Let $K$ be a algebraically closed field. Consider the set $$\mathcal {C} = \left \{(l^p,l^q) \in K^2 \mid p,q\ \text {fixed in}\ \Bbb Z,p,q>0,l\ \text {varies over}\ K \right \}.$$ Prove that $\mathcal {C}$ is an affine algebraic $K$-variety.
Let $\mathcal {Z} \left ( \left \{x^q - y^p \right \} \right )$ denote the zero set of the polynomial $x^q-y^p.$ Then I observed that $\mathcal {C} \subseteq \mathcal {Z} \left ( \left \{x^q - y^p \right \} \right ).$ But how do I prove the other way round? I tried by taking a zero let's say $(a,b)$ of the polynomial $x^q-y^p.$ So we have $a^q = b^p.$Then since $K$ is algebraically closed there exists $\alpha \in K$ such that ${\alpha}^p = a.$ If we can show that ${\alpha}^q = b$ then we are through. Now $$\begin{align} {\alpha}^q & = {\alpha}^{\frac {qp} {p}} = \left ({\alpha}^p \right )^{\frac q p} = a^{\frac q p} = b. \end{align}$$
So we get exactly what we wanted. I have given this proof in the exam. But I don't know why our instructor marked it as wrong and gave me half credit for the answer.
Would anybody please help me by finding the mistakes in the above answer (if any) and give me some suggestions as to how do I improve my answer? Any help will be highly appreciated.
Thank you very much.
First of all, if $\gcd(p,q)>1$, then depending on $\text{char}(K)$, you might run into trouble. For example, if $K = \mathbb{C}$ and $p=q=2$, then $\mathcal C = \mathcal Z(x-y)$, but $\mathcal C \ne \mathcal Z(x^2-y^2)$. If $d = \gcd(p,q)$, you should be taking $p' = p/d$ and $q' = q/d$ and using the polynomial $x^{q'}-y^{p'}$.
Aside from this, the main problem in your argument is taking "the" $p^{\small\text{th}}$ root of $a$ -- there is no distinguished one. There are in fact $p$ values of $\alpha$ (with multiplicity) so that $\alpha^p = a$. If you pick one at random, you are not guaranteed that $\alpha^q = b$. For example, take $K = \mathbb{C}$, $p = 2$, $q = 3$; then the point $(1,1)$ is in $\mathcal C$, but if I pick $\alpha = -1$ as the square root of $a = 1$ in your argument, then $\alpha^3 = -1 \ne b$. You have to pick the right $\alpha$. (It doesn't make sense to write $a^{q/p}$ or anything like that in a generic field.)
So now let's assume $(a,b)$ is a root of $x^{q'}-y^{p'}$. If $(a,b) = (0,0)$, we are done, so we assume this is not the case, i.e. that $a,b \ne 0$. We also assume without loss of generality that $\text{char}(K) \not | \;\; p'$. Since $\text{char}(K) \not | \;\; p'$, there are $p'$ distinct $(p')^{\small\text{th}}$ roots of $a$, call them $\alpha_1,\dots,\alpha_{p'}$. Now, for each $i$, we have that $(\alpha_i^{q'})^{p'} = (\alpha_i^{p'})^{q'} =a^{q’}= b^{p'}$, and the $\alpha_i^{q'}$ are distinct, since if $\alpha_i^{q'} = \alpha_j^{q'}$, then $\alpha_i/\alpha_j$ is a $(q')^{\small\text{th}}$ root of unity as well as a $(p')^{\small\text{th}}$ root of unity, and so $\alpha_i/\alpha_j = 1$ since $\gcd(p',q') = 1$. Therefore, $\alpha_1^{q'},\dots,\alpha_{p'}^{q'}$ are the $(p')^{\small\text{th}}$ roots of $b^{p'}$, so we have $\alpha_i^{q'} = b$ for some $i$. Let $\ell$ be any $d^{\small\text{th}}$ root of $\alpha_i$. Then $\ell^p = (\ell^d)^{p'} = \alpha_i^{p'} = a$, and similarly, $\ell^q = (\ell^d)^{q'} = \alpha_i^{q'} = b$.