Uniqueness of Minimal Generating Monomial Subsets of Ideals (9.6.14)

Dummit and Foote Abstract Algebra, section 9.6, exercise 14:

Suppose $I$ is a monomial ideal in $F[x_1,...,x_n]$ minimally generated by $M=\{m_1,...,m_k\}$ for $m_i$ a monomial. Prove this generating subset is unique.

Proof: Note we must necessarily have $m_i \not \mid m_j$ for all $i≠j$ or else it is not a minimally generating set. Now, assume another minimally generating set $N=\{n_1,...,n_l\}$. Since $(M)=(N)=I$ we have by exercise 10 that $LT(n_1)=n_1$ is divisible by, say, $m_1$. Similarly, $m_1$ is divisible by a monomial from $N$. Continue in this fashion so that we must necessarily reach a point in the chain of division such that $m_i$ repeats, so that $m_i \mid ... \mid n_j \mid ... \mid m_i$ and now $m_i=n_j$. Starting again with an element $m$ in $M$ besides $m_i$, we can follow the chain once more which will reveal another equality between elements, which won't be the one between $m_i$ and $n_j$ as then $m \mid m_i$, a contradiction. In this fashion, every element of the minimally-ordered generating set will be paired up with its equal in the opposite set, i.e. will be a subset of the opposite set. By the nature of minimally generated sets, this inclusion cannot be proper, and now $M=N$.$~\square$