# Hilbert's basis theorem

In mathematics, specifically commutative algebra, Hilbert's basis theorem states that every ideal in the ring of multivariate polynomials over a Noetherian ring is finitely generated. 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 aR[X] were a non-finitely generated left-ideal. Then by recursion (using the axiom of countable choice) there is a sequence (fn)nN of polynomials such that if bn is the left ideal generated by f0, ..., fn−1 then fn in a\bn is of minimal degree. It is clear that (deg(fn))nN is a non-decreasing sequence of naturals. Let an be the leading coefficient of fn and let b be the left ideal in R generated by {a0, a1, ...}. Since R is left-Noetherian, we have that b must be finitely generated; and since the an comprise an R-basis, it follows that for a finite amount of them, say {ai : i < N}, will suffice. So for example, $a_N=\sum_{i some ui in R. Now consider

$g \triangleq\sum_{i

whose leading term is equal to that of fN; moreover, gbN. However, fNbN, which means that fNga\bN has degree less than fN, contradicting the minimality.

Second Proof. Let aR[X] be a left-ideal. Let b be the set of leading coefficients of members of a. This is obviously a left-ideal over R, and so is finitely generated by the leading coefficients of finitely many members of a; say f0, ..., fN−1. Let $d\triangleq\max_{i}\deg(f_{i}).\,$ Let bk be the set of leading coefficients of members of a, whose degree is ≤ k. As before, the bk are left-ideals over R, and so are finitely generated by the leading coefficients of finitely many members of a, say $f^{(k)}_{0},\ldots,f^{(k)}_{N^{(k)}-1},\,$ with degrees ≤ k. Now let a* ⊆ R[X] be the left-ideal generated by

$\{f_{i},f^{(k)}_{j}:i

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

Case 1: deg(h) ≥ d. Regardless of this condition, we have ab, so is a left-linear combination $a=\sum_j u_j a_j\,$ of the coefficients of the fj Consider $\tilde{h}\triangleq\sum_{j}u_{j}X^{\deg(h)-\deg(f_{j})}f_{j},\,$ which has the same leading term as h; moreover $\tilde{h}\in\mathfrak{a}^{\ast}\not\ni h\,$ so $h-\tilde{h}\in\mathfrak{a}\setminus \mathfrak{a}^{\ast}\,$ of degree < deg(h), contradicting minimality.
Case 2: deg(h) = k < d. Then abk so is a left-linear combination $a=\sum_j u_j a^{(k)}_j$ of the leading coefficients of the $f^{(k)}_j.$ Considering $\tilde{h}\triangleq\sum_j u_j X^{\deg(h)-\deg(f^{(k)}_{j})}f^{(k)}_{j},$ we yield a similar contradiction as in Case 1.

Thus our claim holds, and a = 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. First, by induction we see that $R[X_{0},X_{1},\ldots,X_{n-1}]\,$ will also be Noetherian. Second, since any affine variety over Rn (i.e. a locus-set of a collection of polynomials) may be written as the locus of an ideal $\mathfrak{a}\subseteq R[X_{0},X_{1},\ldots,X_{n-1}]\,$ 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. Finally, if $\mathcal{A}\,$ is a finitely-generated R-algebra, then we know that $\mathcal{A}\cong R[X_{0},X_{1},\ldots,X_{n-1}]/\langle\mathfrak{a}\rangle\,$ (i.e. mod-ing out by relations), where a a set of polynomials. We can assume that a is an ideal and thus is finitely generated. So $\mathcal{A}\,$ is a free R-algebra (on n generators) generated by finitely many relations $\mathcal{A}\cong R[X_{0},X_{1},\ldots,X_{n-1}]/\langle p_{0},\ldots,p_{N-1}\rangle$.

## Mizar System

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