## Every monomial ideal has a unique inclusion-minimal monomial generating set

Let $F$ be a field, and let $I \subseteq F[x_1, \ldots, x_n]$ be a monomial ideal. Suppose $M = \{m_1, \ldots, m_k\}$ is an inclusion-minimal generating set for $I$. Prove that the $m_i$ are unique (up to associates and a rearrangement).

First, note that if $A \subseteq I$ is a monomial generating set of $I$, then by this previous exercise, some finite subset of $A$ also generates $I$. In particular, every inclusion-minimal monomial generating set of $I$ is finite.

Suppose $T = \{t_1, \ldots, t_\ell\}$ is a second inclusion-minimal generating set for $I$. Now both $T$ and $M$ are Gröbner bases for $I$ by this previous exercise. In particular, we may assume that no $m_i$ divides another, and similarly that no $t_i$ divides another, since otherwise $M$ or $T$ is not inclusion-minimal.

Now $t_a \in (M)$; by this previous exercise, $t_a$ must be divisible by some $m_b$. Likewise, $m_b$ is divisible by $t_c$. Thus $t_a$ is divisible by $t_c$, and by the minimality of $T$, we have $t_a = t_c$. In particular, $t_a$ and $m_b$ are associates.

Define a relation $\sigma \subseteq T \times M$ by $t_i \sigma m_j$ if and only if $t_i$ and $m_j$ are associates. We have showed that for every $t \in T$, there exists $m \in M$ such that $t \sigma m$. Moreover, if $t \sigma m_1$ and $t \sigma m_2$, then $m_1$ divides $m_2$, a violation of the inclusion-minimalness of $M$. So in fact $\sigma$ is well-defined – that is, a function. Similarly, $\sigma$ is injective, and since $M$ and $T$ are finite sets, $\sigma$ is a bijection.

Thus an inclusion-minimal monomial generating set of $I$ is unique up to associates and a rearrangement.