(a) Prove $\mathcal{S}$ has a maximal element.
(b) Show there exist monomials $x,y∉M$ such that $xy∈M$.
(c) Show there exists $M_0 \subseteq M$ with $M_0$ a finitely generated monomial ideal such that $M_0+(x)=M+(x)$ and $M=M_0+(x)(M : (x))$. Deduce $M$ is finitely generated, forcing $\mathcal{S}=\varnothing$.
44. If $\varnothing≠I\subseteq F[x_1,..,x_n]$ is an ideal, prove $LT(I)$ is finitely generated. Conclude $I$ has a Gröbner basis and deduce Hilbert's Basis Theorem.
Proof: (a) Let $\mathcal{C}$ be a typical chain totally ordered by inclusion. Since $M=\bigcup_{N∈\mathcal{C}} N$ is a monomial ideal by 9.6.39, by Zorn's Lemma it suffices to show it is infinitely generated; assume the contrary, so that $M=(n_1,...,n_r)$. Let $n_i$ be maximal with regard to the ordering $n_i∈N_i \geq n_j∈N_j :⇔ N_j \subseteq N_i$, and observe $N_i \subseteq M = (n_1,...,n_r) \subseteq N_i$ so that $N_i=M$ is finitely generated, a contradiction.
(b) By 9.6.42(a), the only prime monomial ideals are finitely generated. Therefore there exist polynomials $x',y'∈F[x_1,...,x_n]$ such that $x',y'∉M$ and $x'y'∈M$. By 9.6.10, there exist monomial terms in $x',y'$ not divisible by any $m_i$ for $M=(m_1,...)$. Showing $x'y'=m∈M$ and expanding $x'$ to subtract the products of $y'$ and monomial terms of $x'$ divisible by monomials of $M$ (which are elements of $M$), we obtain $xy'∈M$ for $x$ composed of terms not divisible by monomials of $M$. Similarly, $xy∈M$. Now $LT(xy)=LT(x)LT(y)∈LT(M) \subseteq M$ (the latter $\subseteq$ being due again to 9.6.10) so that $LT(x)$ and $LT(y)$ are monomials not within $M$ whose product is within $M$.
(c) Fix a monomial ordering $≤$. First note that $m \mid m' ⇒ md=m'$ so that, since $d \geq 1$, we have $md = m' \geq m$, i.e. $m \mid m' ⇒ m ≤ m'$.
Let $x$ be the minimal monomial by $≤$ not within $M$ such that there exists another monomial $y$ not within $M$ fulfilling $xy∈M$. Revisit $M=(m_1,...)$ and let $M^*=(m_1,...,x)$ as before. Since $M \subset M^*$ by the maximality of $M$ we must have $M^*=(T)$ for some finite subset of monomials $T=\{q_1,...,q_t\}$. Reformat this generating subset so that it is minimal, by excluding a generator $n$ if $M^*=(T \setminus \{n\})$. This process is seen to terminate, and in part we are left with $T=\{n_1,...,n_s\}$ where $n_i \not \mid n_j$ for all $i≠j$. Assume $x∉T$: Then by 9.6.10 since $x∈M^*$ we have $n_i \mid x$ so $n_id=x$ for some $i$, and since $n_i∈M⇒x∈M$, yet also $n_i∉M⇒n_i < x \land n_i(dy)∈M$ with $dy∉M$ since $d < x$ as well, we inevitably violate the minimality of $x$, and we must instead have $x∈T$. Rewrite $T=\{n_1,...,n_{s-1},x\}$.
Now let $M_0=(T \setminus \{x\})$. Assume $n_j∈T \setminus \{x\}$ yet $n_j∉M$: Since the elements of $T$ are from $M^*$ generated by $\{m_1,...,x\}$ we must have either $m_i \mid n_j$ for some $i$ or $x \mid n_j$; the latter is excluded by the minimalization of $T$ and the former implies $n_j∈M$. Therefore $M_0 \subseteq M$, and evidently $M_0+(x)=M+(x)=M^*$.
Show $M=M_0+(x)(M : (x))$: ($\supseteq$) By the definition of $(M : (x))$, the direction is clear. ($\subseteq$) We have $m_i∈M \subseteq M_0+(x)$, so that either $n_k \mid m_i ⇒ m_i∈M_0$ or $x \mid m_i ⇒ m_i ∈ (x)(M : (x))$.
Finally, we see that $M_0$ is finitely generated, and also $y∉M$ with $y∈(M : (x))$ so that $(M : (x))$ is finitely generated by the maximality of $M$, and also $(x)(M : (x))$ and now $M_0+(x)(M : (x))=M$ is finitely generated, a contradiction.$~\square$
44. Since $LT(I)$ is a monomial ideal, by the previous it is finitely generated. In the same way as in 9.6.1, we may generate each of the elements of the finite generating set of $LT(I)$ by a finite number of finite combinations of $LT(g_i)$ for $g_i$ elements of $I$, so that by proposition 24 these $g_i$ form a Gröbner basis and as well a finite generating set for the arbitrary ideal $I$, which is Hilbert's Basis Theorem.$~\square$
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