I ran across this when reading another question on math.se, so I posted this proof for verification just to be sure I wasn't missing something.

We wish to characterise prime ideals as precisely those that are disjoint from a multiplicative subset $S \subseteq R$. That is:

• An ideal $P$ is prime iff $P = R \setminus S$, where $S$ is a multiplicative subset that cannot be made larger (ie, is maximal wrt to the $\subseteq$ ordering).

I'll be using the definition of prime as:

• An ideal $P$ is prime if for all $x, y \in R$, $xy \in P \implies x \in P \lor y \in P$.

§ Prime ideal implies complement is maximal multiplicative subset:

Let $S = \equiv R \setminus P$ be the complement of the prime ideal $P \subsetneq R$ in question.

• Since $P \neq R$, $1 \not \in$P. (if $1 \in P$, then every element $x . 1 \in P$since $P$ is an ideal, and must be closed under multiplication with the entire ring). Hence, $1 \in S$.
• For any $x, y \in S$, we need $xy \in S$ for $S$ to be mulitplicative.
• For contradiction, let us say that $x, y \in S$ such that $xy \not \in S$. Translating to $P$, this means that $x, y \not \in P$ such that $xy \in P$. This contradictions the definition of $P$ being prime.

§ Ideal whose complement is maximal multiplicative subset implies ideal is prime.

• Let $I$ be an ideal of the ring $R$ such that its complement $S \equiv R / I$is a maximal multiplicative subset.
• Let $i_1 i_2 \in I$. For $I$ to be prime, we need to show that $i_1 \in I$ or $i_2 \in I$.
• For contradiction, let $i_1, i_2 \not \in I$. Thus, $i_1, i_2 \in S$. Since $S$ is multiplicative, $i_1 i_2 \in S$. That is, $i_1 i_2 \not \in I$ (since $I$ is disjoint from $S$).
• But this violates our assumption that $i_1 i_2 \in I$. Hence, contradiction.