## 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 $M = (x_1, \ldots, x_t)$.

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.