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 a ring homomorphism $R\to F$.
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 $f\colon R\to F$, 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 $T$.]
Here is a question I need to answer. For the first part I can show if $x, y$ are not in $I$ then nor does $xy$. The second part is just the field of fractions.
For the third part, I think I need to find an ideal $J$ containing $I$ such $J$ is prime, so that $R/J$ is an integral domain, and use the second part. To find $J$ prime, I think as suggested by the hint, I should go for a maximal ideal disjoint from $S$ containing $I$, but how can I do that?