= Solution
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 $h\neq0$. <Noetherianity> gives finite generators $I=(f_1,\ldots,f_r)$.
Use the <Rabinowitsch trick>: in the <polynomial ring> with one auxiliary variable, form
$$
J=(f_1,\ldots,f_r,1-th)\subseteq k[x_1,\ldots,x_n,t].
$$
Apply 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 identity
$$
1=\sum_{i=1}^r P_i(x,t)f_i(x)+Q(x,t)(1-th(x)).
$$
Substitute $t=h^{-1}$ 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 $h\in\sqrt I$ and
$$
\boxed{I(V(I))=\sqrt I.}
$$
This proves the strong <Hilbert Nullstellensatz> from the polynomial-ring weak form by converting nonvanishing into an extra equation.
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