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I would like to show that the following result holds:

The equations $x^3+y^3=1$ and $y^2=4x^3-1$ are the same Riemann surface in $\mathbb{CP}^2$ and as a consequence there are two meromorphic functions $f,g$ such that $f^3+g^3=1$.

I first tried using Able's theorem but didn't get very far using this approach.

Using compactification I managed to find the points that needed to be addaad to the surface of each of the equations but I don't how to proceed with this attempt in order to complete my proof of the desired claim.

FD_bfa
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Math if Fun
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  • Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User Jun 15 '22 at 16:03
  • can this help? https://math.stackexchange.com/questions/29935/f3-g3-1-for-two-meromorphic-functions maybe you can find f and g first then prove they are the same – onRiv Jun 16 '22 at 01:57
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    I find your question unclear. To begin with, what do you mean by "the same?" Do you mean "same as subsets" or "biholomorphic" or something else? Next, your equations, as written, do not even define subsets in $P^2$ since the latter has three homogeneous coordinates and you are using only two coordinates. Do you mean that these equations define complex affine curves in the affine patch and then you complete these affine curves by adding points at infinity? You should edit your question to improve clarity. – Moishe Kohan Jun 16 '22 at 14:25

1 Answers1

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I'm thinking if this proof is correct or not:

$x^3+y^3=1$ is the Riemann surface in $\Bbb{CP}^2$ define by:

$$ \{[X,Y,Z] \in \Bbb{C}^3| X^3+Y^3=Z^3, (X, Y, Z) \neq (0, 0, 0)\} $$

Let

$$ \begin{aligned} X &= U \\ Y &= \frac{\sqrt{3}}{6}V -\frac{1}{2}W \\ Z &= \frac{\sqrt{3}}{6}V +\frac{1}{2}W \\ \end{aligned} $$ Then $(X, Y, Z) \mapsto (U, V, W)$ is a biholomorphic automorphism in $\Bbb{CP}^2$ and:

$$ \begin{aligned} X^3+Y^3-Z^3&=U^3+(\frac{\sqrt{3}}{6}V-\frac{1}{2}W)^3-(\frac{\sqrt{3}}{6}V+\frac{1}{2}W)^3 \\ &=U^3-\frac{1}{4}W(V^2+W^2) \end{aligned} $$ Hence $$ \{[X,Y,Z] \in \Bbb{C}^3| X^3+Y^3=Z^3, (X, Y, Z) \neq (0, 0, 0)\} \\ = \{[U, V, W] \in \Bbb{C}^3|U^3-\frac{1}{4}W(V^2+W^2)=0, (U, V, W) \neq (0, 0, 0)\} $$

Setting $\frac{U}{W}=u, \frac{V}{W}=v$, in affine coordinate, it's $v^2=4u^3-1$.

onRiv
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