Hilbert's Nullstellensatz

Step 1 (weak Nullstellensatz, used as input). Since $k$ is algebraically closed, every maximal ideal of $k[x_1,\ldots,x_n]$ has the form $(x_1-a_1,\ldots,x_n-a_n)$ for a point $(a_1,\ldots,a_n) \in k^n$; this is proved via Zariski's Lemma (a field that is finitely generated as an algebra over $k$ is a finite field extension of $k$, hence equals $k$ since $k$ is algebraically closed). Consequently, if $J \ne (1)$ then $J$ lies in some maximal ideal, so $V(J) \ne \emptyset$.

Step 2 (easy inclusion, $\sqrt{J} \subseteq I(V(J))$). If $f^m \in J$ for some $m$, then $f^m$ vanishes at every point of $V(J)$, forcing $f$ itself to vanish there (a product of field elements is $0$ only if a factor is $0$), so $f \in I(V(J))$.

Step 3 (Rabinowitsch trick for the reverse inclusion). Let $f \in I(V(J))$. Introduce a new variable $y$ and form $J' = J + (1 - yf) \subseteq k[x_1,\ldots,x_n,y]$. Any point of $V(J')$ would need to lie in $V(J)$ (to satisfy the generators of $J$) and also satisfy $1-yf=0$; but $f$ vanishes on all of $V(J)$, making $1-yf=1$ there, which is never $0$. Hence $V(J') = \emptyset$, so by Step 1 (contrapositive), $J' = (1)$: there exist polynomials with $1 = h(x,y)(1-yf) + \sum_i g_i(x,y) J_i(x)$ for generators $J_i$ of $J$.

Step 4 (clearing denominators). Substitute $y = 1/f$ formally in this identity (working in $k[x_1,\ldots,x_n][1/f]$) — the term with $h$ vanishes since $1-yf$ becomes $0$, leaving $1 = \sum_i g_i(x,1/f) J_i(x)$. Multiplying through by a sufficiently high power $f^m$ to clear every denominator $f$ introduced by the $g_i$ yields $f^m = \sum_i \tilde g_i(x) J_i(x) \in J$ for polynomials $\tilde g_i$, i.e. $f \in \sqrt{J}$. Combined with Step 2, $I(V(J)) = \sqrt{J}$, and taking radicals of both sides when $J=\sqrt J$ gives the stated corollary $\sqrt{I(V(J))} = J$.