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Hilbert's basis theorem

## Summary

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

## Statement

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

Hilbert's Basis Theorem. If ${\displaystyle R}$  is a Noetherian ring, then ${\displaystyle R[X]}$  is a Noetherian ring.

Corollary. If ${\displaystyle R}$  is a Noetherian ring, then ${\displaystyle R[X_{1},\dotsc ,X_{n}]}$  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.[1]

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 ${\displaystyle R}$  is a left (resp. right) Noetherian ring, then the polynomial ring ${\displaystyle R[X]}$  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 ${\displaystyle {\mathfrak {a}}\subseteq R[X]}$  is a non-finitely generated left ideal. Then by recursion (using the axiom of dependent choice) there is a sequence of polynomials ${\displaystyle \{f_{0},f_{1},\ldots \}}$  such that if ${\displaystyle {\mathfrak {b}}_{n}}$  is the left ideal generated by ${\displaystyle f_{0},\ldots ,f_{n-1}}$  then ${\displaystyle f_{n}\in {\mathfrak {a}}\setminus {\mathfrak {b}}_{n}}$  is of minimal degree. It is clear that ${\displaystyle \{\deg(f_{0}),\deg(f_{1}),\ldots \}}$  is a non-decreasing sequence of natural numbers. Let ${\displaystyle a_{n}}$  be the leading coefficient of ${\displaystyle f_{n}}$  and let ${\displaystyle {\mathfrak {b}}}$  be the left ideal in ${\displaystyle R}$  generated by ${\displaystyle a_{0},a_{1},\ldots }$ . Since ${\displaystyle R}$  is Noetherian the chain of ideals

${\displaystyle (a_{0})\subset (a_{0},a_{1})\subset (a_{0},a_{1},a_{2})\subset \cdots }$

must terminate. Thus ${\displaystyle {\mathfrak {b}}=(a_{0},\ldots ,a_{N-1})}$  for some integer ${\displaystyle N}$ . So in particular,

${\displaystyle a_{N}=\sum _{i

Now consider

${\displaystyle g=\sum _{i

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

### Second Proof

Let ${\displaystyle {\mathfrak {a}}\subseteq R[X]}$  be a left ideal. Let ${\displaystyle {\mathfrak {b}}}$  be the set of leading coefficients of members of ${\displaystyle {\mathfrak {a}}}$ . This is obviously a left ideal over ${\displaystyle R}$ , and so is finitely generated by the leading coefficients of finitely many members of ${\displaystyle {\mathfrak {a}}}$ ; say ${\displaystyle f_{0},\ldots ,f_{N-1}}$ . Let ${\displaystyle d}$  be the maximum of the set ${\displaystyle \{\deg(f_{0}),\ldots ,\deg(f_{N-1})\}}$ , and let ${\displaystyle {\mathfrak {b}}_{k}}$  be the set of leading coefficients of members of ${\displaystyle {\mathfrak {a}}}$ , whose degree is ${\displaystyle \leq k}$ . As before, the ${\displaystyle {\mathfrak {b}}_{k}}$  are left ideals over ${\displaystyle R}$ , and so are finitely generated by the leading coefficients of finitely many members of ${\displaystyle {\mathfrak {a}}}$ , say

${\displaystyle f_{0}^{(k)},\ldots ,f_{N^{(k)}-1}^{(k)}}$

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

${\displaystyle \left\{f_{i},f_{j}^{(k)}\,:\ i

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

Case 1: ${\displaystyle \deg(h)\geq d}$ . Regardless of this condition, we have ${\displaystyle a\in {\mathfrak {b}}}$ , so is a left linear combination
${\displaystyle a=\sum _{j}u_{j}a_{j}}$
of the coefficients of the ${\displaystyle f_{j}}$ . Consider
${\displaystyle h_{0}\triangleq \sum _{j}u_{j}X^{\deg(h)-\deg(f_{j})}f_{j},}$
which has the same leading term as ${\displaystyle h}$ ; moreover ${\displaystyle h_{0}\in {\mathfrak {a}}^{*}}$  while ${\displaystyle h\notin {\mathfrak {a}}^{*}}$ . Therefore ${\displaystyle h-h_{0}\in {\mathfrak {a}}\setminus {\mathfrak {a}}^{*}}$  and ${\displaystyle \deg(h-h_{0})<\deg(h)}$ , which contradicts minimality.
Case 2: ${\displaystyle \deg(h)=k . Then ${\displaystyle a\in {\mathfrak {b}}_{k}}$  so is a left linear combination
${\displaystyle a=\sum _{j}u_{j}a_{j}^{(k)}}$
of the leading coefficients of the ${\displaystyle f_{j}^{(k)}}$ . Considering
${\displaystyle h_{0}\triangleq \sum _{j}u_{j}X^{\deg(h)-\deg(f_{j}^{(k)})}f_{j}^{(k)},}$
we yield a similar contradiction as in Case 1.

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

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

## Applications

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

1. By induction we see that ${\displaystyle R[X_{0},\dotsc ,X_{n-1}]}$  will also be Noetherian.
2. Since any affine variety over ${\displaystyle R^{n}}$  (i.e. a locus-set of a collection of polynomials) may be written as the locus of an ideal ${\displaystyle {\mathfrak {a}}\subset R[X_{0},\dotsc ,X_{n-1}]}$  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 ${\displaystyle A}$  is a finitely-generated ${\displaystyle R}$ -algebra, then we know that ${\displaystyle A\simeq R[X_{0},\dotsc ,X_{n-1}]/{\mathfrak {a}}}$ , where ${\displaystyle {\mathfrak {a}}}$  is an ideal. The basis theorem implies that ${\displaystyle {\mathfrak {a}}}$  must be finitely generated, say ${\displaystyle {\mathfrak {a}}=(p_{0},\dotsc ,p_{N-1})}$ , i.e. ${\displaystyle A}$  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).

## References

1. ^ Hilbert, David (1890). "Ueber die Theorie der algebraischen Formen". Mathematische Annalen. 36 (4): 473–534. doi:10.1007/BF01208503. ISSN 0025-5831. S2CID 179177713.