## 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.