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Let $C(F(x,y))$ denote the zero locus of $F(x,y)$

Let $A=${$(x,y) : y\ge\frac{2\sqrt{3}}{3}$}

Let $B=${$(x,y) : 0\le y\le\frac{2\sqrt{3}}{3}$}

Let $C=${$(x,y) : -\frac{\sqrt{3}}{3}\le y\le0$}

Let $D=${$(x,y) : y\le-\frac{\sqrt{3}}{3}$}

It can easily be checked that:

  • $$C(\sqrt[3]{\frac{x^{2}+\sqrt{x^{4}-\frac{4}{27}}}{2}}+\sqrt[3]{\frac{x^{2}-\sqrt{x^{4}-\frac{4}{27}}}{2}}-y)=C(y^3-x^2-y) \cap A$$
  • $$C(-\frac{\sqrt{3}}{3}\cos\left(\frac{1}{3}\tan^{-1}\left(\frac{\sqrt{12-81x^{4}}}{9x^{2}}\right)\right)+\sin\left(\frac{1}{3}\tan^{-1}\left(\frac{\sqrt{12-81x^{4}}}{9x^{2}}\right)\right)-y)=C(y^3-x^2-y) \cap C$$
  • $$C(-\frac{\sqrt{3}}{3}\cos\left(\frac{1}{3}\tan^{-1}\left(\frac{\sqrt{12-81x^{4}}}{9x^{2}}\right)\right)-\sin\left(\frac{1}{3}\tan^{-1}\left(\frac{\sqrt{12-81x^{4}}}{9x^{2}}\right)\right)-y)=C(y^3-x^2-y) \cap D$$

By the fundamental theorem of algebra, $$y^3-y-x^2=(y-f(x))(y-g(x))(y-h(x))$$ for some $f,g,h$

Thus, I was expecting the zero locus of $y^3-x^2-y$ to be the union of the respective zero loci of 3 expressions of the form $y-f(x)$

What seems to be my misunderstanding?

Additionally, does there exist an expression $y-G(x)$ such that

$$C(G(x)-y)=C(y^3-x^2-y) \cap B$$

?

If so, how can it be derived?

Simon M
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  • The elliptic curve cannot be rationally parametrized, but there's always local stuff like branches and puiseux series. The piecewise thing might fit with that, I'm not sure. – Jan-Magnus Økland Feb 07 '24 at 13:03
  • @Jan-MagnusØkland Only curves of genus zero can be rationally parametrized, right? That means that $f,g$ and $h$ are guaranteed to not be rational functions – Simon M Feb 07 '24 at 17:27
  • @Jan-MagnusØkland “local stuff” is exactly what I’ve done so far. The only thing that confuses me is the fact that the union of the three pieces I’ve come up with so far is not the entire elliptic curve – Simon M Feb 07 '24 at 17:33

1 Answers1

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It turns out that the zero locus of $y^3-x^2-y$ is indeed the union of the respective zero loci of 3 expressions of the form $y-f(x)$

It turns out that there is an expression $y-H(x)$ such that $$C(H(x)-y)=C(y^3-x^2-y) \cap (A\cup B)$$

Indeed,$$C(\frac{2\sqrt{3}}{3}\cos\left(\frac{1}{3}\tan^{-1}\left(\frac{\sqrt{12-81x^{4}}}{9x^{2}}\right)\right)-y)=C(y^3-x^2-y) \cap (A\cup B)$$

However, even though $\frac{2\sqrt{3}}{3}\cos\left(\frac{1}{3}\tan^{-1}\left(\frac{\sqrt{12-81x^{4}}}{9x^{2}}\right)\right)$ is real for $x\in (-\infty,-\sqrt[4]{\frac{4}{27}}]\cup [\sqrt[4]{\frac{4}{27}},\infty)$, we will have an issue with regard to desmos, as desmos cannot handle any expression involving the square root of a negative number.

This can be remedied, though. Note that $$\frac{2\sqrt{3}}{3}\cos\left(\frac{1}{3}\tan^{-1}\left(\frac{\sqrt{12-81x^{4}}}{9x^{2}}\right)\right)=\frac{2\sqrt{3}}{3}\cosh\left(\frac{1}{3}\tanh^{-1}\left(\frac{\sqrt{81x^{4}-12}}{9x^{2}}\right)\right)$$

Thus, in order to get desmos to play nice with us, we can just use both of these, while remembering that in reality either one of these would be all that's needed for the entirety of $C(y^3-x^2-y) \cap (A\cup B)$

Simon M
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