[proofplan] The inclusion $\sqrt{I} \subset I(V(I))$ is a direct consequence of the definitions: if $f^m \in I$ then $f^m$, hence $f$, vanishes on $V(I)$. The reverse inclusion is the substantive content and is proved by Rabinowitsch's trick: given $f$ vanishing on $V(I) = V(g_1, \ldots, g_r)$, adjoin a fresh variable $T$ and form the auxiliary ideal $J = (g_1, \ldots, g_r, 1 - fT)$ in $k[X_1, \ldots, X_n, T]$. The vanishing locus $V(J)$ is empty, so by the Weak Nullstellensatz $J = (1)$, which produces a polynomial identity. Specialising $T = 1/f$ in the field of fractions and clearing denominators yields $f^N \in I$, hence $f \in \sqrt{I}$. This converts the geometric statement "$f$ vanishes on $V(I)$" into the algebraic statement "$f^N \in I$". [/proofplan]

[step:Prove the easy inclusion $\sqrt{I} \subset I(V(I))$] Let $f \in \sqrt{I}$. By definition of the radical, there exists $m \ge 1$ such that $f^m \in I$. Take any $p \in V(I)$. By the definition of the vanishing locus, every element of $I$ vanishes at $p$, so in particular $f^m(p) = 0$. Since $k$ is a field (hence an integral domain), $f^m(p) = (f(p))^m = 0$ in $k$ forces $f(p) = 0$. As $p \in V(I)$ was arbitrary, $f$ vanishes on $V(I)$, so $f \in I(V(I))$. [/step]

[step:Set up the Rabinowitsch construction in $k[X_1, \ldots, X_n, T]$] We now prove the substantive inclusion $I(V(I)) \subset \sqrt{I}$. Let $f \in I(V(I))$, so $f$ vanishes on $V(I)$. We will show $f \in \sqrt{I}$ by constructing an integer $N \ge 1$ and elements of $k[X_1, \ldots, X_n]$ that exhibit $f^N$ as a member of $I$. By the *Hilbert Basis Theorem* (iterating the fact that $R[X]$ is Noetherian whenever $R$ is, starting from the field $k$), $k[X_1, \ldots, X_n]$ is Noetherian, so $I$ is finitely generated. Write $I = (g_1, \ldots, g_r)$ with $g_i \in k[X_1, \ldots, X_n]$. Adjoin a fresh variable $T$ and form the polynomial ring $k[X_1, \ldots, X_n, T]$. Define the ideal \begin{align*} J = (g_1, \ldots, g_r,\, 1 - f T) \subset k[X_1, \ldots, X_n, T], \end{align*} where $g_i$ and $f$ are viewed as elements of $k[X_1, \ldots, X_n] \subset k[X_1, \ldots, X_n, T]$ (constants in $T$). The case $f = 0$ is immediate (then $f \in I$ since $0 \in I$), so we assume $f \neq 0$ in the polynomial ring $k[X_1, \ldots, X_n]$. [/step]

[step:Show that $V(J) = \varnothing$ in $\mathbb{A}^{n+1}_k$] Suppose for contradiction that $V(J) \neq \varnothing$, and let $(a_1, \ldots, a_n, b) \in V(J)$. Set $a := (a_1, \ldots, a_n) \in \mathbb{A}^n_k$. By definition of $V(J)$, every generator of $J$ vanishes at $(a, b)$: \begin{itemize} \item For each $i$, $g_i(a) = g_i(a, b) = 0$ (since $g_i$ does not depend on $T$). Hence $g_i(a) = 0$ for every $i$, so $a \in V(g_1, \ldots, g_r) = V(I)$. \item By hypothesis, $f \in I(V(I))$, so $f(a) = 0$. \item Also $1 - f(a) \cdot b = 0$, so $f(a) \cdot b = 1$ in $k$. \end{itemize} Combining the last two: $0 \cdot b = 1$ in $k$, so $0 = 1$ in $k$, which is false in any field. Hence no such $(a, b)$ exists, and \begin{align*} V(J) = \varnothing. \end{align*} [/step]

[step:Apply the Weak Nullstellensatz to conclude $J = k[X_1, \ldots, X_n, T]$] The polynomial ring $k[X_1, \ldots, X_n, T]$ is the polynomial ring in $n + 1$ variables over the algebraically closed field $k$. By the [Weak Nullstellensatz](/theorems/2123) (taking the contrapositive: if $J$ is proper then $V(J) \neq \varnothing$), the equality $V(J) = \varnothing$ from Step 3 forces $J$ to fail to be proper. So \begin{align*} J = k[X_1, \ldots, X_n, T]. \end{align*} In particular, $1 \in J$, so there exist polynomials $h_0, h_1, \ldots, h_r \in k[X_1, \ldots, X_n, T]$ with \begin{align*} 1 = h_0(X, T)\, (1 - fT) + \sum_{i=1}^r h_i(X, T)\, g_i(X), \tag{$\ast$} \end{align*} where we have written $X = (X_1, \ldots, X_n)$ for brevity. [/step]

[step:Specialise the identity by setting $T = 1/f$ in $k(X_1, \ldots, X_n)$] The identity $(\ast)$ holds in $k[X_1, \ldots, X_n, T]$, hence in any extension ring. We pass to the field of rational functions \begin{align*} K = k(X_1, \ldots, X_n) = \mathrm{Frac}(k[X_1, \ldots, X_n]), \end{align*} and view $K[T]$ as the polynomial ring over $K$. Inside $K$, the element $f \neq 0$ is invertible, so $1/f \in K$ is defined. Specialise $T \mapsto 1/f$, i.e. apply the evaluation homomorphism \begin{align*} \mathrm{ev}_{1/f} : K[T] &\to K \\ T &\mapsto 1/f \end{align*} to both sides of $(\ast)$. The first term on the right becomes $h_0(X, 1/f) \cdot (1 - f \cdot (1/f)) = h_0(X, 1/f) \cdot 0 = 0$. The remaining terms give \begin{align*} 1 = \sum_{i=1}^r h_i(X, 1/f)\, g_i(X) \qquad \text{in } K. \tag{$\ast\ast$} \end{align*} [/step]

[step:Clear denominators to obtain $f^N \in I$]Each $h_i(X, T)$ is a polynomial in $T$ with coefficients in $k[X_1, \ldots, X_n]$; write \begin{align*} h_i(X, T) = \sum_{j=0}^{d_i} h_{i,j}(X)\, T^j \end{align*} with $h_{i,j} \in k[X_1, \ldots, X_n]$ and $d_i \ge 0$. Substituting $T = 1/f$, \begin{align*} h_i(X, 1/f) = \sum_{j=0}^{d_i} h_{i,j}(X)\, f^{-j} = f^{-d_i} \sum_{j=0}^{d_i} h_{i,j}(X)\, f^{d_i - j}. \end{align*} Let $D = \max_{1 \le i \le r} d_i$. Multiplying $(\ast\ast)$ through by $f^D \in k[X_1, \ldots, X_n]$, \begin{align*} f^D = \sum_{i=1}^r f^{D - d_i} \cdot f^{d_i} \cdot h_i(X, 1/f) \cdot g_i(X) = \sum_{i=1}^r \tilde{h}_i(X) \cdot g_i(X), \end{align*} where \begin{align*} \tilde{h}_i(X) := f^{D - d_i}(X) \cdot \sum_{j=0}^{d_i} h_{i,j}(X)\, f(X)^{d_i - j} = \sum_{j=0}^{d_i} h_{i,j}(X)\, f(X)^{D - j} \in k[X_1, \ldots, X_n]. \end{align*} The polynomial $\tilde{h}_i$ has no denominators and lies in $k[X_1, \ldots, X_n]$. We have written \begin{align*} f^D = \sum_{i=1}^r \tilde{h}_i \cdot g_i \in (g_1, \ldots, g_r) = I. \end{align*} We now choose $N \ge 1$. If $D \ge 1$, set $N = D$: then $f^N \in I$ directly, so $f \in \sqrt{I}$. If $D = 0$ — meaning every $h_i$ is constant in $T$, so $h_i(X, 1/f) = h_{i,0}(X)$ — the identity $(\ast\ast)$ reads \begin{align*} 1 = \sum_{i=1}^r h_{i,0}(X) \cdot g_i(X), \end{align*} which is already an identity in $k[X_1, \ldots, X_n]$ and exhibits $1 \in I$. Hence $I = k[X_1, \ldots, X_n]$, so $f \cdot 1 = f \in I$ and therefore $f^1 \in I$, giving $f \in \sqrt{I}$ with $N = 1$. (Geometrically, $I = k[X_1, \ldots, X_n]$ means $V(I) = \varnothing$, and the equality $I(V(I)) = \sqrt{I}$ reduces to $I(\varnothing) = k[X_1, \ldots, X_n] = \sqrt{I}$, which holds directly from the definitions.) As $f \in I(V(I))$ was arbitrary, $I(V(I)) \subset \sqrt{I}$. Combining with Step 1, $I(V(I)) = \sqrt{I}$, completing the proof.[/step]

[guided]We have an identity in $K = k(X_1, \ldots, X_n)$: \begin{align*} 1 = \sum_{i=1}^r h_i(X, 1/f)\, g_i(X). \end{align*} Each $h_i(X, 1/f)$ is a rational function in the $X_j$ — its only denominators are powers of $f$, which arose from substituting $T = 1/f$ into a polynomial in $T$. The goal is to convert this identity in $K$ into an identity in $k[X_1, \ldots, X_n]$ — equivalently, to clear all denominators. \textbf{Identifying the denominators.} Write $h_i(X, T) = \sum_{j=0}^{d_i} h_{i,j}(X) T^j$ as a polynomial in $T$ of degree $d_i$ with $h_{i,j} \in k[X_1, \ldots, X_n]$. Setting $T = 1/f$, \begin{align*} h_i(X, 1/f) = h_{i,0}(X) + h_{i,1}(X)\frac{1}{f} + \cdots + h_{i,d_i}(X)\frac{1}{f^{d_i}}. \end{align*} The maximum power of $f$ in any denominator is $d_i$. \textbf{Choosing the multiplier.} Let $D = \max_i d_i$. Multiplying through by $f^D$ — a single common denominator that absorbs every $1/f^{d_i}$ that appears — gives a relation in $k[X_1, \ldots, X_n]$. Specifically, \begin{align*} f^D \cdot h_i(X, 1/f) = \sum_{j=0}^{d_i} h_{i,j}(X)\, f^{D-j}(X), \end{align*} and each exponent $D - j \ge D - d_i \ge 0$, so this is a polynomial in $X$ — no denominators. Call this polynomial $\tilde{h}_i(X)$. \textbf{The cleared identity.} Multiplying the original identity in $K$ by $f^D$: \begin{align*} f^D = \sum_{i=1}^r f^D \cdot h_i(X, 1/f) \cdot g_i(X) = \sum_{i=1}^r \tilde{h}_i(X)\, g_i(X). \end{align*} The right-hand side is a $k[X_1, \ldots, X_n]$-linear combination of the generators of $I$, hence lies in $I$. Therefore $f^D \in I$. \textbf{Conclusion.} We have $f^D \in I$ with $D \ge 0$. We must produce $N \ge 1$ with $f^N \in I$ to conclude $f \in \sqrt{I}$. There are two cases. \emph{Case $D \ge 1$.} Take $N = D$, so $f^N = f^D \in I$ directly. \emph{Case $D = 0$.} Every $h_i$ is constant in $T$, so $(\ast\ast)$ is already the polynomial identity $1 = \sum_i h_{i,0}(X) g_i(X)$ in $k[X_1, \ldots, X_n]$. This exhibits $1 \in I$, hence $I = k[X_1, \ldots, X_n]$, hence $V(I) = \varnothing$ and every polynomial — in particular $f$ itself — lies in $I$. Taking $N = 1$ gives $f^N = f \in I$, so $f \in \sqrt{I}$. In either case $f \in \sqrt{I}$, as required. \textbf{Why Rabinowitsch works — geometric intuition.} The "magic" element $1 - fT$ encodes geometrically the open set $D(f) = \{p : f(p) \neq 0\}$: a point $(a, b) \in V(J)$ would be a point $a \in V(I)$ together with a number $b = 1/f(a) \in k$, i.e. a point on the graph of $1/f$ over $V(I)$. The hypothesis "$f$ vanishes on $V(I)$" makes this graph empty (no $b$ can satisfy $b \cdot 0 = 1$), so the "lift" to $\mathbb{A}^{n+1}$ has no points. The Weak Nullstellensatz then converts this geometric emptiness into the algebraic statement $1 \in J$, which clears denominators back down to give the polynomial identity $f^N \in I$.[/guided]