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 $f \in R[x]$ , if $f=\sum_{k=0}^na_kx^k$ with $a n$ not equal to 0, then $deg f : = n$ and $a n$ 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 $f 1, f 2,...$ of elements of I such that $f i + 1$ has minimal degree among elements of $I\setminus J_i$ , where $J i$ is the ideal generated by $f 1,..., f i$ . Let $a i$ be the leading coeffecient of $f i$ and let J be the ideal of R generated by the sequence $a 1, a 2,...$ . Since R is Noetherian there exists N such that J is generated by $a 1,..., a N$ . Therefore $a_{N+1} = \sum_{i=1}^Nu_ia_i$ for some $u_1,...,u_N \in R$ . We obtain a contradiction by considering $g = \sum_{i=1}^Nu_if_ix^{n_i}$ where $n i = deg f N + 1 - deg f i$ , because $deg g = deg f N + 1$ and their leading coefficients agree, so that $f N + 1 - g$ has degree strictly less than $deg f N + 1$ , contradicting the choice of $f N + 1$ . 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= $r (1) X d - d (1) + ... + r (n) X d - d (n)$ 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.