The sum and product of two monomial ideals is a monomial ideal

Let F be a field and let R = F[x_1, \ldots, x_m]. Let M = (m_i \ |\ i \in I) and N = (n_j \ |\ j \in J) be monomial ideals in R, where m_i and n_j are monomials. Prove that M+N = (m_i, n_j \ |\ i \in I, j \in J) and MN = (m_in_j \ |\ i \in I, j \in J).


First we show that M+N = (m_i, n_j \ |\ i \in I, j \in J).

(\subseteq) Let x+y \in M+N, where x = \sum_{i \in I} a_im_i and y = \sum_{j \in J} b_jn_j, and all but finitely many of the a_i and b_i are zero. Then x+y = \sum_I a_im_i + \sum_J b_jn_j \in (m_i, n_j \ |\ i \in I, j \in J). (\supseteq) Let x \in (m_i, n_j \ |\ i \in I, j \in J). Collecting the terms in m_i and n_j, respectively, we can write x = \sum_I a_im_i + \sum_J b_jn_j; certainly then x \in M+N.

Now we show that MN = (m_in_j \ |\ i \in I, j \in J).

(\subseteq) Let x \in MN; say x = \sum_K z_kw_k where z_k = \sum_I a_{i,k}m_i \in M and w_k = \sum_J b_{j,k}n_j \in N. Then x = \sum_K \sum_I \sum_J a_{i,k}b_{j,k}m_in_j \in (m_i, n_j \ |\ i \in I, j \in J). (\supseteq) Let x \in (m_i, n_j \ |\ i \in I, j \in J); say x = \sum_{I \times J} c_{i,j} m_i n_j, where all but finitely many c_{i,j} are zero. Certainly then x \in MN.

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