Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 3 / 11F / Solution

Use the standard commutative-ring convention with identity and a nonempty multiplicative subset. An ideal disjoint from that subset is proper. Suppose $ab\in I$ but $a,b\notin I$. Maximality among disjoint ideals means that both $I+(a)$ and $I+(b)$ meet $S$. Thus $s=i+ra\in S$ and $t=j+ub\in S$ for some $i,j\in I$, $r,u\in R$. Their product belongs to $S$ by multiplicative closure, buta contradiction. Consequently $I$ is prime. This is prime ideal avoiding a multiplicative subset.

For an integral domain, form its field of fractions from pairs $(a,b)$ with $b\ne 0$, identifying $(a,b)$ and $(c,d)$ when $ad=bc$. Write the class as $a/b$ and define addition and multiplication byThe domain condition makes the equivalence relation and these operations well-defined. Every nonzero $a/b$ has inverse $b/a$. The map $a\mapsto a/1$ is an injective ring homomorphism because $a/1=0/1$ implies $a=0$.

For the final assertion, pass to $R/I$. The image $\overline{S}$ is multiplicatively closed and does not contain zero, because $S\cap I=\varnothing$. Apply the allowed existence result to obtain an ideal $J$ of $R/I$ maximal among those disjoint from $\overline{S}$. The first proof makes $J$ prime. Its inverse image $P$ in $R$ is therefore prime, contains $I$ and avoids $S$. The quotient $R/P$ is an integral domain; composeThis homomorphism kills every element of $I$. No element of $S$ maps to zero, because the first map kills precisely $P$ and the second is injective. Thus it has both required properties. Commutativity is essential to the general statement; it is the ring convention under which the printed prime-ideal argument is used.