Characterize the prime monomial ideals in a polynomial ring

Let F be a field and let R = F[x_1, \ldots, x_t]. Let M = (m_i \ |\ i \in I) be a monomial ideal (not necessarily finitely generated).

  1. Prove that M is prime if and only if M = (S) for some S \subseteq \{x_1, \ldots, x_t\}. In particular, there are only finitely many prime monomial ideals, each of which is finitely generated.
  2. Prove that (x_1, \ldots, x_t) is the only maximal ideal which is also monomial.

We begin with a lemma.

Lemma: Let M = (m_i \ |\ i \in I) be a monomial ideal. Then we may assume without loss of generality that each m_i is divisibility-minimal with respect to inclusion in M; that is, no proper divisor of m_i is in M. Proof: Note that each m_i has finite degree, and thus has finitely many divisors. Let \{n_{i,j}\}_{j \in J_i} be the proper divisors of m_i which are minimal with respect to inclusion in M. Certainly (n_{i,j} \ |\ i \in I, j \in J_i) \subseteq M = (m_i \ I\ i \in I) \subseteq (n_{i,j} \ |\ i \in I, j \in J_i). Note that if I is finite, then there are only finitely many n_{i,j}, and if I is transfinite, then |\bigcup_{i \in I} J_i| = |I|. 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). \square

Now let M = (m_i \ |\ i \in I) be a prime monomial ideal, and let S \subseteq \{x_1, \ldots, x_t\} consist of precisely those indeterminates which divide some m_i. By the lemma, we may assume that the m_i are divisibiliy minimal- that no proper divisor of m_i is again in M. We claim that M = (S). The (\subseteq) direction is clear. Now let x_k \in S; then m_k = x_kq_k for some m_k. Since M is a prime ideal, either x_k or q_k is in M. But since no proper divisor of m_k is in M, it follows that x_k is not a proper divisor; instead, m_k = ux_k for some unit u. In particular, x_k \in M. Thus (S) \subseteq M, and we have M = (S).

On the other hand, suppose S \subseteq \{x_1,\ldots,x_t\} and consider the ideal M = (S). Let’s say that (with a reordering if necessary) S = \{x_1,\ldots,x_k\}. Now F[x_1,\ldots,x_t]/(S) \cong F[x_k+1,\ldots,x_t], which is a domain (even if k = t). So (S) is prime.

Now let M be a maximal ideal which has a monomial generating set. Since M is also prime, by the above argument we have M = (S) for some S \subseteq \{x_1, \ldots, x_t\}. Suppose x_k \notin M for some k; then M \subsetneq (M,x_k) \subsetneq R, contradicting the maximality of M. Thus M = (x_1, \ldots, x_t).

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  • John Doe  On September 20, 2011 at 7:36 pm

    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).

    • nbloomf  On September 20, 2011 at 9:14 pm

      You’re right. It should be fixed now. (Essentially, for such an ideal, the corresponding quotient R/(S) is isomorphic to either F or to F[x_1,\ldots,x_s], which are both domains, and so (S) is prime.


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