Hilbert's Nullstellensatz

We sketch the weak form first: if $J$ is a proper ideal of $k[x_1,\ldots,x_n]$ then $V(J)$ is nonempty. Pick a maximal ideal $\mathfrak{m}$ containing $J$; the quotient $k[x_1,\ldots,x_n]/\mathfrak{m}$ is a field, and it is a finitely generated $k$-algebra. Zariski's lemma says any such field extension of $k$ is a finite algebraic extension of $k$; since $k$ is algebraically closed, this forces $k[x_1,\ldots,x_n]/\mathfrak{m} = k$. The images of $x_1,\ldots,x_n$ under this quotient map give a point $a \in k^n$ with $g(a)=0$ for every $g \in J$, so $a \in V(J)$.

For the strong form $I(V(J)) = \sqrt{J}$, the inclusion $\sqrt{J} \subseteq I(V(J))$ is immediate: if $f^r \in J$ then $f$ vanishes wherever every element of $J$ vanishes. The reverse inclusion uses the Rabinowitsch trick: to show $f \in I(V(J))$ implies $f \in \sqrt{J}$, introduce a new variable $y$ and consider the ideal generated by $J$ together with $1 - yf$ in $k[x_1,\ldots,x_n,y]$.

This enlarged ideal has empty zero set (any common zero would need every generator of $J$ to vanish, forcing $f$ to vanish there too, which contradicts $1-yf=0$). By the weak Nullstellensatz just proved, the enlarged ideal must be the whole ring, so $1$ is a polynomial combination of the generators. Substituting $y = 1/f$ and clearing denominators produces an explicit expression showing some power $f^r$ lies in $J$, which is exactly $f \in \sqrt{J}$.