Hilbert's Basis Theorem

Ambitious goals seem to pop up whenever I have free time, but I have a poor track record of hitting them. To fill the extra time I have over the summer, I started reading Eisenbud's Commutative Algebra texbook as a side project. Let's hope I get far enough to learn something useful.

In this post, I want to write a bit about a result known as Hilbert's Basis Theorem (HBT). We'll motivate why we should care about it through a use case prevalent in algebraic geometry.

Motivating example

In algebraic geometry, one defines "shapes" and "curves" by taking a set of polynomials $S \subset k[x_1, \ldots, x_r]$ over an algebraically closed field $k$ and studying the set of points in $k^r$ where all the polynomials in $S$ are equal to zero: $$Z(S) = \{ p \in k^r : f(p) = 0~\text{for all}~f \in S\}$$ We call such sets algebraic sets.

For example, if $f = x^2 + y^2 - 1 \in \mathbb{R}[x,y]$ then $Z(\{f\})$ is the algebraic set consisting of points where $x^2 + y^2 = 1$, which is precisely the unit circle.

Sets of polynomials seem pretty daunting; they can easily be uncountably large. Hence, at first glance characterizing these algebraic sets seems like a difficult task. However, Hilbert's basis theorem gives us a a very useful result which allows us to restrict our attention from arbitrary subsets of $k[x_1, \ldots, x_r]$ to just finitely generated ideals.

Corollary of Hilbert's basis theorem: Any algebraic set can be written as $Z(I)$ where $I \subset k[x_1, \ldots, x_r]$ is a finitely generated ideal.

This makes the problem significantly easier: since any $f \in I$ can be represented using a finite basis $f = \sum_{i=1}^n k_i f_i$ we only need to concern ourselves with the zero locus of a finite number of polynomials!

Statement of the theorem

To state the theorem in it's full generality, we will need the following definition. Let $R$ be a commutative ring.

Definition: $R$ is Noetherian if any ideal $I \subset R$ is finitely generated.

An equivalent definition is the ascending chain condition: for any ascending chain of ideals of $R$ $$I_1 \subset I_2 \subset \cdots \subset I_k \subset I_{k+1} \subset \cdots$$ there exists a $n$ after which $$I_n = I_{n+1} = I_{n+2} = \cdots$$

Hilbert's basis theorem says that adjoining elements to a Noetherian ring preserves the Noetherian property.

Theorem (Hilbert's basis theorem): If $R$ is Noetherian, then so is $R[x]$.

Consequences

By induction, we get that:

Corollary: If $R$ is Noetherian, then so is $R[x_1, \ldots, x_r]$.

Noting that the only ideals of a field $k$ are $\{0\} = \langle 0 \rangle$ and $k = \langle 1 \rangle$ (if $a \in I$ is non-zero, then since $a^{-1} \in k$ because $I$ is an ideal we must have $a^{-1} a = 1 \in I$, hence $I \supset \langle 1 \rangle = k$). As both of these are finitely generated:

Lemma: Any field $k$ is Noetherian

Together with Hilbert's basis theorem and the above corollary, we have

Corollary: $k[x_1, \ldots, x_r]$ is Noetherian; all of its ideals are finitely generated.

Our discussion up until now has focused only on ideals, whereas the motivating example was about algebraic sets defined by arbitrary subsets of $k[x_1, \ldots, x_r]$. The following proposition ties the two together.

Proposition: For any $S \subset R[x_1, \ldots, x_r]$, $Z(S) = Z(\langle S \rangle)$ where $\langle S \rangle$ is the ideal generated by $S$.

Proof: Since $S \subset \langle S \rangle$, $Z(S) \supset Z(\langle S \rangle)$ (aside: $Z$ is a contravariant functor from the category consisting subsets of $R[x_1, \ldots, x_r]$ to the category consisting of algebraic sets on $R^r$, both with inclusions as arrows).

Conversely, if $p \in Z(S)$ then evaluating any $f = \sum_i r_i f_i \in \langle S \rangle$ at $p$ yields $$f(p) = \sum_i r_i f_i(p) = \sum_i r_i 0 = 0$$ since the $f_i \in S$ are equal to zero for $p \in S$. $\square$

Proof of corollary in motivating example

Armed with these result, the proof of the corollary in the motivating example is swift.

Proof: An algebraic set is of the form $Z(S)$ where $S \subset k[x_1, \ldots, x_r]$. By the previous proposition, $Z(S) = Z(I)$ where $I = \langle S \rangle) \subset k[x_1, \ldots, x_r]$ is an ideal. By the above corollary, $k[x_1, \ldots, x_r]$ is Noetherian hence $I$ must be finitely generated. $\square$

Proving Hilbert's basis theorem

We've almost tied up all the loose ends in this discussion; all that remains is proving Hilbert's basis theorem itself. In this section, we will complete this last step.

Before we get there, we will need an alternate characterization of Noetherian rings.

Lemma: $R$ is Noetherian $\iff$ $R$ satisfies the ascending chain condition: for any ascending chain of ideals $$I_1 \subset I_2 \subset I_3 \subset \cdots$$ there exists $n \in \mathbb{N}$ after which the chain stabilizes, i.e. $$I_n = I_{n+1} = \cdots$$

Proof: Since $I = \cup_i I_i$ is an ideal of $R$, it has a finite set of generators $\{a_i\}_{i=1}^k$. Each $a_i \in I_{j_{i}}$ for some $j_{i} \in \mathbb{N}$, so taking $n = \max_{1 \leq i \leq k} j_i$ suffices.

Conversely, define the ascending chain $I_k = \langle a_i \rangle_{i=1}^k$ where $a_0 \in I$ and $a_i \in I \setminus I_{i-1}$ are chosen arbitrarily. This chain stabilizes at some $n$ where $I = I_n = \langle a_i \rangle_{i=1}^k$. $\square$

Proof of Hilbert's basis theorem: Let $I \subset R[x]$ be an ideal. We will construct a finite set of generators inductively. Choose $f_0 \in I$ to be any element with least degree and for $i \geq 1$ pick $f_i \in I \setminus \langle f_k \rangle_{k=1}^{i-1}$ any element with least degree. It remains to show that $R[x] = \langle f_k \rangle_{k=1}^{i}$ for some $i \in \mathbb{N}$. For each $f_i$, we can write

$$f_i = \sum_{j=0}^{\deg f_i} a_{i,j} x^{j}$$

Consider the ascending chain of ideals $J_k = (a_{i, \deg f_i})_{i=1}^k$ generated by all the initial coefficients of the selected $f_i$. This is an ascending chain, so $J = \langle a_i \rangle_{i=1}^m$ for some $m$. We claim that $I = \langle f_i \rangle_{i=1}^m$.

To see this, let $f_{m+1} \in I \setminus \langle f_k \rangle_{k=1}^{i-1}$. Then since $a_{m+1,\deg f_{m+1}} \in J$, we have

$$a_{m+1,\deg f_{m+1}} = \sum_{j=1}^m r_j a_{j,\deg f_j}$$

for some coefficients $r_j \in R$. Since $\deg f_{m+1} \geq \deg f_m \geq \deg f_{m-1} \geq \cdots$ by how we chose these elements, we have that $\deg f_{m+1} - \deg f_j \geq 0$ and therefore can define

$$g = \sum_{j=1}^m \underbrace{u_j x^{\deg f_{m+1} - \deg f_j}}_{\in R[x]} f_j$$

By the definition of ideals, $g \in \langle f_k \rangle_{k=1}^{m}$. Now notice $f_{m+1} - g \in I$ (as the sum of two elements in $I$) but $f_{m+1} - g \not\in \langle f_k \rangle_{k=1}^{m}$ (otherwise adding $g$ would give that $f_{m+1} \in \langle f_k \rangle_{k=1}^{m}$), so $f_{m+1} - g \in I \setminus \langle f_k \rangle_{k=1}^{m}$

What's more, the initial coefficient of $g$ is given by (notation following Sedgewick and Flajolet)

$$[z^{\deg g}] g = [z^{\deg f_{m+1}}] g = \sum_{j=1}^m u_j [z^{\deg f_{m+1}}] \sum_{k=1}^{\deg f_j} a_{j,k} x^{\deg f_{m+1} - \deg f_j + k} = \sum_{j=1}^m u_j a_{j,\deg f_j} = a_{m+1,\deg f_{m+1}}$$

Since the degree and initial coefficients of $f_{m+1}$ and $g$ match, we have that $\deg(f_{m+1} - g) < \deg f_{m+1}$, contradicting the choice of $f_{m+1}$ having minimal degree amongst all polynomials in $I \setminus \langle f_k \rangle_{k=1}^{m}$. $\square$

Closing remarks

Our proof of Hilbert's basis theorem is almost identical to that in Eisenbud with some additional commentary and explanation. The technique of matching the initial term between two polynomials and arguing that the degree of the difference is strictly less is a common proof method when working with polynomials.

Most of our work was possible when polynomial coefficients were over a ring $R$. We only needed coefficients to be from a field $k$ so that we could use $k$ being Noetherian to conclude the motivating example.

While we only showed that $Z(S) = Z(\langle S \rangle)$ for any subset $S \subset k[x_1, \ldots, x_r]$, the connection between algebraic sets (our geometric objects) and ideals (our algebraic objects) is much deeper. In fact, it's not terribly hard to show that $Z(I(X)) = \overline{X}$ for any $X \subset k^r$, where $I(X) = \{f \in k[x_1, \ldots, x_r] : f(p) = 0~\text{for all}~p \in X\}$ (exercise: prove this is an ideal) and $\overline{X}$ is the smallest algebraic set containing $X$ (i.e. the closure of $X$ under the Zariski topology).

Here, one may be tempted to conclude that algebraic sets of $k^r$ and ideals of $k[x_1, \ldots, x_r]$ are in correspondence, but this is false. The precise statement here is Hilbert's Nullstellensatz (which we will likely have more to say about later):

The correspondence $I \mapsto Z(I)$ and $X \mapsto I(X)$ induces a bijection between the algebraic sets of $k^r$ and the radical ideals of $k[x_1, \ldots, x_r]$.