Past exam of the algebraic Geometry course of the University of Oxford / 2011 / 1 / c

Part (b) immediately gives $\sqrt{I}\subseteq I(V(I))$. For the reverse inclusion, let $h\in I(V(I))$. The case $h=0$ is immediate, so assume $h\ne 0$. Noetherianity gives finite generators $I=(f_1,\dots ,f_r)$.

Use the Rabinowitsch trick: in the polynomial ring with one auxiliary variable, formApply the assumed maximal-ideal description to polynomial rings, including this ring in $n+1$ variables. If $J$ were proper, it would lie in the maximal ideal of some point $(a,b)$. Then $a\in V(I)$ and $1-bh(a)=0$, impossible because $h$ vanishes on $V(I)$. Therefore $J$ is the whole ring, and there is an identitySubstitute $t=h^{-1}$ in the localization $R_h$. The last term vanishes; clearing the finitely many powers of $h$ from the remaining denominators yields $h^N\in I$ for some $N$. Thus $h\in \sqrt{I}$ andThis proves the strong Hilbert Nullstellensatz from the polynomial-ring weak form by converting nonvanishing into an extra equation.