is $\rad(I) = \{r \in R : r^n \in I \text{ for some
positive integer } n \}.$ We say that $I$ is a {\it radical ideal } if
$\rad(I) = I$. A subset $S$ of the ideal $I$ {\em generates $I$ up to
radical} if $\rad(I)=\rad(SR)$.  The ideal $I$ is {\it radically finite
} if it is generated up to radical by a finite set.

We recall that a ring $R$ is said to have {\it Noetherian spectrum }
if the set $\Spec R$ of prime ideals of $R$ with the Zariski topology
satisfies the descending chain condition on closed subsets. In
ideal-theoretic terminology, $R$ has Noetherian spectrum if and only
if $R$ satisfies the ascending chain condition (a.c.c.) on radical
ideals. Thus a Noetherian ring has Noetherian spectrum and each ring
having only finitely many prime ideals has Noetherian spectrum. As
shown in \cite[Prop. 2.1]{OP}, $\Spec R$ is Noetherian if and only if
each ideal of $R$ is radically finite.  It is well known that $R$ has
Noetherian spectrum if and only if $R$ satisfies the two properties:
(i) a.c.c. on prime ideals, and (ii) every ideal of $R$ has only
finitely many minimal prime ideals \cite{Mo}, \cite[Theorem 88,
page~59 and Ex.~25, page~65]{Ka}.

In analogy with the result of Cohen that a ring $R$ is Noetherian if
each prime ideal of $R$ is finitely generated, it is shown in
\cite[Corollary~2.4]{OP} that $R$ has Noetherian spectrum if each
prime ideal of $R$ is radically finite.  It is shown in
\cite[Theorem~2.5]{OP} that Noetherian spectrum is preserved under
polynomial extension in finitely many indeterminates. Thus finitely
generated algebras over a ring with Noetherian spectrum again have
Noetherian spectrum.


In Section 1 we prove that if $R$ is a $\Gamma$-graded ring, where
$\Gamma$ is a monoid with finitely generated group of quotients, and
if $\h-Spec R$ is Noetherian, then $\Spec R$ is Noetherian (Theorem
\ref{homog}).  In Section 2 we deal with monoid rings.  It turns out
that if $M$ is a monoid with finitely generated group of quotients and
$k$ is a field, then the homogeneous spectrum of the monoid ring
$k[M]$ is not necessarily Noetherian (Example \ref{notnoeth}).  On the
positive side, $\h-Spec A[M]$ is Noetherian if $A$ is a ring with
Noetherian spectrum and $M$ is a monoid of torsion-free rank $\le2$.

We thank David Rush for pointing out to us several errors in an
earlier version of this paper.

\section{The homogeneous spectrum }
The {\it homogeneous spectrum, } $\h-Spec R$, of a graded ring $R =
\bigoplus_{\gamma \in \Gamma}R_\gamma$ is the set of homogeneous prime
ideals of $R$. The most common choices for the commutative monoid
$\Gamma$ are the monoid $\mathbb N$ of nonnegative integers or its
group of quotients $\mathbb Z$. A standard technique using homogeneous
localization shows the following: if $R = \bigoplus_{n \in \mathbb
Z}R_n$ is a $\mathbb Z$-graded integral domain, if $t$ is a nonzero
element of $R_1$, and if $H$ is the multiplicative set of nonzero
homogeneous elements of $R$, then the localization $R_H$ of $R$ with
respect to $H$ is the graded Laurent polynomial ring $K_0[t, t^{-1}]$,
where $K_0$ is a field \cite[page~157]{ZS}.  This implies the
following remark.


\begin{rem} {\em
Suppose $R = \bigoplus_{n \in \mathbb Z}R_n$ is a graded integral
domain and $P$ is a nonzero prime ideal of $R$. If zero is the only
homogeneous element contained in $P$, then the localization $R_P$ is
one-dimensional and Noetherian. }
\end{rem}

If $R = \bigoplus_{n \in \mathbb Z}R_n$ is a graded ring with no
nonzero homogeneous prime ideals, then $R_0$ is a field and either $R
= R_0$, or $R$ is a Laurent polynomial ring $R_0[x, x^{-1}]$
\cite[page~83]{E}.


Every ring can be viewed as a graded ring with the trivial gradation
that assigns degree zero to every element of the ring.  Thus Nagata in
\cite[Section 8]{N} develops primary decomposition for graded ideals
in a graded Noetherian ring.  It is not surprising that there is an
interrelationship among the Noetherian properties of $\Spec R$, $\h-Spec
R$, $\Spec R[X]$ and $\h-Spec R[X]$.

Proposition \ref{pnoe} is useful in considering the Noetherian
property of spectra. It follows by induction from
\cite[Prop.~2.2~(ii)]{OP}, but we prefer to prove it directly.

\begin{prop}
\label{pnoe}
Let $I$ be an ideal of a ring $R$. Let $J$ be an ideal of $R$ and $S$
a subset of $J$ such that $J=\rad SR$. If $I + J$ is radically finite
and if for each $s\in S$, the ideal $IR[1/s]$ is radically finite,
then $I$ is radically finite.
\end{prop}

\begin{proof}
Since $I + J$ is radically finite and since $J=\rad SR$, there exist
finite sets $F\subseteq I$ and $G\subseteq S$ such that
$\rad(I+J)=\rad((F, G)R)$.  For each $g\in G$ there exists a finite
subset $T_g$ of $I$ such that $\rad(IR[1/g])=\rad (T_gR[1/g])$. Let
$I' = (F\cup\bigcup_{g\in G}T_g)R$, thus $I'\subseteq I$.  Suppose $P
\in \Spec R$ and $I' \subseteq P$. If $G \subseteq P$, then $I
\subseteq P$ since $\rad(I' + GR) = \rad(I + J)$.  Otherwise, we have
$g\notin P$ for some element $g\in G$.  Therefore $\rad(I'R[1/g]) =
\rad(IR[1/g]) \subseteq PR[1/g]$.  Since $P$ is the preimage in $R$ of
$PR[1/g]$, we have $\rad(I) \subseteq P$.  Therefore $\rad(I') =
\rad(I)$, so $I$ is radically finite.
\end{proof}

For Corollary \ref{snoe}, we use that the (homogeneous) spectrum of a
graded ring $R$ is Noetherian iff each (homogeneous) ideal of $R$ is
radically finite.


\begin{cor}\label{snoe}
\begin{enumerate}
\item%(1)
Let $S$ be a finite subset of a ring $R$.  If $\Spec(R/SR)$ is
Noetherian and for each $s \in S$, $\Spec(R[1/s])$ is Noetherian, then
$\Spec R$ is Noetherian.
\item%(2)
Let $S$ be a finite set of homogeneous elements of a graded ring $R$.  If
$\h-Spec(R/SR)$ is Noetherian and for each $s \in S$,
$\h-Spec(R[1/s])$ is Noetherian, then $\h-Spec R$ is Noetherian.
\end{enumerate}
\end{cor}


The hypotheses in Proposition \ref{pnoe} and Corollary \ref{snoe}
concerning the set $S$ may be modified as follows and still give the
same conclusion:
\begin{prop}\label{var}
Let $I$ be an ideal of a ring $R$, and let $S$ be a finite subset of
$R$. Let $U$ be the multiplicatively closed subset of $R$ generated by
$S$.

\begin{enumerate}

\item
If $I + sR$ is radically finite for each $s\in S$ and $IR_U$ is
radically finite, then $I$ is radically finite.

\item
If $\Spec(R/sR)$ is Noetherian for each $s\in S$ and $\Spec R_U$ is
Noetherian, then $\Spec R$ is Noetherian.

\item
If $R$ is a $\Gamma$-graded ring for some monoid $\Gamma$, each $s\in
S$ is homogeneous with $\h-Spec (R/sR)$ Noetherian and if $\h-Spec
R_U$ is Noetherian, then $\h-Spec R$ is Noetherian.

\end{enumerate}

\end{prop}

The next Corollary is a special case of Proposition \ref{pnoe}.


\begin{cor}\label{qnoe}
Suppose $S$ is a subset of a ring $R$ that generates $R$ as an ideal
and let $I$ be an ideal of $R$.  If $IR[1/s]$ is radically finite for
each $s \in S$, then $I$ is radically finite.

If $R[1/s]$ has Noetherian spectrum for each $s \in S$, then $R$ has
Noetherian spectrum.
\end{cor}


In analogy with Corollary \ref{qnoe}, it is a standard result in
commutative algebra that if $SR = R$ and $R[1/s]$ is a Noetherian ring
for each $s \in S$, then $R$ is a Noetherian ring.  However, the
analogue of Corollary \ref{snoe} for the Noetherian property of a ring
is false:  There exists a non-Noetherian ring $R$ and an element $s
\in R$ such that $R/sR$ and $R[1/s]$ are Noetherian.  For example, let
$X, Y$ be indeterminates over a field $k$, let $R := k[X,
\{Y/X^n\}_{n=0}^\infty]$ and let $s = X$. Then
$P=(\{Y/X^n\}_{n=0}^\infty )$ is a nonfinitely generated prime ideal
of $R$,  so $R$ is not Noetherian,  although both $R/XR = k$ and
$R[1/X] = k[X, Y, 1/X]$ are Noetherian.
Incidentally, both the ideal $(P+XR)/XR=(0)$ of $R/XR$ and
the ideal $PR[1/X]$ of $R[1/X]$ are principal.


Proposition \ref{op} is the graded analogue of \cite[Theorem~2.5]{OP}.

\begin{prop}\label{op}
Suppose $\Gamma$ is a torsion-free cancellative commutative monoid
with group of quotients $G$ and $R = \bigoplus_{\gamma \in
\Gamma}R_\gamma$ is a $\Gamma$-graded commutative ring. Fix $g \in G$,
and consider the polynomial ring $R[X]$ as a graded extension ring of
$R$ uniquely determined by defining $X$ to be a homogeneous element of
degree $g$.  If $\h-Spec R$ is Noetherian, then $\h-Spec R[X]$ is
Noetherian.
\end{prop}

\begin{proof}
Assume that $\h-Spec R$ is Noetherian, but $\h-Spec R[X]$ is not
Noetherian.  Then there exists a homogeneous prime ideal $P$ of $R[X]$
that is maximal with respect to not being radically finite.  Since $P
\cap R = p$ is a homogeneous prime ideal of $R$ and $\h-Spec R$ is
Noetherian, we may pass from $R[X]$ to $R[X]/p[X] \cong
(R/p)[X]$ and assume that $P \cap R = (0)$.  Then $R$ is a graded domain
and $\h-Spec R$ is Noetherian.  Choose an  element $f \in P$ having
minimal degree $d$ as a polynomial in $R[X]$.  By replacing $f$ by one
of its nonzero homogeneous components, we may assume that $f = a_dX^d
+ a_{d-1}X^{d-1} + \dots + a_0$, where the elements $a_i \in R$ are
homogeneous elements of $R$ with $a_d \ne 0$.  Since $P \cap R = (0)$,
we have $d > 0$ and $a_d \not\in P$.  The maximality of $P$ with
respect to not being radically finite implies $(P,a_d)R[X]$ is
radically finite.  Since $a_d^{-1}f$ is a polynomial of minimal degree
in $PR[1/a_d][X]$ and since this polynomial is monic in $R[1/a_d][X]$,
we see that $PR[1/a_d][X]=(f)$.  But Proposition \ref{var} (1) then
implies that $P$ is radically finite, a contradiction.  \end{proof}

We use Proposition \ref{op} in the proof of Theorem \ref{homog}.

\begin{thm}\label{homog}
Let $R = \bigoplus_{\gamma \in \Gamma}R_\gamma$ be a $\Gamma$-graded
commutative ring, where $\Gamma$ is a nonzero torsion-free
cancellative commutative monoid such that its group of quotients $G$
is finitely generated.  If $\h-Spec R$ is Noetherian, then $\Spec R$
is also Noetherian.
\end{thm}

\begin{proof}
Up to a group isomorphism, we have $G \cong \mathbb Z^d$ for some
positive integer $d$.  Hence we may assume $G = \mathbb Z^d$. For $1
\le i \le d$, let $g_i$ be the element of $G$ having 1 as its $i$-th
coordinate and zeros elsewhere.  Consider the graded polynomial
extension ring $R[\mathbf X] := R[X_1, \dots, X_d]$ obtained by
defining $X_i$ to be a homogeneous element of degree $g_i$ for $i = 1,
\dots, d$.  Proposition \ref{op} implies that $\h-Spec R[\mathbf X]$
is Noetherian.  We associate with each nonzero element $r \in R$ a
homogeneous element $\widetilde r \in R[\mathbf X]$ such that
$\deg(\widetilde r) = (c_1, \dots, c_d)$, where $c_i$ is the maximum
of the $i$-th coordinates of the degrees of the nonzero homogeneous
components of $r \in R$ as follows: let $r=r_0+\dots+r_k$ be the
homogeneous decomposition of $r$; set $\widetilde
r=\sum_{i=1}^kr_i\mathbf X^{m_i}$, where $m_i=(c_1,\dots,c_d)-\deg
r_i$ for each $i$ and $\mathbf
X^{(a_1,\dots,a_d)}=\prod_{i=1}^dX_i^{a_i}$ for each sequence
$(a_1,\dots,a_d)$ in $\mathbb Z^d$. We define $\widetilde 0 = 0$. With
each ideal $I$ of $R$, let $\widetilde I$ denote the homogeneous ideal
of $R[\mathbf X]$ generated by $\{ \widetilde r : r \in I \}$
($\widetilde r$ is the {\em homogenization} of $r$ and $\widetilde I$
is the homogenization of $I$).

Let $\phi : R[\mathbf X] \to R$ denote the $R$-algebra homomorphism
defined by $\phi(X_i) = 1$ for $i = 1, \dots, n$. Since $\phi$ is an
$R$-algebra homomorphism and $\phi(\widetilde r) = r$ for each $r \in
R$, for each ideal $I$ of $R$, we have $\phi(\widetilde I) = I$ (the
meaning of $\phi$ is {\em dehomogenization}).  Therefore the map $I
\to \widetilde I$ is a one-to-one inclusion preserving correspondence
of the set of ideals of $R$ into the set of homogeneous ideals of
$R[\mathbf X]$.


Let $I$ be an ideal of $R$. Since $\h-Spec R[\mathbf X]$ is Noetherian
there exists a finite set $S$ such that $\rad \widetilde I=\rad
(SR[\mathbf X])$.  We have $\rad I=\rad \phi (\widetilde I)=\rad
(\phi(S))R$, thus $I$ is radically finite. Therefore $\Spec R$ is
Noetherian.
\end{proof}

The following corollary is immediate from Theorem \ref{homog}.

\begin{cor}
Let $R$ be an $\mathbb N$-graded or a $\mathbb Z$-graded ring. If
$\h-Spec R$ is Noetherian, then $\Spec R$ is Noetherian.
\end{cor}

Without the assumption in Theorem \ref{homog} that the group of
quotients of $\Gamma$ is finitely generated, it is possible to have
$\h-Spec R$ is Noetherian and yet $\Spec R$ is not Noetherian. For
example, if $K$ is an algebraically closed field of characteristic
zero and $\Gamma = \mathbb Q$, then $(0)$ is the only homogeneous
prime ideal of the group ring $R := K[\mathbb Q]$ so $\h-Spec R$ is
Noetherian, but as we note in Theorem \ref{monoidspec} below, $\Spec
R$ is not Noetherian.

\section{The Noetherian spectra of monoid rings }

Suppose $A$ is a ring and $M$ is a cancellative torsion-free
commutative monoid.  We consider the monoid ring $A[M]$ as a graded
ring with its natural $M$-grading where the nonzero elements of $A$
are of degree zero.  The monoid $M$ is naturally identified with a
subset of $A[M]$. We write $X^m$ for $m\in M\subseteq A[M]$. Note that
$0\in M$ is identified with $1\in A[M]$.

A {\em $\mathbb Q$-monoid} in a $\mathbb Q$-vector space $V$ is an
additive submonoid of $V$ that is closed under multiplication by
positive rationals.  A subset of a $\mathbb Q$-monoid $W$ is called a
{\em $\mathbb Q$-ideal} of the $\mathbb Q$-monoid $W$ if it is an
ideal of the monoid $W$ that is closed under multiplication by
positive (that is, strictly positive) rationals.

If $M$ is a cancellative torsion-free monoid with group of quotients
$G$, we denote by $M^{(\mathbb Q)}$ the $\mathbb Q$-monoid generated
by $M$ in $G\otimes_\mathbb Z \mathbb Q$; thus $M=\{qm\,|\, q>0\text
{ in }\mathbb Q, m\in M\}$.


\begin{rem}\label{qid} {\em
Let $S$ be a subset of a monoid $M$, let $R$ be a ring, and let $I$ be
a homogeneous ideal of $R[M]$ containing $S$ and generated
by monomials in $M$. Then $S$ generates $I$
up to radical iff $S$ generates the $\mathbb Q$-ideal $(I\cap
M)^{(\mathbb Q)}$ of $M^{(\mathbb Q)}$. }
\end{rem}


\begin{rem}\label{correspond} {\em
Suppose $M$ is a cancellative torsion-free commutative monoid and $k$
is a field.  There is a natural one-to-one inclusion preserving
correspondence between the homogeneous radical ideals of the monoid
domain $k[M]$ and the $\mathbb Q$-ideals of the $\mathbb Q$-monoid
$M^{(\mathbb Q)}$.
Indeed, to each $\mathbb Q$-ideal $L$ of $M^{(\mathbb Q)}$ (which is
generated by $L\cap M$) we make correspond the ideal of $k[M]$
generated by $L\cap M$. }
\end{rem}


\begin{lem}\label{qideal}
Suppose $M$ is a torsion-free cancellative commutative monoid, $A$ is
a ring with Noetherian spectrum, and $P$ is a homogeneous prime ideal
of the monoid ring $A[M]$. Then the following two conditions are
equivalent:
\begin{enumerate}
\item
The prime ideal $P$ is radically finite in $A[M]$.
\item
The $\mathbb Q$-ideal $(P\cap M)^{(\mathbb Q)}$ of $M^{(\mathbb Q)}$
is finitely generated.
\end{enumerate}
\end{lem}
\begin{proof}
Since $P$ is prime and homogeneous, $P$ is generated by $(P \cap
A)\cup (P\cap M)$.  Since $\Spec A$ is Noetherian, we see that $P$
is radically finite iff the ideal in $A[M]$ generated by $P\cap M$ is
radically finite iff the $\mathbb Q$-ideal $(P\cap M)^{(\mathbb Q)}$
of $M^{(\mathbb Q)}$ is finitely generated (Remark \ref{qid}).  This
proves Lemma \ref{qideal}.
\end{proof}

The following is an immediate corollary to Lemma \ref{qideal}.

\begin{cor}\label{qqideal}
Let $M$ be a torsion-free cancellative commutative monoid and let $A$
be a ring with Noetherian spectrum. Then the following two conditions
are equivalent:
\begin{enumerate}
\item
The monoid ring $A[M]$ has  Noetherian homogeneous spectrum.

\item
Each $\mathbb Q$-ideal in the $\mathbb Q$-monoid $M^{(\mathbb Q)}$ is
finitely generated.
\end{enumerate}
\end{cor}


We denote the torsion-free rank of a monoid $M$ by $\rank M$.


\begin{prop}\label{monhom}
Suppose $A$ is a ring and $M$ is a cancellative torsion-free
commutative monoid.

\begin{enumerate}

\item

If $\Spec A[M]$ is Noetherian, then $\Spec A$ is Noetherian and
$\rank M$ is finite.

\item

If $\Spec A$ is Noetherian and if $\rank M\le2$, then $\h-Spec A[M]$
is Noetherian.

\end{enumerate}
\end{prop}


\begin{proof}
\begin{enumerate}
\item
$\Spec A$ is Noetherian since every ideal $I$ of $A$ satisfies
$I=IA[M]\cap A$, and if $I$ is a radical ideal of $A$, then $IA[M]$ is
a radical ideal in $A[M]$.  If $\rank M$ is infinite, let
$B$ be an infinite set of elements in $M$ which are linearly
independent over $\mathbb Q$ in the $\mathbb Q$-vector space
$G\otimes_{\mathbb Z} \mathbb Q$, where $G$ is the group of quotients
of $M$. Then the ideal of $A[M]$ generated by  
$\{X^b - X^c : b, c \in B \}$ is not radically finite. Therefore
$\Spec A[M]$ is not Noetherian.


\item
By Lemma \ref{qideal} it suffices to show each $\mathbb Q$-ideal in
$M^{(\mathbb Q)}$ is finitely generated.  We may assume that
$M\subseteq \mathbb Q^2$. Let $W$ be a nonempty $\mathbb Q$-ideal of
$M^{(\mathbb Q)}$.  We show that $W$ is a finitely generated ideal of
$\widetilde W:=W\cup \{\mathbf 0\}$.  If $W$ spans a one-dimensional
subspace and $\mathbf v$ is a nonzero element of $W$, then the
$\mathbb Q$-ideal $W$ is generated by $\mathbf 0$ if $-\mathbf v\in
W$, and by $\mathbf v$ otherwise. If $W$ spans $\mathbb Q^2$, then
choose two linearly independent vectors in $W$. By changing
coordinates, we may assume that these vectors are $(1,0)$ and
$(0,1)$. If $W$ contains a vector $\mathbf v$ with both coordinates
strictly negative, then $W$ is generated by $\mathbf 0$ as a $\mathbb
Q$-ideal. Otherwise, define vectors $\mathbf u$ and $\mathbf v$ as
follows: if $a=\min \{y\,|\, (1,y)\in W\}$ exists, let $\mathbf
u=(1,a)$; if the minimum does not exist, let $\mathbf
u=(1,0)$. Similarly, define a vector $\mathbf v$ with second
coordinate $1$. Then $\mathbf u$ and $\mathbf v$ generate $W$ as a
$\mathbb Q$-ideal of $\widetilde W$.
\end{enumerate}\end{proof}


\begin{thm}\label{monoidspec}
Let $A$ be a ring with Noetherian spectrum and $M$ be a cancellative
torsion-free commutative monoid.  If the group of quotients of $M$ is
finitely generated and if $\rank M\le2$, then the monoid ring $A[M]$
has Noetherian spectrum.

On the other hand, if $A[M]$ has Noetherian spectrum and if $A$
contains an algebraically closed field of zero characteristic, then
the group of quotients of $M$ is finitely generated.
\end{thm}

\begin{proof}
Assume that the group of quotients of $M$ is finitely generated and
that $\rank M\le2$. By Proposition \ref{monhom} (2), $A[M]$ has
Noetherian homogeneous spectrum. By Theorem \ref{homog}, $\Spec A[M]$
is Noetherian.


For the second statement, assume that the group of quotients of $M$ is
not finitely generated. By Proposition \ref{monhom} (1), we may assume
that $M$ has finite rank. It follows that there exists an element
$s\in M$ that is divisible by infinitely many positive integers. Since
$A$ contains all roots of unity and they are distinct, we obtain that
over the element $X^s-1$ of $A[M]$ there are infinitely many minimal
primes. Therefore $\Spec A[M]$ is not Noetherian.
\end{proof}

With regard to \ref{monoidspec}, if the monoid $M$ is finitely
generated, then it follows from \cite[Theorem 2.5]{OP},
that $\Spec A[M]$ is Noetherian if $\Spec A$ is Noetherian.



\begin{exmp} {\em
Over a field $k$ of characteristic $p > 0$, there exists a monoid $M$
for which the group of quotients is not finitely generated and yet the
monoid domain $k[M]$ has Noetherian spectrum. For example, if $M :=
\mathbb Z[\{1/p^n\}_{n=1}^\infty]$, then $k[M]$ is an integral purely
inseparable extension of $k[\mathbb Z]$ and $\Spec (k[M])$ is
Noetherian. }
\end{exmp}



A {\em prime} $\mathbb Q$-ideal of a $\mathbb Q$-monoid $M$ is a
$\mathbb Q$-ideal $Q$ of $M$ that is a prime ideal, that is, if
$a+b\in Q$, then either $a \in Q$ or $b \in Q$.


Let $S$ be a subset of a vector space over $\mathbb Q$.  $S$ is {\em
$\mathbb Q$-convex} if for any points $p,q$ in $S$ and rational $0\le
t\le 1$ we have $tp+(1-t)q\in S$.


\begin{rem}\label{prime} {\em
Let $M$ be a $\mathbb Q$-monoid in a vector space over $\mathbb Q$,
and let $I$ be a subset of $M$ that is closed under addition and under
multiplication by positive rationals; thus $I$ is a $\mathbb Q$-convex
set. Then $I$ is an ideal of $M$ iff for any two points $p\in I$ and
$q\in M$ and any rational $0<t<1$, we have $tp+(1-t)q\in I$. Moreover,
for $I$ as above, if $I$ is an ideal, then $I$ is prime iff the set
$M\setminus I$ is $\mathbb Q$-convex. }
\end{rem}


We denote by $C$ the unit circle in $\mathbb R^2$, that is,
$C=\{(x,y)\in \mathbb R^2\,:\, x^2+y^2=1\}$. We let $C_\mathbb Q=C\cap
\mathbb Q^2$. For any subset $S$ of $\mathbb R^n$ we denote by $S^+$
the set of points in $S$ with nonnegative coordinates.


\begin{exmp}\label{notnoeth} {\em
Let $n\ge3$. Then there exists a cancellative torsion-free commutative
monoid $M$ of rank $n$ such that the group of quotients of $M$ is
finitely generated, but for any ring $A$ the homogeneous spectrum of
$A[M]$ is not Noetherian. Furthermore, the monoid $M$ is completely
integrally closed. Hence, if $A$ is an integrally closed (completely
integrally closed) domain, then $A[M]$ is an integrally closed
(completely integrally closed) domain. }
\end{exmp}

First let $n=3$. Let $W$ be the $\mathbb Q$-submonoid of $\mathbb Q^3$
generated by the set $\left\{(x,y,1)\,:\, (x,y)\in C_\mathbb
Q\right\}$. We claim that the $\mathbb Q$-ideal $W\setminus\{\mathbf
0\}$ of $W$ is not finitely generated; moreover, if 
$(p,1)\in C_\mathbb Q\times \{1\}$, then
$(p,1)$ does not belong to the $\mathbb Q$-ideal of $W$ generated by
$C_\mathbb Q\times \{1\}\setminus \{( p,1)\}$.  Indeed, by Remark
\ref{prime}, the set of points $(x,y,z)\in W$ such that $\dfrac
1z(x,y)\ne p$ is a $\mathbb Q$-ideal of $W$ which does not contain
$(p,1)$.  Set $M=W\cap \mathbb Z^3$. More explicitly, since the convex
hull of $C_\mathbb Q$ equals the rational unit disk, we see that
$M=\left\{X^aY^bZ^c\,|\, (a,b,c)\in \mathbb Z^3, c\ge 0\text{ and
}a^2+b^2\le c^2\right\}$.

Now let $A$ be any ring. Since the $\mathbb Q$-ideal generated by
$W\setminus\{\mathbf 0\}$ in $W$ is not finitely generated, we obtain
by Lemma \ref{qideal} that the ideal in $A[M]$ generated by the
nonzero elements of $M$  is not radically finite; thus $\h-Spec A[M]$ is
not Noetherian.

If $n>3$ let $\widetilde M=M\oplus \mathbb Z^{n-3}$, where $M$ is the
monoid defined above. Then  $\rank \widetilde M=n$ and $\widetilde
M$ satisfies our requirements.

Clearly, $M$ is a completely integrally closed monoid. Thus the
assertions on $A[M]$ follow from \cite[Corollary 12.7 (2) and
Corollary 12.11 (2)]{G}.  \qed

We now elaborate on Example \ref{notnoeth}, but with $W$ replaced by
$W^+$.  As seen in Example \ref{notnoeth}, $R$ is a completely
integrally closed domain, and $\h-Spec R$ is not Noetherian. Moreover,
$R = k[M]$ is a subring of the polynomial ring $k[X,Y,Z]$ and has
fraction field $k(X,Y,Z)$.  By \cite[Theorem 21.4]{G}, $\dim R=3$.  It
is interesting that the maximal homogeneous ideal $N$ of $R$ has
height $3$, but its homogeneous height (defined using just homogeneous
prime ideals) is $2$.  Indeed, let $P\ne N$ be a nonzero prime
homogeneous ideal of $R$. Let $Q$ be the $\mathbb Q$-ideal of $W$
generated by the points $(a,b,c)$ in $\mathbb Q^3$ such that
$X^aY^bZ^c\in P$. Since $Q$ is a prime $\mathbb Q$-ideal of $W$ and
since $C_\mathbb Q^+$ is dense in $C^+$, by Remark \ref{prime} we
easily obtain that $Q$ contains $C_\mathbb Q^+\times \{1\}$ except one
point.  Thus the homogeneous height of $N$ is at most $2$. Since the
$\mathbb Q$-ideal of $W$ generated by $C_\mathbb Q^+\times \{1\}$ with
one point removed is prime, we see that the homogeneous height of $N$
is $2$.

On the other hand, $\hgt N =3 $.  More generally, if $R$ is a
$k$-subalgebra of a polynomial ring $k[\mathbf X]:= k[X_1,\dots,X_n]$
over a field $k$ with the same fraction field $k(\mathbf X)$, then
$\hgt( (\mathbf X)k[\mathbf X]\cap R) =n$.  Indeed, this prime ideal
has height at most $n$ since $k(\mathbf X)$ has transcendence degree
$n$ over $k$.  Moreover, each nonzero ideal of $k[\mathbf X]$ has a
nonzero intersection with $R$.  Since the primes of height $n-1$ of
$k[\mathbf X]$ contained in $(\mathbf X)k[\mathbf X]$ intersect in
zero, there exists such a prime ideal $P_{n-1}$ of $k[\mathbf X]$ such
that $P_{n-1} \cap R \subsetneq (\mathbf X)k[\mathbf X] \cap R$.
Repeating this argument, we find a strictly descending chain of prime
ideals contained in $R$:  $(\mathbf X)k[\mathbf X] \cap R\supsetneq
(P_{n-1} \cap R) \supsetneq\dots\supsetneq P_0=(0)$.

This behavior where the dimension of the homogeneous spectrum of a
graded integral domain $R$ is less than $\dim R$ also occurs in the
case where $R$ is an $\mathbb N$-graded integral domain.  For example,
if $A$ is a one-dimensional quasilocal integral domain such that the
polynomial ring $A[X]$ has dimension three \cite{S}, then the
homogeneous spectrum of $A[X]$ in its natural $\mathbb N$-grading has
dimension two.

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\end{document}

