# 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 {f0, f1, ...}. 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(f0), deg(f1), ...}, 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 {a0, ..., aN−1}, will suffice. So for example,

$a_N=\sum_{i

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 be the maximum of the set {deg(f0), ..., deg(fN−1)}, and 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

$f^{(k)}_{0}, \cdots, f^{(k)}_{N^{(k)}-1},$

with degrees k. Now let a* ⊆ R[X] be the left-ideal generated by

$\left \{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
$h_0 \triangleq\sum_{j}u_{j}X^{\deg(h)-\deg(f_{j})}f_{j},$
which has the same leading term as h; moreover h0a* while ha*. Therefore hh0a\a* and deg(hh0) < deg(h), which contradicts 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
$h_0 \triangleq\sum_j u_j X^{\deg(h)-\deg(f^{(k)}_{j})}f^{(k)}_{j},$
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.

1. By induction we see that R[X0, ..., Xn−1] will also be Noetherian.
2. 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 aR[X0, ..., Xn−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 AR[X0, ..., Xn−1]/(a), where a a set of polynomials. We can assume that a is an ideal and thus is finitely generated. So A is a free R-algebra (on n generators) generated by finitely many relations AR[X0, ..., Xn−1]/(p0, ..., pN−1).

## Mizar System

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