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.

Contents

Proof

The following more general statement will be proved.

Theorem. If R\, is a left- (respectively right-) Noetherian ring, then the polynomial ring R[X]\, is also a left- (respectively right-) Noetherian ring.

It suffices to consider just the "Left" case.

Proof (Theorem)

Suppose per contra that \mathfrak{a}\subseteq R[X]\, were a non-finitely generated left-ideal. Then it would be that by recursion (using the axiom of countable choice) that a sequence (f_{n})_{n\in\mathbb{N}}\, of polynomials could be found so that, letting \mathfrak{b}_{n}\triangleq\langle f_{0},\ldots,f_{n-1}\rangle,\, f_{n}\in \mathfrak{a}\setminus \mathfrak{b}_{n}\, of minimal degree. It is clear that (\deg(f_{n}))_{n\in\mathbb{N}}\, is a non-decreasing sequence of naturals. Now consider the left-ideal \mathfrak{b}\triangleq\langle a_{n}:n\in\mathbb{N}\rangle\, over R,\, where the a_{n}\, are the leading coefficients of the f_{n}.\,. Since R\, is left-Noetherian, we have that \mathfrak{b}\, must be finitely generated; and since the a_{n}\, comprise an R\,-basis, it follows that for a finite amount of them, say \{a_{i}:i<N\},\, will suffice. So for example, a_{N}=\Sigma_{i<N}u_{i}a_{i}\, some u_{i}\in R.\, Now consider \tilde{f}(X)\triangleq\Sigma_{i<N}u_{i}X^{\deg(f_{N})-\deg(f_{i})}f_{i}(X),\, whose leading term is equal to that of f_{N}\,; moreover \tilde{f}\in\mathfrak{b}_{N}\not\ni f_{N}\, so f_{N}-\tilde{f}\in\mathfrak{a}\setminus\mathfrak{b}_{N}\, of degree <\deg(f_{N}),\, contradicting minimality.

 

 

 

 

\blacksquare\, (Thm)

A constructive proof (not invoking the axiom of choice) also exists.

Proof (Theorem):

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\triangleq\max_{i}\deg(f_{i}).\, 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^{(k)}_{0},\ldots,f^{(k)}_{N^{(k)}-1},\, with degrees \leq k.\, Now let \mathfrak{a}^{\ast}\subseteq R[X]\, be the left-ideal generated by \{f_{i},f^{(k)}_{j}:i<N,j<N^{(k)},k<d\}.\, We have \mathfrak{a}^{\ast}\subseteq\mathfrak{a}\, and claim also \mathfrak{a}\subseteq\mathfrak{a}^{\ast}.\,

Suppose per contra this were not so. Then let h\in\mathfrak{a}\setminus\mathfrak{a}^{\ast}\, 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=\Sigma_{j}u_{j}a_{j}\, of the coefficients of the f_{j}.\, Consider \tilde{h}\triangleq\Sigma_{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 a\in \mathfrak{b}_{k}\, so is a left-linear combination a=\Sigma_{j}u_{j}a^{(k)}_{j}\, of the leading coefficients of the f^{(k)}_{j}.\, Considering \tilde{h}\triangleq\Sigma_{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 \mathfrak{a}=\mathfrak{a}^{\ast}\, which is finitely generated.

 

 

 

 

\blacksquare\, (Thm)

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 Nötherian 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 Nötherian. Second, 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}\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}\, were 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 \mathfrak{a}\, a set of polynomials. We can assume that \mathfrak{a}\, is an ideal and thus is finitely generated. So \mathcal{A}\, would be 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.

References