 Solution

Use the Rabinowitsch trick. In the polynomial ring $k[x_1,\dots ,x_n,t]$, considerIf $I$ were proper, it would lie in a maximal ideal. Part (a), together with algebraic closedness of $k$, is the Weak Hilbert Nullstellensatz, so that maximal ideal would be evaluation at a point $(a_1,\dots ,a_n,b)\in k^{n+1}$. We would then haveBut the hypothesis gives $f(a)=0$, a contradiction. Therefore $I$ is the unit ideal.

There are polynomials $g_j$ and $h$ such thatSubstitute $t=f^{-1}$ in the localization $k[x_1,\dots ,x_n{]}_f$. This gives $1\in J_f$; clearing a power of $f$ yieldsThis is the Strong Hilbert Nullstellensatz inclusion $I(Z(J))\subseteq \sqrt{J}$.