Proposition. 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 $$a=m_1+r_1x\in S;\ \ \ \ b=m_2+r_2y\in S;$$ where $m_1,m_2\in P$ and $r_1,r_2\in R$ . Note that $ab\in S$ . We calculate $$ab=(m_1+r_1x)(m_2+r_2y)=m_1m_2+m_2r_1x+m_1r_2y+xyr_1r_2.$$ Of course $m_1m_2, m_2r_1x, m_1r_2y\in P$ , because $m_1,m_2\in P$ and $xyr_1r_2\in P$ 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.