I need to find the radicals of the following ideals:
i) $\mathfrak{a} = (xy^3, x(x-y))$
ii) $\mathfrak{b} = (xy^3, x^2(y-3))$
iii) $\mathfrak{c} = (x^2(y-z), xy(y-z), xz(y-z)^2)$
Can I just use the Nullstellensatz? My working below seems a bit too easy, which makes me think I'm doing something horrendously awful.
Let $k$ be an algebraically closed field.
i) It's pretty obvious that $Z(\mathfrak{a}) = \{ (0,t) \ | \ t \in k \} = Z(x) $. So by the Nullstellensatz, $\sqrt{\mathfrak{a}} = I(Z(\mathfrak{a})) = I(Z(x)) = (x) $.
ii) Isn't this the same as above?
iii) $Z(\mathfrak{c}) = \{ (0,s,t) \ | \ s,t \in k \}\cup \{(s,t,t) \ | \ s,t \in k\} = Z(x) \cup Z(y-z)$. So $I(Z(\mathfrak{c})) = I(Z(x)) \cap I(Z(y-z)) = (x) \cap (y-z) = (x(y-z))$
Am I breaking any laws?
Thanks!