§ Every ideal that is maximal wrt. being disjoint from a multiplicative subset is prime

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⊆RS \subseteq R. That is:
  • An ideal PP is prime iff P=R∖SP = R \setminus S, where SS 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 PP is prime if for all x,y∈Rx, y \in R, xy∈P  ⟹  x∈P∨y∈Pxy \in P \implies x \in P \lor y \in P.

§ Prime ideal implies complement is maximal multiplicative subset:

Let S=≡R∖PS = \equiv R \setminus P be the complement of the prime ideal P⊊RP \subsetneq R in question.
  • Since P≠RP \neq R, 1∉1 \not \in P. (if 1∈P1 \in P, then every element x.1∈Px . 1 \in Psince PP is an ideal, and must be closed under multiplication with the entire ring). Hence, 1∈S1 \in S.
  • For any x,y∈Sx, y \in S, we need xy∈Sxy \in S for SS to be mulitplicative.
  • For contradiction, let us say that x,y∈Sx, y \in S such that xy∉Sxy \not \in S. Translating to PP, this means that x,y∉Px, y \not \in P such that xy∈Pxy \in P. This contradictions the definition of PP being prime.

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

  • Let II be an ideal of the ring RR such that its complement S≡R/IS \equiv R / Iis a maximal multiplicative subset.
  • Let i1i2∈Ii_1 i_2 \in I. For II to be prime, we need to show that i1∈Ii_1 \in I or i2∈Ii_2 \in I.
  • For contradiction, let i1,i2∉Ii_1, i_2 \not \in I. Thus, i1,i2∈Si_1, i_2 \in S. Since SS is multiplicative, i1i2∈Si_1 i_2 \in S. That is, i1i2∉Ii_1 i_2 \not \in I (since II is disjoint from SS).
  • But this violates our assumption that i1i2∈Ii_1 i_2 \in I. Hence, contradiction.