= Solution
The <radical of an ideal> is
$$
\boxed{\sqrt I=\{f\in R:f^N\in I\text{ for some integer }N\geq1\}.}
$$
The <ideal> is <radical> if $I=\sqrt I$. Its <zero set> is $V(I)=\{a\in k^n:f(a)=0\text{ for every }f\in I\}$, an <affine variety> by definition.
Every $f\in I$ vanishes at every <point> of $V(I)$, so $I\subseteq I(V(I))$. If $g^N\in I(V(I))$, then $g(a)^N=0$ in the <field> $k$ for every $a\in V(I)$, so $g(a)=0$ for each such <point>. Hence $g\in I(V(I))$ and
$$
\boxed{I(V(I))\text{ is a radical ideal containing }I.}
$$
Back to article page
