proof of Hilbert basis theorem

Let R be a noetherian ringMathworldPlanetmath and let f(x)=anxn+an-1xn-1++a1x+a0R[x] with an0. Then call an the initial coefficientMathworldPlanetmath of f.

Let I be an ideal in R[x]. We will show I is finitely generatedMathworldPlanetmathPlanetmathPlanetmath, so that R[x] is noetherian. Now let f0 be a polynomialMathworldPlanetmathPlanetmath of least degree in I, and if f0,f1,,fk have been chosen then choose fk+1 from I(f0,f1,,fk) of minimal degree. Continuing inductively gives a sequence (fk) of elements of I.

Let ak be the initial coefficient of fk, and consider the ideal J=(a1,a2,a3,) of initial coefficients. Since R is noetherian, J=(a0,,aN) for some N.

Then I=(f0,f1,,fN). For if not then fN+1I(f0,f1,,fN), and aN+1=k=0Nukak for some u1,u2,,uNR. Let g(x)=k=0Nukfkxνk where νk=deg(fN+1)-deg(fk).

Then deg(fN+1-g)<deg(fN+1), and fN+1-gI and fN+1-g(f0,f1,,fN). But this contradicts minimality of deg(fN+1).

Hence, R[x] is noetherian.

Title proof of Hilbert basis theorem
Canonical name ProofOfHilbertBasisTheorem
Date of creation 2013-03-22 12:59:27
Last modified on 2013-03-22 12:59:27
Owner bwebste (988)
Last modified by bwebste (988)
Numerical id 6
Author bwebste (988)
Entry type Proof
Classification msc 13E05