Hilbert's basis theorem

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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 f1, f2, ... of elements of I such that fi+1 has minimal degree in I \ Ji, where Ji is the ideal generated by f1, ..., fi. Let ai be the leading coefficient of fi and let J be the ideal of R generated by a1, a2, ... Since R is Noetherian there exists an integer N such that J is generated by a1, ..., aN, so in particular aN+1 = u1a1 + ... + uNaN for some u1, ..., uN in R. Now consider g = u1f1xn1 + ... + uNfNxnN where ni = deg fN+1 − deg fi. Because deg g = deg fN+1 and the leading coefficients of g and fN+1 agree, the difference fN+1g has degree strictly less than the degree of fN+1, contradicting the choice of fN+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 a1,... ,an in L, and pick f1,... ,fn in I such that the leading coefficient of fi is ai. Let di be the degree of fi and let N be the maximum of the di. Now for each k = 0, ..., N1 let Lk be the set of leading coefficients of elements of I with degree at most k. Then again, Lk is an ideal in R, so is finitely generated by ak1,... ,akmk say. As before, let fki in I have leading coefficient aki. Let H be the ideal in R[X] generated by the fi and the fki. 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 = r1a1 + ... + rnan and g = r1Xdd1f1 + ... + rnXddnfn is of the same degree as f and has the same leading coefficient. Since g is in H, fg is not, which contradicts the minimality of f. If on the other hand d is strictly smaller than N, then a is in Ld, so a = r1ad1 + ... + rmdadmd. A similar construction as above gives the same contradiction. Thus, I = H, which is finitely generated, and this finishes the proof.