Past exam of the algebraic Geometry course of the University of Oxford / 2011 / 1 / b / Solution

The radical of an ideal isThe 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