Hilbert's basis theorem

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In mathematics, Hilbert's basis theorem, first proved by David Hilbert in 1888, states that, if k is a field, then every ideal in the ring of multivariate polynomials k[x₁, x₂, ..., x_n] is finitely generated. This can be translated into algebraic geometry as follows: every variety over k can be described as the set of common roots of finitely many polynomial equations.

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.

[edit] Proof

A slightly more general statement of Hilbert's basis theorem is: if R is a left (respectively right) Noetherian ring, then the polynomial ring R[X] is also left (respectively right) Noetherian.

Proof: For <math>f \in R[x]</math>, if <math>f=\sum_{k=0}^na_kx^k</math> with <math>a_n</math> not equal to 0, then <math>\deg f := n</math> and <math>a_n</math> is the leading coefficient of f. Let I be an ideal in R[x] and assume I is not finitely generated. Then inductively construct a sequence <math>f_1,f_2,...</math> of elements of I such that <math>f_{i+1}</math> has minimal degree among elements of <math>I\setminus J_i</math>, where <math>J_i</math> is the ideal generated by <math>f_1,...,f_i</math>. Let <math>a_i</math> be the leading coeffecient of <math>f_i</math> and let J be the ideal of R generated by the sequence <math>a_1,a_2,...</math>. Since R is Noetherian there exists N such that J is generated by <math>a_1,...,a_N</math>. Therefore <math>a_{N+1} = \sum_{i=1}^Nu_ia_i</math> for some <math>u_1,...,u_N \in R</math>. We obtain a contradiction by considering <math>g = \sum_{i=1}^Nu_if_ix^{n_i}</math> where <math>n_i = \deg f_{N+1} - \deg f_i</math>, because <math>\deg g = \deg f_{N+1}</math> and their leading coefficients agree, so that <math>f_{N+1} - g</math> has degree strictly less than <math>\deg f_{N+1}</math>, contradicting the choice of <math>f_{N+1}</math>. Thus I is finitely generated. Since I was an arbitrary ideal in R[x], every ideal in R[x] is finitely generated and R[x] is therefore Noetherian.

or:

Given an ideal J of R[X] let L be the set of leading coefficients of the elements of J. Then L is clearly an ideal in R so is finitely generated by a(1),...,a(n) in L, and there are f(1),...,f(n) in J with a(i) being the leading coefficient of f(i). Let d(i) be the degree of f(i) and let N be the maximum of the d(i)'s. Now for each k=0,...,N-1 let L(k) be the set of leading coeficients of elements of J with 0. Then again, L(k) is clearly an ideal in R so is finitely generated by a(k,1),...,a(k,m(k)) say. As before, let f(k,i) in J have leading coefficient a(k,i). Let H be the ideal in R[X] generated by the f(i)'s and f(k,i)'s. Then surely H is contained in J and assume there is an element f in L not belonging to H, of least degree d, and leading coefficient a. If d is larger or equal to N then a is in L so, a=r(1)a(1)+...+r(1)a(n) and g= <math>r(1)X^{d-d(1)}+...+r(n)X^{d-d(n)}</math> is a polynomial of same degree as f with the same leading coefficient, hence f-g is in J and by minimality of f, f-g is in H so f is in H, a contradiction. If, on the other hand, d is strictly smaller than N, then a is in L(d) so a=r(1)a(d,1)+...+r(m(d))a(d,m(d)) and g=r(1)f(d,1)+...+r(m(d))f(d,m(d)) is a polynomial in H of the same degree as f and with the same leading coefficient. Hence, f-g is in H by minimality of f, and so f is in H, a contradiction. Then H=J and so J is finitely generated, and consequently R[X] is Noetherian.