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Let $E: y^2 = 4x^3 - b$ an affine equation of an elliptic curve in $\mathbb{P}^2_{\mathbb{C}}$. Let $b$ be chosen such that the map $f: \mathbb{C} \rightarrow \mathbb{P}^2$ given by $z \mapsto (\wp(z):\wp'(z):1)$ where $\wp$ is the Weierstrass $\wp$-function for the lattice $\Lambda = \{n+m\omega \mid n,m \in \mathbb{Z}, \omega^3=1, \omega \not= 1\}$, has image $E$. Now, I have proven that $(\wp(\omega z):\wp'(\omega z):1) = (\omega\wp(z):\wp'(z):1)$, and I need to show that $ f\left(\dfrac{1-\omega}{3}\right)=(0,\pm\sqrt{b})$. How could I go about doing this?

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You are almost there. Show that $f((1-\omega)/3) = f(\omega(1-\omega)/3)$, because the difference $(1-\omega)/3 - \omega(1-\omega)/3\in \Lambda$.

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  • $\begingroup$ sorry, I don't see how $(1-\omega)/3 - \omega(1-\omega)/3$ could be in $n+m \omega$ for $n,m \in \mathbb{Z}$ $\endgroup$
    – user198182
    Commented May 4, 2015 at 8:10
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    $\begingroup$ @user198182: $\omega^2+\omega+1=0$. $\endgroup$ Commented May 4, 2015 at 13:12

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