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 ...
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 ...
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