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.

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