Hilbert's basis theorem

Short description: Polynomial rings are Noetherian rings

In mathematics, specifically commutative algebra, Hilbert's basis theorem says that a polynomial ring over a Noetherian ring is Noetherian.

Statement

If $R$ is a ring, let $R[X]$ denote the ring of polynomials in the indeterminate $X$ over $R$. Hilbert proved that if $R$ is "not too large", in the sense that if $R$ is Noetherian, the same must be true for $R[X]$. Formally,

Hilbert's Basis Theorem. If

$R$

is a Noetherian ring, then

$R[X]$

is a Noetherian ring.

Corollary. If

$R$

is a Noetherian ring, then

$R[X_1,\dots ,X_n]$

is a Noetherian ring.

This can be translated into algebraic geometry as follows: every algebraic set over a field can be described as the set of common roots of finitely many polynomial equations. Hilbert proved the theorem (for the special case of polynomial rings over a field) in the course of his proof of finite generation of rings of invariants.

Hilbert produced an innovative proof by contradiction using mathematical induction; his method does not give an algorithm to produce the finitely many basis polynomials for a given ideal: it only shows that they must exist. One can determine basis polynomials using the method of Gröbner bases.

Proof

Theorem. If $R$ is a left (resp. right) Noetherian ring, then the polynomial ring $R[X]$ is also a left (resp. right) Noetherian ring.

Remark. We will give two proofs, in both only the "left" case is considered; the proof for the right case is similar.

First proof

Suppose $\mathfrak{𝔞}\subseteq R[X]$ is a non-finitely generated left ideal. Then by recursion (using the axiom of dependent choice) there is a sequence of polynomials $\{f_0,f_1,\dots \}$ such that if ${\mathfrak{𝔟}}_n$ is the left ideal generated by $f_0,\dots ,f_{n-1}$ then $f_n\in \mathfrak{𝔞}\setminus {\mathfrak{𝔟}}_n$ is of minimal degree. By construction, $\{\mathrm{deg}(f_0),\mathrm{deg}(f_1),\dots \}$ is a non-decreasing sequence of natural numbers. Let $a_n$ be the leading coefficient of $f_n$ and let $\mathfrak{𝔟}$ be the left ideal in $R$ generated by $a_0,a_1,\dots$. Since $R$ is Noetherian the chain of ideals

$(a_0)\subset (a_0,a_1)\subset (a_0,a_1,a_2)\subset \cdots$

must terminate. Thus $\mathfrak{𝔟}=(a_0,\dots ,a_{N-1})$ for some integer $N$. So in particular,

$a_N=\sum _{i<N}{u}_i{a}_i,u_i\in R.$

Now consider

$g=\sum _{i<N}{u}_i{X}^{\mathrm{deg}(f_N)-\mathrm{deg}(f_i)}{f}_i,$

whose leading term is equal to that of $f_N$; moreover, $g\in {\mathfrak{𝔟}}_N$. However, $f_N\notin {\mathfrak{𝔟}}_N$, which means that $f_N-g\in \mathfrak{𝔞}\setminus {\mathfrak{𝔟}}_N$ has degree less than $f_N$, contradicting the minimality.

Second proof

Let $\mathfrak{𝔞}\subseteq R[X]$ be a left ideal. Let $\mathfrak{𝔟}$ be the set of leading coefficients of members of $\mathfrak{𝔞}$. This is obviously a left ideal over $R$, and so is finitely generated by the leading coefficients of finitely many members of $\mathfrak{𝔞}$; say $f_0,\dots ,f_{N-1}$. Let $d$ be the maximum of the set $\{\mathrm{deg}(f_0),\dots ,\mathrm{deg}(f_{N-1})\}$, and let ${\mathfrak{𝔟}}_k$ be the set of leading coefficients of members of $\mathfrak{𝔞}$, whose degree is $\le k$. As before, the ${\mathfrak{𝔟}}_k$ are left ideals over $R$, and so are finitely generated by the leading coefficients of finitely many members of $\mathfrak{𝔞}$, say

$f_0^{(k)},\dots ,f_{N^{(k)}-1}^{(k)}$

with degrees $\le k$. Now let ${\mathfrak{𝔞}}^{*}\subseteq R[X]$ be the left ideal generated by:

$\{f_i,f_j^{(k)}:i<N,j<N^{(k)},k<d\}.$

We have ${\mathfrak{𝔞}}^{*}\subseteq \mathfrak{𝔞}$ and claim also $\mathfrak{𝔞}\subseteq {\mathfrak{𝔞}}^{*}$. Suppose for the sake of contradiction this is not so. Then let $h\in \mathfrak{𝔞}\setminus {\mathfrak{𝔞}}^{*}$ be of minimal degree, and denote its leading coefficient by $a$.

Case 1: $\mathrm{deg}(h)\ge d$. Regardless of this condition, we have $a\in \mathfrak{𝔟}$, so $a$ is a left linear combination

$a=\sum _j{u}_j{a}_j$

of the coefficients of the $f_j$. Consider

$h_0=\sum _j{u}_j{X}^{\mathrm{deg}(h)-\mathrm{deg}(f_j)}{f}_j,$

which has the same leading term as $h$; moreover $h_0\in {\mathfrak{𝔞}}^{*}$ while $h\notin {\mathfrak{𝔞}}^{*}$. Therefore $h-h_0\in \mathfrak{𝔞}\setminus {\mathfrak{𝔞}}^{*}$ and $\mathrm{deg}(h-h_0)<\mathrm{deg}(h)$, which contradicts minimality.

Case 2: $\mathrm{deg}(h)=k<d$. Then $a\in {\mathfrak{𝔟}}_k$ so $a$ is a left linear combination

$a=\sum _j{u}_j{a}_j^{(k)}$

of the leading coefficients of the $f_j^{(k)}$. Considering

$h_0=\sum _j{u}_j{X}^{\mathrm{deg}(h)-\mathrm{deg}(f_j^{(k)})}{f}_j^{(k)},$

we yield a similar contradiction as in Case 1.

Thus our claim holds, and $\mathfrak{𝔞}={\mathfrak{𝔞}}^{*}$ which is finitely generated.

Note that the only reason we had to split into two cases was to ensure that the powers of $X$ multiplying the factors were non-negative in the constructions.

Applications

Let $R$ be a Noetherian commutative ring. Hilbert's basis theorem has some immediate corollaries.

1. By induction we see that $R[X_0,\dots ,X_{n-1}]$ will also be Noetherian.
2. Since any affine variety over $R^n$ (i.e. a locus-set of a collection of polynomials) may be written as the locus of an ideal $\mathfrak{𝔞}\subset R[X_0,\dots ,X_{n-1}]$ and further as the locus of its generators, it follows that every affine variety is the locus of finitely many polynomials — i.e. the intersection of finitely many hypersurfaces.
3. If $A$ is a finitely-generated $R$-algebra, then we know that $A\simeq R[X_0,\dots ,X_{n-1}]/\mathfrak{𝔞}$, where $\mathfrak{𝔞}$ is an ideal. The basis theorem implies that $\mathfrak{𝔞}$ must be finitely generated, say $\mathfrak{𝔞}=(p_0,\dots ,p_{N-1})$, i.e. $A$ is finitely presented.

Formal proofs

Formal proofs of Hilbert's basis theorem have been verified through the Mizar project (see HILBASIS file) and Lean (see ring_theory.polynomial).

Further reading

• Cox, Little, and O'Shea, Ideals, Varieties, and Algorithms, Springer-Verlag, 1997.

• {{citation | last=Roman | first=Stephen