# Hilbert's basis theorem

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

## Statement

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

Hilbert's Basis Theorem.If is a Noetherian ring, then is a Noetherian ring.

Corollary.If is a Noetherian ring, then 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. Template:Harvs 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 is a left (resp. right) Noetherian ring, then the polynomial ring 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 were a non-finitely generated left-ideal. Then by recursion (using the axiom of dependent choice) there is a sequence of polynomials such that if is the left ideal generated by then in is of minimal degree. It is clear that is a non-decreasing sequence of naturals. Let be the leading coefficient of and let be the left ideal in generated by . Since is left-Noetherian, we have that must be finitely generated; and since the comprise an -basis, it follows that for a finite amount of them, say , will suffice. So for example,

Now consider

whose leading term is equal to that of ; moreover, . However, , which means that has degree less than , contradicting the minimality.

### Second Proof

Let be a left-ideal. Let be the set of leading coefficients of members of . This is obviously a left-ideal over , and so is finitely generated by the leading coefficients of finitely many members of ; say . Let be the maximum of the set , and let be the set of leading coefficients of members of , whose degree is . As before, the are left-ideals over , and so are finitely generated by the leading coefficients of finitely many members of , say

with degrees . Now let be the left-ideal generated by

We have and claim also . Suppose for the sake of contradiction this is not so. Then let be of minimal degree, and denote its leading coefficient by .

- we yield a similar contradiction as in Case 1.

Thus our claim holds, and which is finitely generated.

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

## Applications

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

- By induction we see that will also be Noetherian.
- Since any affine variety over (i.e. a locus-set of a collection of polynomials) may be written as the locus of an ideal 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.
- If is a finitely-generated -algebra, then we know that , where is an ideal. The basis theorem implies that must be finitely generated, say , i.e. is finitely presented.

## Mizar System

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

## References

- Cox, Little, and O'Shea,
*Ideals, Varieties, and Algorithms*, Springer-Verlag, 1997. - {{#invoke:citation/CS1|citation

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