Let be a field and let . Let be a monomial ideal (not necessarily finitely generated).
- Prove that is prime if and only if for some . In particular, there are only finitely many prime monomial ideals, each of which is finitely generated.
- Prove that is the only maximal ideal which is also monomial.
We begin with a lemma.
Lemma: Let be a monomial ideal. Then we may assume without loss of generality that each is divisibility-minimal with respect to inclusion in ; that is, no proper divisor of is in . Proof: Note that each has finite degree, and thus has finitely many divisors. Let be the proper divisors of which are minimal with respect to inclusion in . Certainly . Note that if is finite, then there are only finitely many , and if is transfinite, then . Thus, given a monomial generating set, we may replace it with one whose elements are divisibility minimal without affecting its cardinality (if transfinite) or finiteness (if finite).
Now let be a prime monomial ideal, and let consist of precisely those indeterminates which divide some . By the lemma, we may assume that the are divisibiliy minimal- that no proper divisor of is again in . We claim that . The direction is clear. Now let ; then for some . Since is a prime ideal, either or is in . But since no proper divisor of is in , it follows that is not a proper divisor; instead, for some unit . In particular, . Thus , and we have .
On the other hand, suppose and consider the ideal . Let’s say that (with a reordering if necessary) . Now , which is a domain (even if ). So is prime.
Now let be a maximal ideal which has a monomial generating set. Since is also prime, by the above argument we have for some . Suppose for some ; then , contradicting the maximality of . Thus .