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

  1. Jan 11, 2014 #1
    I'm at a bit of a loss how to do this. I suspect it's the set [itex] \left\{t_1^3, t_1t_2, t_2^2) \mid t \in {\mathbb C}\right\} [/itex]. Certainly the polynomial [itex] x^2z^3 - y^6 [/itex] 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 [itex] (a,b,c) [/itex] then write [itex] a = t_1^3 [/itex] for some [itex] t_1 \in {\mathbb C} [/itex] and [itex] c = t_2^2 [/itex] for some [itex] t_2 \in {\mathbb C} [/itex]. Then

    [itex] t_1^6 t_2^6 - b^6 = 0 [/itex]

    [itex] b^6 = t_1^6 t_2^6 [/itex]

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

    The b in the case above would surely be [itex] b = t_1t_2 e^{i2\pi k/6} [/itex] for some [itex] k = 0,1,2,3,4,5 [/itex], so it's multiplied by some arbitrary phase factor. I get 6 possibilities. What do I do now?
     
  2. jcsd
  3. Jan 11, 2014 #2

    PeroK

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    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
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