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单词 ProofOfHilbertsNullstellensatz
释义

proof of Hilbert’s Nullstellensatz


Let K be an algebraically closed field, let n0, and let I be an ideal of the polynomial ringMathworldPlanetmath K[x1,,xn]. Let fK[x1,,xn] bea polynomialMathworldPlanetmath with the property that

f(a1,,an)=0 for all (a1,,an)V(I).

Suppose that frI for all r>0; in particular, I is strictly smaller than K[x1,,xn] and f0. Consider the ring

R=K[x1,,xn,1/f]K(x1,,xn).

The R-ideal RI is strictly smaller than R, since

RI=r=0f-rI

does not contain the unit element. Let y be an indeterminate overK[x1,,xn], and let J be the inverse image of RI underthe homomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath

ϕ:K[x1,,xn,y]R

acting as the identityPlanetmathPlanetmathPlanetmathPlanetmath on K[x1,,xn] and sending y to1/f. Then J is strictly smaller than K[x1,,xn,y], sothe weak Nullstellensatz gives us an element (a1,,an,b)Kn+1 such that g(a1,,an,b)=0 for all gJ. Inparticular, we see that g(a1,,an)=0 for all gI. OurassumptionPlanetmathPlanetmath on f therefore implies f(a1,,an)=0. However,J also contains the element 1-yf since ϕ sends this elementto zero. This leads to the following contradictionMathworldPlanetmathPlanetmath:

0=(1-yf)(a1,,an,b)=1-bf(a1,,an)=1.

The assumption that frI for all r>0 is therefore false,i.e. there is an r>0 with frI.

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更新时间:2025/5/4 10:16:24