Prime and Maximal Monomial Ideals (9.6.42)

Dummit and Foote Abstract Algebra, section 9.6, exercise 42:

(a) Show $M$ is a monomial prime ideal if and only if $M=(S)$ for some $S \subseteq \{x_1,...,x_n\}$.
(b) Show $(x_1,...,x_n)$ is the only maximal monomial ideal.

Proof: (a) ($\Leftarrow$) Letting $I$ be such that $S=\{x_i~|~i∈I\}$, we have $F[x_1,...,x_n]/M ≅ F[x_j~|~j∉I]$ which is again a polynomial ring in several variables over a field and is thus in particular an integral domain. ($⇒$) Since $M$ is a monomial ideal, let $M=(m_1,...)$ with the monomials countably ordered by $m_1 ≤ m_2$ if $\text{deg}(m_2) < \text{deg}(m_1)$ and decided by lexicographic ordering if $\text{deg}(m_2)=\text{deg}(m_1)$. Trim this generating list by a process of eliminating $m_i$ if it is not generatable by linear combinations of previous monomials. Choose $m∉\{x_1,...,x_n\}$ in the generating set. letting $x_i$ be a variable dividing $m$, we must have $x_i∉M$ else by exercise 10 $x_i$ is divisible by a term in the generating list, forcing $x_i$ in the generating list so that by the ordering $x_i$ appears before $m$ and now $m$ could not have been included. Since $x_i$ was arbitrary, we have the product of the variables dividing $m$ violating the primality of $M$. Therefore $m$ does not exist, and this generating subset is a subset of $\{x_1,...,x_n\}$.

(b) Since maximal ideals are prime ideals, if $M=(S)$ with $S≠\{x_1,...,x_n\}$, by part (a) we have $M \subset (x_1,...,x_n) \subset F[x_1,...,x_n]$, a contradiction.$~\square$