Dickson’s Lemma: Every monomial ideal is finitely generated

Let F be a field and let R = F[x_1, \ldots, x_t]. Prove that every monomial ideal in R is finitely generated.

Let \mathcal{S} denote the class of all monomial ideals in R which are not finitely generated by monomials. Note that \mathcal{S} is partially ordered by inclusion. Suppose by way of contradiction that \mathcal{S} is nonempty. Let \{M_i\}_{i \in I} be a linearly ordered collection of ideals in \mathcal{S}; by this previous exercise, \bigcup_I M_i is a monomial ideal. Suppose now that \bigcup_I M_i is finitely generated; say by A. Now each a \in A is contained in some M_k; since A is finite, we may choose the largest such M_k. Then M_k \subseteq (A) \subseteq M_k, so that M_k is in fact finitely generated- a contradiction. Thus \bigcup_I M_i is not finitely generated, and so is in \mathcal{S}. Thus every chain in \mathsf{S} has an upper bound in \mathsf{S}. By Zorn’s lemma, \mathsf{S} contains a maximal element M.

Now by this previous exercise, we know that the prime ideals of R are all finitely generated. In particular, M is not prime. Thus, there exist polynomials p,q \in R such that pq \in M but p,q \notin M. By this previous exercise, we know that some monomial terms x of p and y of q are not in M, but that (since xy is a monomial term of pq) xy \in M.

Since x \notin M, we have the proper containment M \subsetneq M + (x). By this previous exercise, M + (x) is a monomial ideal. Since M is maximal in the class of non-finitely generated monomial ideals, M + (x) must be finitely generated- say by G = \{g_1, \ldots, g_s\}. Note that each of these g_i is in either M or (x); relabeling if necessary, say that g_1, \ldots, g_\ell \in M and g_{\ell+1}, \ldots g_s \in (x). In particular, note that \ell < s since otherwise x \in M. Let M_0 = (g_1, \ldots, g_\ell). Then we have M + (x) = (g_1, \ldots, g_s) = M_0 + (g_{\ell+1}, g_s) \subseteq M_0 + (x) \subseteq M + (x). In particular, M+(x) = M_0 + (x).

We claim that M = M_0 + (x)(M : (x)). To see the (\supseteq) direction, let a \in (x) and b \in (M : (x)). Certainly then ab \in M, and so (x)(M : (x)) \subseteq M. Thus M_0 + (x)(M : (x)) \subseteq M. Next, let p \in M. Then p+x \in M + (x) = M_0 + (x); thus we have p + x = q + ax where q \in M_0 and a \in R. Rearranging, we have (a-1)x = p-q \in M, so that a-1 \in (M : (x)). Thus p = q + (a-1)x \in M_0 + (x)(M : (x)), and so M \subseteq M_0 + (x)(M : (x)). Thus we have M = M_0 + (x)(M : (x)).

Next, we note that xy \in M. Then y(x) \subseteq M, and so y \in (M : (x)). Certainly M \subseteq (M : (x)), and this containment is proper since y \notin M. By this previous exercise, (M : (x)) is a monomial ideal, and by the maximality of M it is finitely generated. Then by this previous exercise, M_0 + (x)(M : (x)) = M is finitely generated, a contradiction.

And so our assumption that \mathcal{S} is nonempty was faulty; in fact, every monomial ideal in R is finitely generated by monomials.

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