# multiplicative sets in rings and prime ideals

Let $R$ be a commutative ring, $S\subseteq R$ a mutliplicative subset of $R$ such that $0\not\in S$. Then there exists prime ideal $P\subseteq R$ such that $P\cap S=\emptyset$.

Proof. Consider the family $\mathcal{A}=\{I\subseteq R\ |\ I\mbox{ is an ideal and }I\cap S=\emptyset\}$. Of course $\mathcal{A}\neq\emptyset$, because the zero ideal $0\in\mathcal{A}$. We will show, that $\mathcal{A}$ is inductive (i.e. satisfies Zorn’s Lemma’s assumptions) with respect to inclusion.

Let $\{I_{k}\}_{k\in K}$ be a chain in $\mathcal{A}$ (i.e. for any $a,b\in K$ either $I_{a}\subseteq I_{b}$ or $I_{b}\subseteq I_{a}$). Consider $I=\bigcup_{k\in K}I_{k}$. Obviously $I$ is an ideal. Furthermore, if $x\in I\cap S$, then there is $k\in K$ such that $x\in I_{k}\cap S=\emptyset$. Thus $I\cap S=\emptyset$, so $I\in\mathcal{A}$. Lastely each $I_{k}\subseteq I$, which completes this part of proof.

By Zorn’s Lemma there is a maximal element $P\in\mathcal{A}$. We will show that this ideal is prime. Let $x,y\in R$ be such that $xy\in P$. Assume that neither $x\not\in P$ nor $y\not\in P$. Then $P\subset P+(x)$ and $P\subset P+(y)$ and these inclusions are proper. Therefore both $P+(x)$ and $P+(y)$ do not belong to $\mathcal{A}$ (because $P$ is maximal). This implies that there exist $a\in(P+(x))\cap S$ and $b\in(P+(y))\cap S$. Thus

 $a=m_{1}+r_{1}x\in S;\ \ \ \ b=m_{2}+r_{2}y\in S;$

where $m_{1},m_{2}\in P$ and $r_{1},r_{2}\in R$. Note that $ab\in S$. We calculate

 $ab=(m_{1}+r_{1}x)(m_{2}+r_{2}y)=m_{1}m_{2}+m_{2}r_{1}x+m_{1}r_{2}y+xyr_{1}r_{2}.$

Of course $m_{1}m_{2},m_{2}r_{1}x,m_{1}r_{2}y\in P$, because $m_{1},m_{2}\in P$ and $xyr_{1}r_{2}\in P$ by our assumption that $xy\in P$. This shows, that $ab\in P$. But $ab\in S$ and $P\in\mathcal{A}$. Contradiction. $\square$

Corollary. Let $R$ be a commutative ring, $I$ an ideal in $R$ and $S\subseteq R$ a multiplicative subset such that $I\cap S=\emptyset$. Then there exists prime ideal $P$ in $R$ such that $I\subseteq P$ and $P\cap S=\emptyset$.

Proof. Let $\pi:R\to R/I$ be the projection. Then $\pi(S)\subseteq R/I$ is a multiplicative subset in $R/I$ such that $0+I\not\in\pi(S)$ (because $I\cap S=\emptyset$). Thus, by proposition, there exists a prime ideal $P$ in $R/I$ such that $P\cap\pi(S)=\emptyset$. Of course the preimage of a prime ideal is again a prime ideal. Furthermore $I\subseteq\pi^{-1}(P)$. Finaly $\pi^{-1}(P)\cap S=\emptyset$, because $P\cap\pi(S)=\emptyset$. This completes the proof.

Title multiplicative sets in rings and prime ideals MultiplicativeSetsInRingsAndPrimeIdeals 2013-03-22 19:03:58 2013-03-22 19:03:58 joking (16130) joking (16130) 5 joking (16130) Theorem msc 16U20 msc 13B30