BREAKING NEWS

## Summary

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 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 $\{f_{0},f_{1},\ldots \}$  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 natural numbers. Let $a_{n}$  be the leading coefficient of $f_{n}$  and let ${\mathfrak {b}}$  be the left ideal in $R$  generated by $a_{0},a_{1},\ldots$ . 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 proofs

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