[proofplan] We prove the containment by unpacking the definitions. An element $f\in \sqrt I$ has a positive power $f^m$ lying in $I$. Every polynomial in $I$ vanishes on every point of $V(I)$, so evaluating at any $a\in V(I)$ gives $f(a)^m=0$. Since a field has no nonzero nilpotent elements, $f(a)=0$, and therefore $f$ lies in the vanishing ideal of $V(I)$. [/proofplan] [step:Choose an arbitrary element of the radical] Let $R=k[x_1,\dots,x_n]$, and let $f\in \sqrt I$ be arbitrary. By the definition of the radical of an ideal, there exists an integer $m\in\mathbb N$ such that \begin{align*} f^m\in I. \end{align*} It is enough to prove that $f\in I(V(I))$. [/step] [step:Evaluate the radical power at an arbitrary point of the vanishing set] Let $a\in V(I)$ be arbitrary. Since $V(I)$ is the set of common zeros of all polynomials in $I$, and since $f^m\in I$, we have \begin{align*} (f^m)(a)=0. \end{align*} Evaluation of products in the [polynomial ring](/page/Polynomial%20Ring) gives \begin{align*} (f(a))^m=0. \end{align*} Because $k$ is a field, the only element of $k$ whose positive power is $0$ is $0$: if $c\in k$ and $c\ne 0$, then $c^{-1}\in k$ and multiplying $c^m=0$ by $(c^{-1})^m$ would give $1=0$, impossible in a field. Applying this to $c=f(a)$ gives \begin{align*} f(a)=0. \end{align*} [guided] We now use the point $a\in V(I)$ to turn ideal membership into a vanishing statement. By definition, \begin{align*} V(I)=\{b\in \mathbb A_k^n: g(b)=0 \text{ for every } g\in I\}. \end{align*} Thus, because $a\in V(I)$ and $f^m\in I$, the polynomial $f^m$ vanishes at $a$: \begin{align*} (f^m)(a)=0. \end{align*} Evaluation at $a$ is compatible with multiplication of polynomials, so evaluating the product power $f^m$ gives the corresponding power of the scalar $f(a)\in k$: \begin{align*} (f(a))^m=0. \end{align*} The remaining point is purely field-theoretic. Let $c\in k$ satisfy $c^m=0$ for some $m\in\mathbb N$. If $c\ne 0$, then $c$ has an inverse $c^{-1}\in k$. Multiplying the equality $c^m=0$ by $(c^{-1})^m$ gives \begin{align*} 1=0, \end{align*} which contradicts the field axiom that $1\ne 0$. Therefore $c=0$. Taking $c=f(a)$, we obtain \begin{align*} f(a)=0. \end{align*} [/guided] [/step] [step:Conclude that the polynomial lies in the vanishing ideal] The point $a\in V(I)$ was arbitrary, and the previous step proves that $f(a)=0$ for every $a\in V(I)$. By the definition of the vanishing ideal of a subset of [affine space](/page/Affine%20Space), \begin{align*} f\in I(V(I)). \end{align*} Since every arbitrary element $f\in \sqrt I$ lies in $I(V(I))$, we conclude that \begin{align*} \sqrt I\subseteq I(V(I)). \end{align*} [/step]