## Characterize the prime monomial ideals in a polynomial ring

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 .

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## Comments

You set out to prove that a monomial ideal is prime if and only if it is of this form. Yet it seems you proved only one direction. Maybe I am missing something but there is no proof that an ideal (S) is indeed prime, just that if an ideal IS prime than it’s of the form (S).

You’re right. It should be fixed now. (Essentially, for such an ideal, the corresponding quotient is isomorphic to either or to , which are both domains, and so is prime.

Thanks!