Let $k$ be a field, $J$ an ideal of $k\left[x_{1}, \ldots, x_{n}\right]$, and let $R=k\left[x_{1}, \ldots, x_{n}\right] / J$. Define the radical $\sqrt{J}$ of $J$ and show that it is also an ideal.

The Nullstellensatz says that if $J$ is a maximal ideal, then the inclusion $k \subseteq R$ is an algebraic extension of fields. Suppose from now on that $k$ is algebraically closed. Assuming the above statement of the Nullstellensatz, prove the following.

(i) If $J$ is a maximal ideal, then $J=\left(x_{1}-a_{1}, \ldots, x_{n}-a_{n}\right)$, for some $\left(a_{1}, \ldots, a_{n}\right) \in k^{n}$.

(ii) If $J \neq k\left[x_{1}, \ldots, x_{n}\right]$, then $Z(J) \neq \emptyset$, where

$$Z(J)=\left\{a \in k^{n} \mid f(a)=0 \text { for all } f \in J\right\}$$

(iii) For $V$ an affine subvariety of $k^{n}$, we set

$$I(V)=\left\{f \in k\left[x_{1}, \ldots, x_{n}\right] \mid f(a)=0 \text { for all } a \in V\right\}$$

Prove that $J=I(V)$ for some affine subvariety $V \subseteq k^{n}$, if and only if $J=\sqrt{J}$.

[Hint. Given $f \in J$, you may wish to consider the ideal in $k\left[x_{1}, \ldots, x_{n}, y\right]$ generated $b y J$ and $y f-1$.]

(iv) If $A$ is a finitely generated algebra over $k$, and $A$ does not contain nilpotent elements, then there is an affine variety $V \subseteq k^{n}$, for some $n$, with $A=k\left[x_{1}, \ldots, x_{n}\right] / I(V)$.

Assuming $\operatorname{char}(k) \neq 2$, find $\sqrt{J}$ when $J$ is the ideal $\left(x(x-y)^{2}, y(x+y)^{2}\right)$ in $k[x, y]$.