Take the tour ×  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 constrcut 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 integ