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:


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

  1. $\deg f_{mj}=m$;

  2. $\lambda(f_{mj})=a_{mj}$.

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


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

  1. $Xf \in I$ (since $I$ is an ideal),

  2. $\deg Xf = m+1$,

  3. $\lambda(Xf)=a$,

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!

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    I think it would help if you summarize the proof here and tell us which steps should be motivated for you. Also remark that mathematicians often have put a great effort for finding proofs, even if they don't tell you, and that it is also a matter of experience and skills to find these proofs. – Martin Brandenburg Oct 12 '14 at 21:43
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    "Having typed it out in this way and really had a good think about it, it doesn't seem too unnatural I suppose" - this is what I was expecting! :-) I think that you have already done a good job in explaining the ideas behind the proof. I hope that now others will comment on your remaining doubts that this is a natural proof. – Martin Brandenburg Oct 13 '14 at 15:33
  • Thank you for your help! :) – user183870 Oct 13 '14 at 15:35
  • I'd just like to point out one little detail - you should write $J_m = (a_{m1}, \ldots a_{mr_n})$. That is, the number of generators of $J_m$ doesn't necessarily coincide with $m$. – silvascientist May 13 '17 at 00:01
  • Is not $J$ also an ideal of $A$. If the leading coefficients come from different degree polynomial, we can then always multiply the lower degree polynomial by sufficient power of $x$ (which will still be in I) and then add up those two get the required polynomial whose leading coefficient is the sum of the given leading coefficients. @MartinBrandenburg Am I making sense? – Babai Feb 03 '18 at 12:52
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    I see that someone has already pointed out the fact I mentioned. May the OP might want to edit the question to make it perfect. But then arises the question, is not introduucing $J_m$ made the bookkeeping more difficult. – Babai Feb 03 '18 at 12:54

1 Answers1


I came across this old post and, even if the following remark isn't of great importance, I just wanted to point out that the statement below is wrong:

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 $A$.

It is an ideal of $A$.

If $a,b\in J-\{0\}$ with $a\neq b$, then there exists $f(x),g(x)\in I$ such that $a=\lambda(f)$ and $b=\lambda(g)$. Let $r=\deg f$ and $s=\deg g$ with say $r<s$. Then $h=x^{s-r}f-g\in I$ since $I$ is an ideal of $A[X]$, therefore $a-b=\lambda(h)$, so $J$ is an additive group.

[STEIN W] Algebraic Number Theory, p.21.

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