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] be a 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


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


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


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


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

Title proof of Hilbert’s Nullstellensatz
Canonical name ProofOfHilbertsNullstellensatz
Date of creation 2013-03-22 15:27:46
Last modified on 2013-03-22 15:27:46
Owner pbruin (1001)
Last modified by pbruin (1001)
Numerical id 4
Author pbruin (1001)
Entry type Proof
Classification msc 13A10