# How do I find the set of solution to x^2z^3 - y^6 over C?

1. Jan 11, 2014

### Silversonic

I'm at a bit of a loss how to do this. I suspect it's the set $\left\{t_1^3, t_1t_2, t_2^2) \mid t \in {\mathbb C}\right\}$. Certainly the polynomial $x^2z^3 - y^6$ vanishes on these points, but I'm not sure how to show the other inclusion.

The only thing I can think of gets me close, but not totally there.

If I had a solution $(a,b,c)$ then write $a = t_1^3$ for some $t_1 \in {\mathbb C}$ and $c = t_2^2$ for some $t_2 \in {\mathbb C}$. Then

$t_1^6 t_2^6 - b^6 = 0$

$b^6 = t_1^6 t_2^6$

But can I just straight up take the 6th root if we're over the complex numbers? For instance if I had $b^2 = t_1^2t_2^2$ then $b = \pm t_1 t_2$.

The b in the case above would surely be $b = t_1t_2 e^{i2\pi k/6}$ for some $k = 0,1,2,3,4,5$, so it's multiplied by some arbitrary phase factor. I get 6 possibilities. What do I do now?

2. Jan 11, 2014

### PeroK

I think you just need to apply a bit of logic.

What you've shown is that if you take any two complex numbers t1 and t2, then you have a solution of the form you've given.

Next, you need to show that if you have a solution, then you can find t1 and t2. Hint: for any x, y and z, you need to find t1 and t2, so think of fixing y and z and solving for x.

These two things together will prove the result.

Last edited: Jan 11, 2014