## The ideal quotient of a monomial ideal by a finitely generated monomial ideal is a monomial ideal

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 in $R$, and let $n \in R$ be a monomial. For each $i \in I$, let $d_i$ be a greatest common divisor of $m_i$ and $n$, and write $m_i = d_iq_i$.

1. Prove that $(M : (n)) = (q_i \in I)$.
2. Deduce that if $N = (n_j \ |\ j \in J)$ is a finitely generated monomial ideal, then $(M : N)$ is a monomial ideal.

We begin with a lemma.

Lemma: (Euclid) Let $R$ be a unique factorization domain, and let $a,b,c \in R$ such that $a$ and $b$ are relatively prime. Then $\mathsf{gcd}(a,bc) = \mathsf{gcd}(a,c)$. Proof: If $d = \mathsf{gcd}(a,c)$, then certainly $d|a$ and $d|bc$. Suppose now that $e|a$ and $e|bc$. If $e$ is a unit, then certainly $e|d$. Suppose $e$ is not a unit, and that $q$ is an irreducible factor of $e$ of multiplicity $k$. Since $q|a$, we have $q$ not dividing $b$. In particular, $q^k|c$. By induction, $e|c$. Hence $e|d$, and thus $d$ is a greatest common divisor of $a$ and $bc$. Now let $d = \mathsf{gcd}(a,bc)$. If $d$ is a unit, then certainly $d$ is also a greatest common divisor of $a$ and $c$. Suppose $d$ is not a unit. We have $d|a$. If $q$ is an irreducible factor of $d$ of multiplicity $k$, then $q$ does not divide $b$; thus $q^k|c$. Hence $d|c$. Now suppose $e|a$ and $e|c$; then $e|bc$, and so $e|d$. THus $ltex e$ is a greatest common divisor of $a$ and $c$. $\square$

Now we show that $(M : (n)) = (t_i \ |\ i \in I)$.

$(\subseteq)$ Suppose $p \in (M : (n))$. Then $p(n) \subseteq M$, and so $pn \in M$. Say $xn = \sum a_im_i$, with $a_i = \sum b_{i,j}$ and the $b_{i,j}$ are monomials. Then $xn = \sum \sum b_{i,j} m_i = \sum \sum b_{i,j} q_i d_i$. Now write $p = \sum c_k$ as a sum of monomials; then $pn = \sum c_kn$. In particular, each term of $pn$ has the form $c_kn = b_{i,j}q_id_i$ for some $i,j,k$. Recall that $d_i$ is a greatest common divisor of $m_i$ and $n$; say $n = d_is_i$. Then we have $c_ks_i = b_{i,j}q_i$ where $s_i$ and $q_i$ are relatively prime. Since $R$ is a unique factorization domain, by Euclid’s lemma we have $q_i|c_k$. By this previous exercise, we have $p \in (q_i \ |\ i \in I)$.

$\supseteq)$ Suppose $p \in (t_i \ | i \in I)$; say $p = \sum a_it_i$. For each $i$, say $n = r_id_i$. Now $xn = \sum a_it_ir_id_i = \sum a_i m_i r_i$, and so $xn \in M$. Thus $x(n) \subseteq M$, and we have $x \in (M : (n))$.

Now suppose $N = (n_j \ |\ j \in J)$ is a finitely generated monomial ideal. By this previous exercise, $(M : N) = \bigcap_{j \in J} (M : (n_j))$ is a finite intersection of monomial ideals. By this previous exercise, $(M : N)$ is a monomial ideal.