## The intersection of two monomial ideals is a monomial ideal

Let $F$ be a field, let $R = F[x_1, \ldots, x_t]$, and let $M = (m_i \ |\ i \in I)$ and $N = (n_j \ |\ j \in J)$ be monomial ideals in $R$ (not necessarily finitely generated). Prove that $M \cap N$ is also a monomial ideal.

Let $e_{i,j}$ be a least common multiple of $m_i$ and $n_j$ for each $i \in I$ and $j \in J$. We wish to show that $M \cap N = (e_{i,j} \ |\ i \in I, j \in J)$.

$(\subseteq)$ Suppose $p \in M \cap N$. By this previous exercise, every term in $p$ is divisible by some $m_i$ and also by some $n_j$; then every term is divisible by some $e_{i,j}$. Thus $p \in (e_{i,j}\ |\ i \in I, j \in J)$.

$(\supseteq)$ Suppose $p \in (e_{i,j} \ |\ i \in I, j \in J)$. Then each term in $p$ is divisible by some $e_{i,j}$; hence each term is divisible by some $m_i$ (so that $p \in M$) and also by some $n_j$ (so that $p \in N$). So $p \in M \cap N$.

Note that an easy induction argument shows that in fact any finite intersection of monomial ideals is a monomial ideal.