Consider all <ideals> containing $I$ and disjoint from the nonempty <multiplicative subset> $S$. This collection is nonempty because it contains $I$. The <ascending chain condition> for the <Noetherian ring> gives a maximal member $P$: otherwise repeatedly choosing a strictly larger member would give an infinite strictly ascending chain of ideals. Since $S$ is nonempty and $P$ misses it, $P\ne R$.

Suppose $ab\in P$ but $a,b\notin P$. Maximality implies that $P+(a)$ and $P+(b)$ both meet $S$, so write $s=p+ra\in S$ and $t=q+ub\in S$ with $p,q\in P$. Then
$$
st=pq+pub+qra+ruab\in P,
$$
while closure under products gives $st\in S$, a contradiction. Hence
$$
\boxed{P\supseteq I,\quad P\cap S=\varnothing,\quad P\text{ is prime}.}
$$
This proves the <prime ideal avoiding a multiplicative subset> result without needing $1\in S$. The assumptions themselves rule out $0\in S$, since every ideal contains zero.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 2 / 11E / b / Solution

In a <Noetherian ring>, an <ideal> $I$ disjoint from a nonempty <multiplicative subset> $S$ extends to an ideal $P$ maximal among those disjoint from $S$, by the <ascending chain condition>. It is proper. If $ab\in P$ but neither factor belongs to $P$, both $P+(a)$ and $P+(b)$ meet $S$. Multiplying such representatives puts an element of $S$ in $P$, a contradiction. Thus $P$ is a <prime ideal>. For a general <commutative ring> the same argument follows after applying the <Zorn lemma> to disjoint ideals; the union of a chain is disjoint from $S$.

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