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.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: