Let $S$ be a multiplicatively closed subset of a ring $R$, and let $I$ be an ideal of $R$ which is maximal among ideals disjoint from $S$. Show that $I$ is prime.

If $R$ is an integral domain, explain briefly how one may construct a field $F$ together with an injective ring homomorphism $R \rightarrow F$.

Deduce that if $R$ is an arbitrary ring, $I$ an ideal of $R$, and $S$ a multiplicatively closed subset disjoint from $I$, then there exists a ring homomorphism $f: R \rightarrow F$, where $F$ is a field, such that $f(x)=0$ for all $x \in I$ and $f(y) \neq 0$ for all $y \in S$.

[You may assume that if $T$ is a multiplicatively closed subset of a ring, and $0 \notin T$, then there exists an ideal which is maximal among ideals disjoint from $T$.]