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单词 ProofOfHilbertBasisTheorem
释义

proof of Hilbert basis theorem


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

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

Let ak be the initial coefficient of fk, andconsider the ideal J=(a1,a2,a3,) of initialcoefficients. Since R is noetherian, J=(a0,,aN) forsome N.

Then I=(f0,f1,,fN). For if not thenfN+1I(f0,f1,,fN), andaN+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 andfN+1-g(f0,f1,,fN). But this contradictsminimality of deg(fN+1).

Hence, R[x] is noetherian.

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