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.

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