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, ..., xn] is finitely generated. This can be translated into algebraic geometry as follows: every algebraic set 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 , if with an not equal to 0, then degf: = n and an 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 f1,f2,... of elements of I such that fi + 1 has minimal degree among elements of , where Ji is the ideal generated by f1,...,fi. Let ai be the leading coeffecient of fi and let J be the ideal of R generated by the sequence a1,a2,.... Since R is Noetherian there exists N such that J is generated by a1,...,aN. Therefore for some . We obtain a contradiction by considering where ni = degfN + 1 − degfi, because degg = degfN + 1 and their leading coefficients agree, so that fN + 1 − g has degree strictly less than degfN + 1, contradicting the choice of fN + 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 a constructive proof:
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 degree atmost k. 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 J 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)Xd − d(1)f(1) + ... + r(n)Xd − d(n)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(m(d))a(d,m(d)). A similar construction as above again gives the same contradiction. Thus, J=H, which is finitely generated. QED.
[edit] Other
The Mizar project has completely formalized and automatically checked a proof of Hilbert's basis theorem in the HILBASIS file.
[edit] References
- Cox, Little, and O'Shea, Ideals, Varieties, and Algorithms, Springer-Verlag, 1997.