multiplicative sets in rings and prime ideals

Proposition. Let $R$ be a commutative ring, $S\subseteq R$ a mutliplicative subset of $R$ such that $0\notin S$. Then there exists prime ideal $P\subseteq R$ such that $P\cap S=\mathrm{\varnothing }$.

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

Let ${\{I_k\}}_{k\in K}$ be a chain in $𝒜$ (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 $x\in I_k\cap S=\mathrm{\varnothing }$. Thus $I\cap S=\mathrm{\varnothing }$, so $I\in 𝒜$. Lastely each $I_k\subseteq I$, which completes this part of proof.

By Zorn’s Lemma there is a maximal element $P\in 𝒜$. We will show that this ideal is prime. Let $x,y\in R$ be such that $xy\in P$. Assume that neither $x\notin P$ nor $y\notin P$. 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 $𝒜$ (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

$$ab=(m_1+r_{1}x)(m_2+r_{2}y)=m_1{m}_2+m_2{r}_{1}x+m_1{r}_{2}y+xy{r}_1{r}_2.$$

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 $xy{r}_1{r}_2\in P$ by our assumption that $xy\in P$. This shows, that $ab\in P$. But $ab\in S$ and $P\in 𝒜$. Contradiction. $\mathrm{\square }$

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

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 $0+I\notin \pi (S)$ (because $I\cap S=\mathrm{\varnothing }$). Thus, by proposition, there exists a prime ideal $P$ in $R/I$ such that $P\cap \pi (S)=\mathrm{\varnothing }$. Of course the preimage of a prime ideal is again a prime ideal. Furthermore $I\subseteq {\pi }^{-1}(P)$. Finaly ${\pi }^{-1}(P)\cap S=\mathrm{\varnothing }$, because $P\cap \pi (S)=\mathrm{\varnothing }$. This completes the proof.