## Noetherian Rings and the Hilbert Basis Theorem.

### October 16, 2010

The Hilbert Basis theorem is probably one of the easiest-to-state theorems that I know of in commutative algebra.  The last time I posted about it, I really butchered the proof; not that it was long, but it doesn’t really do anything for me.  Reading back on it now, it doesn’t seem at all intuitive to me.  The proof came through a long line of telephoning: the professor was reading from his notes, I was copying from the board, and then I was copying from my notes.  Now that I have a bit more time, I’d like to go through the proof again, but this time I’d like to motivate the theorem and proof.  Not just because the proof is a common proof-type (there are a ton of proofs that go a similar way in the commutative algebra book I’m going through) but because it’s not nearly as difficult as it looks at first glance.

To jump into this, we really don’t need all that much.  It should suffice for you to know what a ring is, what a polynomial ring is, and what noetherian means.  This last one is probably a good one to describe.

## What makes something Noetherian?

Most texts introduce this concept as follows: a ring $R$ is noetherian if and only if for every chain of ideals $I_{1}\subseteq I_{2} \subseteq \cdots$ there is some $k\in {\mathbb N}$ such that for every $i\geq k$ we have $I_{i} = I_{i+1}$.  That is, every chain eventually “stops” and has a “biggest” ideal that just keeps repeating.  This is called the ascending chain condition.

A prototypical example of this is the ring of integers.  If we have something like

$(16) \subseteq (8) \subseteq (4) \subseteq (2) \subseteq (2) \subseteq \cdots$

we note that either the sequence stops here, or we must include the ideal $(1)$ which is the entire ring, and so it certainly stops there.  In fact, because we can decompose any integer into primes, the integers always stop in this way.  For example:

$(21) \subseteq (7) \subseteq (7) \subseteq (7) \subseteq \cdots$

or

$(21) \subseteq (3) \subseteq (3) \subseteq (3) \subseteq \cdots$

and so on.  This makes the ring of integers noetherian.  Notice that the integers do not have the descending chain condition; namely,

$(2) \supseteq (4) \supseteq (8) \supseteq \cdots$

can go on forever without repeating the same term twice.  This does not affect the fact that it is noetherian, as to be noetherian we only require the Ascending Chain Condition be satisfied.

On the other hand, there is another way to characterize noetherian rings.  We have that a ring $R$ is noetherian if and only if every ideal $I$ of $R$ is finitely generated; that is, if there exists $a_{1}, \dots, a_{n}\in R$ such that $I = (a_{1}, a_{2}, \dots, a_{n})$.

I find this formulation a bit more intuitive for some reason, though these two formulations are exactly the same.  Let’s just prove this quickly.

Theorem. If a ring $R$ has the ascending chain property, then every ideal of $R$ is finitely generated.  Conversely, if every ideal of $R$ is finitely generated, then $R$ has the ascending chain property.

Proof. First, let’s assume that $R$ has the ascending chain property, and let’s suppose that there exists an ideal $I$ of $R$ which is not finitely generated.  Then we have that there exists a sequence of elements $a_{1}, a_{2}, a_{3}, \dots$ which are all contained in $I$ and such that $a_{i}$ is not a linear combination of any of the previous $a_{j}$ for $j < i$.  Then consider

$(a_{1}) \subseteq (a_{1}, a_{2}) \subseteq (a_{1}, a_{2}, a_{3})\subseteq \cdots$

But this is an ascending chain which does not stop.  This is a contradiction.

Conversely, let’s suppose every ideal of $R$ is finitely generated.  Now, suppose we have a chain

$I_{1} \subseteq I_{2}\subseteq \cdots$

Let’s consider the union of all these things.  You’ll see why in a second.  Let’s call $J = \bigcup_{i} I_{i}$.  Since every ideal of $R$ is finitely generated and $J$ is an ideal (check this!  it’s routine, but it’s always nice to check.) we have that $J$ is finitely generated.  Let’s say that $J = (a_{1}, a_{2}, \dots, a_{n})$.  Note that because this is a union, we have that there must be some $I_{N}$ such that $a_{i}\in I_{N}$ for each $i\in 1, \dots, n$.  Thus, there is some $I_{N}$ which contains all the generators for $J$, and so it must be equal to $J$.  Thus, we have that our chain “stops” at this particular $I_{N}$.  Since the chain was arbitrary, this shows that $R$ has the ascending chain condition.  $\hfill \Box$

(Note: Thanks to Brooke who graciously pointed out that the end of this proof was terribly, terribly wrong.  It is fixed now!)

Before stating the theorem, I want to make a note regarding the proof.  The proof that I’m adapting is the standard proof, and specifically, the one from Reid.  The idea of the proof is as follows: we’re going to take an ideal $I$ and take the polynomials in $I$ and group them together if they have the same order.  So, we’re going to define, for example, $P_{n}$ which is the set of all polynomials of order $n$, and we’re going to construct another set $J_{n}$ which contains only the leading coefficients (the coefficient in front of the highest degree term) of each element in $P_{n}$.  It’s going to turn out that this is going to make a nice chain of ideals in $R$.  This is going to let us finitely generate some of the lower $J_{n}$‘s which is going to lead us towards a way to make a generating set for $I$ which is finite.  This is the general idea of the proof.  So, let’s just dig right into it.

Theorem (Hilbert Basis Theorem).  If $R$ is a noetherian ring, then $R[x]$ is also noetherian.

Proof. We’re going to start by taking an ideal $I\subseteq R[x]$.  We want to prove this is finitely generated.  Construct a set $P_{n}$ which contains every polynomial of order $n$ which is in $I$.  Construct the set $J_{n}$ from $P_{n}$ by letting $J_{n}$ be the leading coefficient for every polynomial in $P_{n}$.  It is clear that $J_{n}$ is an ideal (since $I$ is an ideal; you can feel free to check this.) and we note that $J_{n} \subseteq J_{n+1}$.  This last part seemed strange to me for a bit, but just note that if some polynomial $f(x)\in P_{n}$ then because $I$ is an ideal, we have that $x*f(x) \in P_{n+1}$ which means that the leading coefficient of $f(x)$ is also in $J_{n+1}$.  Thus, we have

$J_{1}\subseteq J_{2}\subseteq \cdots$

and since $R$ is noetherian, we have that $J_{n} = J_{n+1} = \cdots$ for some $n$.  Let $n$, in fact, be the smallest such $n$ that this happens.

For every $m\leq n$, we have that the ideal $J_{m}$ is finitely generated (as $R$ is noetherian) and so let $J_{m} = (a_{m,1}, \dots, a_{m, r_{m}})$.  Note that this is Reid’s notation.  Kind of messy, but there’s not really a way around it.  For each of the $a_{m,i}$‘s in this finite generating set, we have a polynomial $f_{m,i}$ of degree $m$ which corresponds to that $a_{m,i}$; there may be more than one, but just pick one.

Now, we take all of those $f_{m,i}$‘s for every $m\leq n$ and every $i$ in the finite indexing set for each $m$ and we get a finite set:

$\{f_{m,i}\}^{m\leq n}_{1\leq j\leq r_{m}}$

and each of the elements in this set is an element of $I$.  We want to claim that, in fact, these elements generate $I$.  This is the last leg of the proof, so pay attention!

Suppose we have some arbitrary polynomial $g\in I$ of degree $d$.  If $d \geq n$ then its leading coefficient $a$ is in $J_{d} = J_{d-1} = \cdots = J_{n}$.  Therefore, $a = \sum b_{i}a_{n,i}$ where $b_{i}\in R$ and $a_{n,i}$ were the generating elements of $J_{n}$.  We then note that $g - \sum b_{i}x^{d-n}f_{n,i}$ has degree less than $d$.  This is because our $f_{n,i}$‘s have leading coefficients $a_{n,i}$ and when we multiply them by the corresponding $b_{i}$ and sum them up, we get exactly $a$.  Thus, we essentially are eliminating the leading term in $g$ to get something of lesser degree.

Similarly, if we have that $d\leq n$ then $a\in J_{d}$ which means that $a = \sum b_{i}a_{d,i}$ with $b_{i}\in R$.  We do a similar trick, but now we don’t need the extra $x$ terms, so we have just that $g - \sum b_{i}f_{d,i}$ has degree less than $d$ for the same reasons as in the end of the last paragraph.

By induction on the degree of the polynomial, we can write $f$ as the linear combination of finitely many elements; in fact, at MOST, we’d use every element in our finite set

$\{f_{m,i}\}^{m\leq n}_{1\leq j\leq r_{m}}$

but we might not even need that many!  Either way, this proves that $I$ is finitely generated, and since $I$ was an arbitrary ideal of $R[x]$, this implies that $R[x]$ is noetherian.  Neato.  $\hfill \Box$

This result is used quite a bit in algebraic geometry.  But generally, we will use it iteratively, as the following (easy!) corollary shows:

Corollary. If $R$ is a noetherian ring, then $R[x_{1}, \dots, x_{n}]$ is noetherian for any $n\in {\mathbb N}$.

The proof of this is really easy to see if you just note that if $R$ is a noetherian ring, then $R[x_{1}]$ is also a noetherian ring.  Therefore, apply the Hilbert Basis Theorem to $R[x_{1}]$ to prove that $R[x_{1}, x_{2}]$ is a noetherian ring.  Rinse and repeat as needed.