A prime containing an ideal and avoiding a multiplicative set

Let $\mathrm{\Sigma }$ be the set of ideals $J⊴R$ with $I\subseteq J$ and $J\cap S=\mathrm{\varnothing }$, ordered by inclusion. The ideal $I$ lies in $\mathrm{\Sigma }$, so $\mathrm{\Sigma }\mathrm{\ne }\mathrm{\varnothing }$.

If $\mathcal{C}\subseteq \mathrm{\Sigma }$ is a chain, then $J=\bigcup \mathcal{C}$ is an ideal containing $I$. Moreover $J\cap S=\mathrm{\varnothing }$, because if $s\in J\cap S$ then $s\in C$ for some $C\in \mathcal{C}$, contradicting $C\cap S=\mathrm{\varnothing }$. Thus every chain in $\mathrm{\Sigma }$ has an upper bound.

Zorn's lemma yields a maximal member $\mathfrak{p}$ of $\mathrm{\Sigma }$. Because $1\in S$ by [L2] and $\mathfrak{p}\cap S=\mathrm{\varnothing }$, the ideal $\mathfrak{p}$ is proper.

Suppose $ab\in \mathfrak{p}$ while $a\mathrm{\notin }\mathfrak{p}$ and $b\mathrm{\notin }\mathfrak{p}$. By maximality of $\mathfrak{p}$ in $\mathrm{\Sigma }$, the larger ideals $\mathfrak{p}+(a)$ and $\mathfrak{p}+(b)$ must meet $S$. Choose $\nu =m+ra\in S\cap (\mathfrak{p}+(a))$ and $\omega =n+tb\in S\cap (\mathfrak{p}+(b))$ with $m,n\in \mathfrak{p}$ and $r,t\in R$. Then $\nu \omega =mn+mtb+nra+rtab\in \mathfrak{p}$, because $m,n,ab\in \mathfrak{p}$. But $\nu \omega \in S$ by [L2], contradicting $\mathfrak{p}\cap S=\mathrm{\varnothing }$. Therefore $\mathfrak{p}$ is prime by [L1].

The ideal $\mathfrak{p}$ is prime, contains $I$, and is disjoint from $S$, exactly as required.