§ 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
We wish to characterise prime ideals as precisely those that are disjoint from
a multiplicative subset . That is:
I'll be using the definition of prime as:
- An ideal is prime iff , where is a multiplicative subset that cannot be made larger (ie, is maximal wrt to the ordering).
- An ideal is prime if for all , .
§ Prime ideal implies complement is maximal multiplicative subset:
Let be the complement of the prime ideal
- Since , P. (if , then every element since is an ideal, and must be closed under multiplication with the entire ring). Hence, .
- For any , we need for to be mulitplicative.
- For contradiction, let us say that such that . Translating to , this means that such that . This contradictions the definition of being prime.
§ Ideal whose complement is maximal multiplicative subset implies ideal is prime.
- Let be an ideal of the ring such that its complement is a maximal multiplicative subset.
- Let . For to be prime, we need to show that or .
- For contradiction, let . Thus, . Since is multiplicative, . That is, (since is disjoint from ).
- But this violates our assumption that . Hence, contradiction.