UTM Ideals Varieties and Algorithm - Chapter 1 Section 4 Exercise 8

Problem:

Solution:

For part (a), if $f^m \in I(V)$, then for all points $p$ in $V$, $f^m(p) = 0 \implies f(p) = 0 \implies f \in I(V)$. Therefore $I(V)$ is radical.

For part (b), $x^2 \in \langle x^2, y^2 \rangle$ but $x \notin \langle x^2, y^2 \rangle$. That shows the $\langle x^2, y^2 \rangle$ is not radical and therefore cannot be an ideal of a variety.

Look forward to Nullstellensatz, what a long word to type ...