multiplicatively closedMultiplicativelyClosed2007-08-18 17:03:532007-10-13 09:49:22Definitionmultiplicative setsaturated multiplicatively closed\usepackage{amssymb,amscd}
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\newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2}}
\newcommand{\mpdiff}[3]{\frac{\partial^#1 #2}{\partial #3^#1}}Let $R$ be a ring. A subset $S$ of $R$ is said to be \emph{multiplicatively closed} if $S\ne \varnothing$, and whenever $a,b\in S$, then $ab\in S$. In other words, $S$ is a multiplicative set where the multiplication defined on $S$ is the multiplication inherited from $R$.
For example, let $a\in R$, the set $S:=\lbrace a^i, a^{i+1}, \cdots, a^n, \cdots \rbrace$ is multiplicatively closed for any positive integer $i$. Another simple example is the set $\lbrace 1\rbrace$, if $R$ is unital.
\textbf{Remarks}. Let $R$ be a commutative ring.
\begin{itemize}
\item
If $P$ is a prime ideal in $R$, then $R-P$ is multiplicatively closed.
\item
Furthermore, an ideal maximal with respect to the being disjoint from a multiplicative set not containing $0$ is a prime ideal.
\item
In particular, assuming $1\in R$, any ideal maximal with respect to being disjoint from $\lbrace 1\rbrace$ is a maximal ideal.
\end{itemize}
A multiplicatively closed set $S$ in a ring $R$ is said to be \emph{saturated} if for any $a\in S$, every divisor of $a$ is also in $S$.
In the example above, if $i=1$ and $a$ has no divisors, then $S$ is saturated.
\textbf{Remarks}.
\begin{itemize}
\item
In a unital ring, a saturated multiplicatively closed set always contains $U(R)$, the group of units of $R$ (since it contains $1$, and therefore, all divisors of $1$). In particular, $U(R)$ itself is saturated multiplicatively closed.
\item Assume $R$ is commutative. $S\subseteq R$ is saturated multiplicatively closed and $0\notin S$ iff $R-S$ is a union of prime ideals in $R$.
\begin{proof}
This can be shown as follows: if let $T$ be a union of prime ideals in $R$ and $a,b\in R-T$. if $ab\notin R-T$, then $ab\in P\subseteq T$ for some prime ideal $P$. Therefore, either $a$ or $b\in P\subseteq T$. This contradicts the assumption that $a,b\notin T$. So $R-T$ is multiplicatively closed. If $ab \in R-T$ with $a\notin R-T$, then $a\in P\subseteq T$ for some prime ideal $P$, which implies $ab\in P\subseteq T$ also. This contradicts the assumption that $ab\notin T$. This shows that $R-T$ is saturated. Of course, $0\notin R-T$, since $0$ lies in any ideal of $R$.
Conversely, assume $S$ is saturated multiplicatively closed and $0\notin S$. For any $r\notin S$, we want to find a prime ideal $P$ containing $r$ such that $P\cap S=\varnothing$. Once we show this, then take the union $T$ of these prime ideals and that $S=R-T$ is immediate. Let $\langle r\rangle$ be the principal ideal generated by $r$. Since $S$ is saturated, $\langle r\rangle\cap S=\varnothing$. Let $M$ be the set of all ideals containing $\langle r\rangle$ and disjoint from $S$. $M$ is non-empty by construction, and we can order $M$ by inclusion. So $M$ is a poset and Zorn's lemma applies. Take any chain $C$ in $M$ containing $\langle r\rangle$ and let $P$ be the maximal element in $C$. Then any ideal larger than $P$ must not be disjoint from $S$, so $P$ is prime by the second remark in the first set of remarks.
\end{proof}
\item The notion of multiplicative closure can be generalized to be defined over any non-empty set with a binary operation (multiplication) defined on it.
\end{itemize}
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\bibitem{ik} I. Kaplansky, {\it Commutative Rings}. University of Chicago Press, 1974.
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