## 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 $x \in MN$.

Post a comment or leave a trackback: Trackback URL.