So I am studying commutative algebra at the moment and I have come across the proof of the Hilbert Basis Theorem (the proof I have is the same as the one in Reid's Undergraduate Commutative Algebra). I can't see how I would ever have thought of such a proof and I can't find anywhere which gives a good motivation for it. I wondered if anyone here could help me out?
EDIT: Following Martin Brandenburg's advice I will outline the proof and explain where I get the feeling: "why would you do that?"
Theorem. If $A$ is a Noetherian ring (commutative with 1), then $A[X]$ is also Noetherian.
Proof/Discussion. First, we pick any ideal $I$ in $A[X]$. We aim to find a finite set of generators for it.
We only have data about ideals in $A$, so we need to pass from the ideal $I$ in $A[X]$ to ideals in $A$.
Given any polynomial $f \in I$, a natural way to obtain elements of $A$ is to look at its coefficients.
The most "obvious" coefficients to look at are the constant term and the leading coefficient.
Just looking at the constant terms discards a lot of information about $I$ and so we don't go this way.
Let $\lambda(f)$ denote the leading coefficient of $f \in A[X]$.
We can define $J=\{a \in A : \exists f \in I \text{ such that } a=\lambda(f)\}$ but this is not necessarily an ideal of $A$.
For example, if $a,b \in J$, then $\exists f,g \in I$ with $a=\lambda(f)$ and $b=\lambda(g)$. It is then natural to say,
"well $f+g \in I$ (since $I$ is an ideal) and $\lambda(f+g)=a+b$ so we have $a+b \in J$"
but this only works if $\deg f = \deg g$. So we are led to defining, for each $m\in\mathbb{N},$
$$J_m=\{a \in A : \exists f \in I \text{ with } \deg f = m \text{ such that } a = \lambda(f)\}\cup \{0\}.$$
Then, indeed, each $J_m$ is indeed an ideal of $A$.
Now we can use that $A$ is Noetherian to obtain that each $J_m$ is finitely generated:
$$J_m=(a_{m1},\ldots,a_{mr_m})$$
for some $a_{ij} \in A$. In particular, given $m \in \mathbb{N}$, we have, for each $1\leq j \leq r_m$, some $f_{mj}\in I$ with
$\deg f_{mj}=m$;
$\lambda(f_{mj})=a_{mj}$.
Define, for each $m\in \mathbb{N}$,
$$S_m = \{f_{mj} : 1\leq j \leq r_m\}.$$
We claim that the (infinite) set $S=\bigcup_{m\in\mathbb{N}}S_m$ generates $I$.
Suppose $f \in I$ with $\deg f = n$ and $\lambda(f)=a$.
Then $a \in J_n$ and so there exists $s_1,\ldots,s_{r_n} \in A$ such that
$$a=s_1a_{n1}+\ldots+s_{r_n}a_{nr_n}.$$
Consequently, if we define $g=\sum_{k=1}^{r_n}s_kf_{nk} \in (S_n)$, we have that $\lambda(g)=a$.
In particular, $f-g \in I$ with $\deg(f-g)<n$.
Proceeding inductively, we obtain $h \in (S_0,S_1,\ldots,S_n)$ such that $f=h$.
This shows that $S$ is an infinite set of generators for $I$.
However, we now observe that $J_m\subseteq J_{m+1}$ for each $m \in \mathbb{N}$.
Indeed, if $a \in J_m$, then we take $f \in I$ with $\deg f=m$ such that $\lambda(f)=a$ and see that
$Xf \in I$ (since $I$ is an ideal),
$\deg Xf = m+1$,
$\lambda(Xf)=a$,
whence it follows that $a \in J_{m+1}$.
As $A$ is Noetherian it satisfies the ACC on ideals and so there exists $N \in \mathbb{N}$ such that
$$J_{N}=J_{N+k} \text{ for all } k \in \mathbb{N}$$
and so the set $S$, which generates $I$, is in fact finite! This concludes the proof. //
Having typed it out in this way and really had a good think about it, it doesn't seem too unnatural I suppose (although, of course, I'm still not at all convinced I could have come up with it).
I suppose the bit where we notice that the $J_m$'s form an ascending chain seems like a "bit of good luck" rather than having any reason to expect that to be the case...
What are your thoughts about my interpretation?
Many thanks!
