The Hilbert Basis Theorem

David Hilbert, 1862-1943 (photo 1912)

Hilbert first proved a form of the basis theorem in 1890.^[1] It was so revolutionary at the time that Paul Gordan reportedly exclaimed, “This is not mathematics, it is theology!”.^[2] van der Waerden gave an updated and generalized proof in Moderne Algebra in 1931, crediting Hilbert for the basic idea and Emil Artin for the specifics.^[3] The proof here is updated still more, though still retaining van der Waerden's degree reduction strategy.

Hilbert Basis Theorem. Let $R$ be a commutative ring with $1$. If $R$ is Noetherian, then $R[x]$ is Noetherian as well.

Proof. Recall that a ring is Noetherian if its ideals satisfy the Ascending Chain Condition, or equivalently, if every ideal is finitely generated.

Suppose $R$ is Noetherian and $U$ is an ideal in $R[x].$ The objective is to prove that $U$ is finitely generated. For $m = 0, 1, 2, \cdots,$ let

$$\begin{align} L_m &= \text{ set of leading coefficients of polynomials of degree } \leq m \text{ in } U, \text{ together with }0 \end{align}$$

Every nonzero $\alpha \in L_m$ is associated with some polynomial $a(x) = \alpha x^s + \cdots \in U$ with $\deg{a} = s \leq m.$

Each $L_m$ is an ideal in $R.$ For suppose $\alpha, \beta \in L_m.$ Then $\alpha$ and $\beta$ are the leading coefficients of some polynomials $a(x), b(x) \in U$ of degree less than or equal to $m:$

$$\begin{align} a(x) &= \alpha x^s + \cdots \in U, \;\; \text{ where } \deg{a} = s \leq m,\\ b(x) &= \beta x^t + \cdots \in U, \;\; \text{ where } \deg{b} = t \leq m. \end{align}$$

Assume $s \geq t$ and consider $c(x) = a(x) - x^{s-t} b(x).$ Multiplying $b(x)$ by $x^{s-t}$ or indeed by any other polynomial keeps the result in $U$ since $U$ is an ideal in $R[x],$ and subtracting that from another element of $U, a(x),$ keeps the final result $c(x)$ in $U$ also, so

$$\begin{align} c(x) &= a(x) - x^{s-t} b(x) \in U\\ &= (\alpha x^s + \cdots ) - (\beta x^s + \cdots)\\ &= (\alpha - \beta) x^s + \cdots. \end{align}$$

Since $c(x) \in U,$ this implies that either $\alpha - \beta = 0$ or $\alpha - \beta$ is the coefficient of the highest power of a polynomial of degree $s \leq m$ in $U.$ In any event, $\alpha - \beta \in L_m.$ Similarly $\lambda \alpha \in L_m$ for any $\lambda \in R$ and so $L_m$ is an ideal in $R$. Also:

$$\begin{align} L_0 \subseteq L_1 \subseteq L_2 \subseteq \cdots, \end{align}$$

so by the ACC in $R,$ the chain stabilizes; that is, there is some $n \geq 0$ such that

$$\begin{align} L_n = L_{n+1} = L_{n+2} = \cdots. \end{align}$$

Assume that $n$ is the smallest such value. Since $R$ is Noetherian, all its ideals are finitely generated, and in particular, each $L_m$ has a finite set of generating values $\alpha \in R$:

$$\begin{align} L_m = (\alpha_{m1}, \alpha_{m2}, \cdots, \alpha_{mk_m}), \hspace{20pt} m \leq n. \end{align}$$

By the definition of the $L_m,$ each $\alpha_{mj}$ is associated with a corresponding polynomial $a_{mj}(x) \in U$ of degree less than or equal to $m.$ The goal is to show that all the $a_{mj}(x) \in U$ taken together generate $U$ in $R[x].$ To this end, build ideals in $R[x]$ paralleling the $L_m:$

$$\begin{align} A_m = (a_{m1}(x), a_{m2}(x), \cdots, a_{mk_m}(x)), \hspace{30pt} m \leq n. \end{align}$$

First we establish the following:

For the sake of illustration, suppose $g(x) = \beta x^{n+3} + \cdots,$ where $\beta \neq 0,$ so $\beta \in L_{n+3} = L_n.$ Assume further that:

$$\begin{align} L_n &= \{\alpha_1, \alpha_2, \alpha_3\}, \text{ putting } \alpha_{j} = \alpha_{n,j},\\ A_n &= \{a_1(x), a_2(x), a_3(x)\}, \text{ putting } a_j(x) = a_{n,j}(x),\\ &= \{\alpha_1 x^{n-1} + \cdots, \;\; \alpha_2 x^n + \cdots, \;\; \alpha_3 x ^{n-2} + \cdots\},\\ \end{align}$$

Note in the above that $\alpha_1 \in L_n$ means that there is a polynomial $a_1(x) \in A_n$ of degree less than or equal to $n$ with $\alpha_1$ as the leading coefficient, say $a_1(x) = \alpha_1 x^{n-1} + \cdots,$ where $\alpha_1 \neq 0$ and the ellipsis indicates terms in lower powers of $x.$ In fact, the degree of $a_1(x)$ could have been anything less than or equal to $n$ and $a_2(x)$ and $a_3(x)$ exemplify that possibility.

Since $\beta \in L_n$:

$$\begin{align} \beta = \lambda_1 \alpha_1 + \lambda_2 \alpha_2 + \lambda_3 \alpha_3, \hspace{20pt} \lambda_i \in R. \end{align}$$

Put:

$$\begin{align} g_1(x) = g(x) - \left(\lambda_1 \color{red}{x^4} a_1(x) + \lambda_2 \color{red}{x^3} a_2(x) + \lambda_3 \color{red}{x^5} a_2(x)\right),\\ \end{align}$$

where the highlighted weighting powers of $x$ are chosen to scale the $a_i (x)$ in question to have the same degree as $g(x).$ Note that the parenthesized expression is in $A_n,$ since it consists of sums of polynomials in $A_n$ multiplied by elements of $R[x].$ Continuing the calculation:

$$\begin{align} g_1(x) &= g(x) - \bigl(\lambda_1 \color{red}{x^4} \underbrace{(\alpha_1 x^{n-1} + \cdots)}_{\Large a_1(x)} + \lambda_2 \color{red}{x^3} \underbrace{(\alpha_2 x^n + \cdots)}_{\Large a_2(x)} + \lambda_3 \color{red}{x^5} \underbrace{(\alpha_3 x^{n-2} + \cdots)}_{\Large a_3(x)}\bigl)\\ &= \beta x^{n+3} + \cdots - \left((\lambda_1 \alpha_1 x^{n+3} + \cdots) + (\lambda_2 \alpha_2 x^{n+3} + \cdots) + (\lambda_3 \alpha_3 x^{n+3} + \cdots)\right)\\ &= \bigl(\underbrace{\beta - (\lambda_1 \alpha_1 + \lambda_2 \alpha_2 + \lambda_3 \alpha_3)}_{\Large 0}\bigl)x^{n+3} + \cdots\\ &= \text{ a polynomial of degree } \leq n+2 = m-1, \end{align}$$

so $g_1(x)$ is the polynomial promised in (1) in this example, and a general proof of (1) proceeds accordingly.

Keeping in mind that $n$ is the point at which the ideals $L_m$ in $R$ stabilize (that is, $L_m = L_n$ for $m \geq n$), repeated applications of (1) produce a series of polynomials $f_i(x) \in U$ for any given $f(x) \in U$ with $\deg{f} = p > n:$

$$\begin{alignat}{3} f(x) &= f_1(x) + b'(x), \hspace{5pt} \quad && b'(x) \in A_n, \quad &&&\deg{f_1} \leq p-1,\\ f_1(x) &= f_2(x) + b''(x), \hspace{5pt} \quad && b''(x) \in A_n, \quad &&&\deg{f_2} \leq p-2,\\ &\vdots\\ f_{r-1}(x) &= f_r(x) + b^{(r)}(x), \hspace{5pt} \quad && b^{(r)}(x) \in A_n, \quad &&&\deg{f_r} \leq n,\\ \end{alignat}$$

where $r \leq p - n,$ the inequality being strict if the degree is reduced by more than one at any step. Unrolling this sequence:

$$\begin{equation}{f(x) = f_r(x) + b(x), \quad b(x) \in A_n, \quad \deg{f_r} \leq n.}\tag{2} \end{equation}$$

That is, for every $f(x) \in U$ of degree greater than $n,$ there is an associated $f_r(x) \in U$ of degree less than or equal to $n$ such that:

$$\begin{equation}{f(x) \in \left(f_r(x), A_n\right), \quad \deg{f_r} \leq n.}\tag{3} \end{equation}$$

A similar degree reduction strategy is used on $f_r(x)$ by applying this principle:

(1') is proven just like (1) except that $g(x) - g_1(x)$ in (1') is in $A_{m-1},$ the next ideal down, instead of $A_n.$ Applying (1') to $f_r(x)$ produces polynomials $h(x)$ and $c(x)$ in $U$ such that:

$$\begin{equation}{f_r(x) = h(x) + c(x), \quad c(x) \in A_{n-1}, \quad \deg{h} \leq n-1.}\tag{4} \end{equation}$$

Combining (3) and (4):

$$\begin{align} f(x) \in \left(h(x), A_n, A_{n-1}\right), \quad \deg{h} \leq n-1. \end{align}$$

Continuing in this fashion:

$$\begin{align} f(x) \in \left(A_n, A_{n-1}, A_{n-2}, \cdots, A_0\right). \end{align}$$

$f(x)$ was assumed to be a polynomial in $U$ of degree greater than $n.$ If $\deg{f} \leq n,$ the latter part of the argument shows that $f(x)$ is in an ideal generated by some subset of the $A_i,$ so in either case:

$$\begin{align} U = \left(A_n, A_{n-1}, A_{n-2}, \cdots, A_0\right). \end{align}$$

Since each of the $A_i$ is finitely generated in $R[x]$ and $n$ is a fixed value depending on $U,$ it follows that $U$ is finitely generated; and since this is true of every ideal in $R[x], R[x]$ is Noetherian.QED.

Mike Bertrand

March 1, 2020

^ 1. "Ueber die Theorie der algebraischen Formen", by David Hilbert, Mathematische Annalen, 36 (4) (1890), pp. 473–534.

^ 2. A History of Abstract Algebra: From Algebraic Equations to Modern Algebra, by Jeremy Gray, Springer (2018), ISBN 978-3-319-94772-3. “This is not mathematics": p. 266. Gray translated the pertinent part of Hilbert's paper into English and discusses it helpfully in Chapter 25, pp. 263-273.

^ 3. See "On the Sources of My Book Moderne Algebra", by B. L. van der Waerden, Historia Mathematica, 2 (1975), pp. 31-40. "Emil Artin for the specifics" : p. 35.