Finding radical of an ideal using Nullstellensatz 
Find $V(J)=\{(x,y)\in\Bbb{C}^2:x^2-y^7=0, x^4-y^5=0\}$ and $\sqrt{J}$, where $J=\langle x^2-y^7,x^4-y^5\rangle$

My attempt:
$(0,0)$ is obviously a solution.
Moreover, I can deduce from the first equation that $x^4=y^{14}$, therefore $y^5=y^{14}$, so $y^9=1$ is a solution. So $y=\{(e^{\frac{2\pi i}{9}})^j\}_{j=1}^9$. I can use the first equation to deduce that $(\pm (e^{\frac{7\pi i}{9}})^j,(e^{\frac{2\pi i}{9}})^j)$ are solutions.
Furthermore, one can see that also $x=\{(e^{\frac{2\pi i}{18}})^j\}_{j=1}^{18}$ are also solutions upon finding appropriate $y'$s (some coincide with solutions I've already found). Basically finding $V(J)$ is not too difficult. But how do I find $\sqrt J$?
Any help would be appreciated.
 A: First note that $(0,0)\in\mathcal{V}(J)$. Now suppose $(a,b)\in\mathcal{V}(J)$ and $(a,b)\neq (0,0)$. Then $a^2=b^7$, so we must have $b\neq 0$, and you have successfully shown that $b$ is then a $9$-th root of unity. Again since $a^2=b^7$, this means that $a$ is an $18$-th root of unity, as you note. On the other hand, for any $k\in\{0,\dots,17\}$, note that $(\zeta^k_{18},\zeta_{9}^{4k})\in \mathcal{V}(J)$. Similarly, for every $l\in\{0,\dots,8\}$, we have $(\zeta^{7l}_{18},\zeta_9^l)\in\mathcal{V}(J)$. Thus, combining all of the above shows that the set of $x$-coordinates of the elements of $\mathcal{V}(J)$ is precisely $\{0,1,\zeta_{18},\dots,\zeta_{18}^{17}\}$ and the set of $y$-coordinates is precisely $\{0,1,\zeta_9,\dots,\zeta_9^{8}\}$. In particular, defining $f=x\prod_{k=0}^{17}(x-\zeta_{18}^k)=x(x^{18}-1)$ and $g=y\prod_{k=0}^{17}(y-\zeta_9^k)=y(y^{9}-1)$, by the lemma below we have $$\sqrt{J}=J+\langle f,g\rangle=\langle x^2-y^7,x^4-y^5,f,g\rangle.$$
If you wish, you can simplify this further by noting that $$x^4-y^5=(x^2+y^7)(x^2-y^7)+y^4f,$$ so in fact $\sqrt{J}=\langle x^2-y^7, f,g\rangle$.

Lemma: Let $J$ be any ideal of $\mathbb{C}[x,y]$ such that $\mathcal{V}(J)=\{(a_1,b_1),\dots,(a_n,b_n)\}$ is a finite collection of points, and let $m,m'\leqslant n$ and $c_i,d_j\in\mathbb{C}$ be such that
\begin{align}&\{c_1,\dots,c_m\}=\{a_1,\dots,a_n\} \\
&\{d_1,\dots,d_{m'}\}=\{b_1,\dots,b_n\}
\end{align} and $c_i\neq c_j$ for any $i< j\leqslant m$ and $d_i\neq d_j$ for any $i< j\leqslant m'$. Write $f=\prod_{k=1}^{m'}(x-c_k)$ and $g=\prod_{k=1}^m(y-d_k)$, and finally denote $K=J+\langle f,g\rangle$. Then $K=\sqrt{J}$.
Proof:

*

*For each $k\in\{1,\dots,n\}$, let $P_k=\langle x-a_k,y-b_k\rangle$. By the Nullstellensatz, the $P_k$ are precisely the maximal ideals of $\mathbb{C}[x,y]$ that contain $J$.

*Since $\mathbb{C}[x,y]$ is a Jacobson ring, also $\sqrt{J}=\bigcap_{k=1}^n P_k$.

*In particular, $K\subseteq\sqrt{J}$. Indeed, by $(2)$, it suffices to show $f,g\in P_k$ for every $k$. But for any polynomial $h$, we have $h\in P_k$ if and only if $h(a_k,b_k)=0$. (Why?)

*To show $\sqrt{J}\subseteq K$, it suffices to show $\sqrt{J}_M\subseteq K_M$ for every maximal ideal $M$ of $\mathbb{C}[x,y]$. (Let me know if you would like a proof of this.) If $M$ is a maximal ideal such that $K\nsubseteq M$, then $K_M=R_M$ and so this fact is trivial; thus, since $J\subseteq K$, by $(1)$ it suffices to show $\sqrt{J}_{P_k}\subseteq K_{P_k}$ for each $k$, so fix $k\in\{1,\dots,n\}$.

*Now, $K_{P_k}$ contains the images of $f$ and $g$ in $R_{P_k}$. By construction of the $c_i$ and $d_j$, let $l$ and $l'$ be such that $c_l=a_k$ and $d_{l'}=b_k$. Note that the images of $x-c_i$ and $y-d_{i'}$ are units in $R_{P_k}$ for each $i\neq l$ and $i'\neq l'$, since by construction $c_i\neq c_l=a_k$ and $d_{i'}\neq c_{l'}=b_k$ (and so $x-c_i\notin P_k$ and $y-d_{i'}\notin P_k$). Thus the images of $f':=\prod_{i\neq l}(x-c_i)$ and $g':=\prod_{i\neq l}(y-d_i)$ are both units in $R_{P_k}$. But $f=(x-c_l)f'$ and $g=(y-d_{l'})f'$, so this means that $K_{P_k}$ contains the images of $x-c_l$ and $y-d_{l'}$ in $R_{P_k}$, and is hence maximal. (Why?)

*By $(3)$, we know that $K_{P_k}\subseteq\sqrt{J}_{P_k}$. Also, since $\sqrt{J}\subseteq P_k$, we know that $\sqrt{J}_{P_k}$ is a proper ideal of $R_{P_k}$, and thus, since $K_{P_k}$ is maximal by $(5)$, this forces $K_{P_k}=\sqrt{J}_{P_k}$, as desired, so we are done. $\blacksquare$

In fact, the natural analogue of this lemma still holds with $\mathbb{C}$ replaced by any algebraically closed field $F$ and $\mathbb{C}[x,y]$ replaced by $F[x_1,\dots,x_n]$ for any $n\in\mathbb{N}$ (the proof is nearly identical), so this gives an easy way of computing the radical of any ideal $J\subseteq F[x_1,\dots,x_n]$ such that $\mathcal{V}(J)$ is finite.
A: Let $J=\langle x^2-y^7, x^4-y^5\rangle$.
Then, $V(J)=\{(x,y)\in \mathbb{C}^2: x^2-y^7, x^4-y^5\}$. The problem from the perspective of algebraic geometry is that this ideal $J$ is not a radical ideal and nullstellensatz gives us a correspondence between radical ideals and the affine algebraic sets.
If you recall what $\sqrt{J}$ is it is the ideal whose elements have powers which lie in $J$.
Observe that $x^4=y^5$ and $x^4=(x^2)^2=y^{14}$. This tells us that $y^{14}-y^5\in J$.  So $y^5(y^9-1)\in J$.
Actually what we have done here is to say we can rewrite $J$ as
$J=\langle x^2-y^7, y^5(y^9-1)\rangle$.
Multiples of this polynomial $y^5(y^9-1)$ are also in $J$ so, $y^5(y^9-1)^5\in J$. Now therefore, we have found that $y(y^9-1)\in \sqrt{J}$ and this is a polynomial not in $J$.
We can do something similar for $x$ to see that $x^{28}-x^{10}=x^{10}(x^{18}-1)\in J$. Thus, $x(x^{18}-1)\in \sqrt{J}$
It therefore turns out that
$\sqrt{J}=\langle x^2-y^7, y(y^9-1), x(x^{18}-1)\rangle$.
How do I know that other polynomials in $J$ aren't magically some power and thus, I might be missing things in the radical $\sqrt{J}$? This is somewhat annoying to show and it's not easy.
P.S. @KReiser: Thanks for spotting the error.
