# Hilbert's basis theorem

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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 $R[X_{1},\dotsc ,X_{n}]$ 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 (1890) 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 $\{f_{0},f_{1},\ldots \}$ of polynomials such that if ${\mathfrak {b}}_{n}$ is the left ideal generated by $f_{0},\ldots ,f_{n-1}$ then $f_{n}\in {\mathfrak {a}}\setminus {\mathfrak {b}}_{n}$ is of minimal degree. It is clear that $\{\deg(f_{0}),\deg(f_{1}),\ldots \}$ is a non-decreasing sequence of naturals. Let $a_{n}$ be the leading coefficient of $f_{n}$ and let ${\mathfrak {b}}$ be the left ideal in $R$ generated by $a_{0},a_{1},\ldots$ . 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 ${\mathfrak {b}}=(a_{0},\ldots ,a_{N-1})$ for some integer $N$ . So in particular,

$a_{N}=\sum _{i Now consider

$g=\sum _{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 $f_{0},\ldots ,f_{N-1}$ . Let $d$ be the maximum of the set $\{\deg(f_{0}),\ldots ,\deg(f_{N-1})\}$ , and let ${\mathfrak {b}}_{k}$ be the set of leading coefficients of members of ${\mathfrak {a}}$ , whose degree is ${}\leq k$ . 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

$f_{0}^{(k)},\ldots ,f_{N^{(k)}-1}^{(k)}$ with degrees ${}\leq k$ . Now let ${\mathfrak {a}}^{*}\subseteq R[X]$ be the left-ideal generated by:

$\left\{f_{i},f_{j}^{(k)}\ :\ i 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: $\deg(h)\geq d$ . Regardless of this condition, we have $a\in {\mathfrak {b}}$ , so is a left-linear combination
$a=\sum _{j}u_{j}a_{j}$ of the coefficients of the $f_{j}$ . Consider
$h_{0}\triangleq \sum _{j}u_{j}X^{\deg(h)-\deg(f_{j})}f_{j},$ 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 $\deg(h-h_{0})<\deg(h)$ , which contradicts minimality.
Case 2: $\deg(h)=k . Then $a\in {\mathfrak {b}}_{k}$ so is a left-linear combination
$a=\sum _{j}u_{j}a_{j}^{(k)}$ of the leading coefficients of the $f_{j}^{(k)}$ . Considering
$h_{0}\triangleq \sum _{j}u_{j}X^{\deg(h)-\deg(f_{j}^{(k)})}f_{j}^{(k)},$ 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 $R[X_{0},\dotsc ,X_{n-1}]$ 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 ${\mathfrak {a}}\subset R[X_{0},\dotsc ,X_{n-1}]$ 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 $A\simeq R[X_{0},\dotsc ,X_{n-1}]/{\mathfrak {a}}$ , where ${\mathfrak {a}}$ is an ideal. The basis theorem implies that ${\mathfrak {a}}$ must be finitely generated, say ${\mathfrak {a}}=(p_{0},\dotsc ,p_{N-1})$ , i.e. $A$ is finitely presented.

## Formal Proof in Mizar

The Mizar project has completely formalized and automatically checked a proof of Hilbert's basis theorem in the HILBASIS file.