In mathematics, 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
The following more general statement will be proved: if R is a left (respectively right) Noetherian ring then the polynomial ring R[X] is also a left (respectively right) Noetherian ring.
Let I be an ideal in R[X] and assume for a contradiction that I is not finitely generated. Inductively construct a sequence f_{1}, f_{2}, ... of elements of I such that f_{i+1} has minimal degree in I \ J_{i}, where J_{i} is the ideal generated by f_{1}, ..., f_{i}. Let a_{i} be the leading coefficient of f_{i} and let J be the ideal of R generated by a_{1}, a_{2}, ... Since R is Noetherian there exists an integer N such that J is generated by a_{1}, ..., a_{N}, so in particular a_{N+1} = u_{1}a_{1} + ... + u_{N}a_{N} for some u_{1}, ..., u_{N} in R. Now consider g = u_{1}f_{1}x^{n1} + ... + u_{N}f_{N}x^{nN} where n_{i} = deg f_{N+1} − deg f_{i}. Because deg g = deg f_{N+1} and the leading coefficients of g and f_{N+1} agree, the difference f_{N+1} − g has degree strictly less than the degree of f_{N+1}, contradicting the choice of f_{N+1}. Therefore I is finitely generated, and the proof is complete.
A constructive proof also exists: Given an ideal I of R[X], let L be the set of leading coefficients of the elements of I. Then L is an ideal in R so is finitely generated by a_{1},... ,a_{n} in L, and pick f_{1},... ,f_{n} in I such that the leading coefficient of f_{i} is a_{i}. Let d_{i} be the degree of f_{i} and let N be the maximum of the d_{i}. Now for each k = 0, ..., N − 1 let L_{k} be the set of leading coefficients of elements of I with degree at most k. Then again, L_{k} is an ideal in R, so is finitely generated by a^{k}_{1},... ,a^{k}_{mk} say. As before, let f^{k}_{i} in I have leading coefficient a^{k}_{i}. Let H be the ideal in R[X] generated by the f_{i} and the f^{k}_{i}. Then surely H is contained in I and assume there is an element f in I not belonging to H, of least degree d, and leading coefficient a. If d is larger than or equal to N then a is in L so, a = r_{1}a_{1} + ... + r_{n}a_{n} and g = r_{1}X^{d−d1}f_{1} + ... + r_{n}X^{d−dn}f_{n} is of the same degree as f and has the same leading coefficient. Since g is in H, f − g is not, which contradicts the minimality of f. 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_{md}a^{d}_{md}. A similar construction as above gives the same contradiction. Thus, I = H, which is finitely generated, and this finishes the proof.
Full article ▸
