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Question: Let $X$ be the Twisted Cubic in $\mathbb{P}^3$, and $\pi_p:X\rightarrow \mathbb{P}^2$, the projection of the Twisted Cubic from $p$. Find the equations of the projection of the twisted cubic from the point $p=(1:0:0:1)$ in the hyperplane $Z_3=0$.

So far I realized that the map $\pi_p(X)$ sends the coordinates $(Z_0:Z_1:Z_2:Z_3)$ to $(Z_0+Z_3:Z_1:Z_2)$. What can I do to try to find the equations?

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I believe the map $\pi_p$ should send $(Z_0:Z_1:Z_2:Z_3)$ to $(Z_0-Z_3:Z_1:Z_2)$.

Now consider the defining equations for the twisted cubic: $$Z_0Z_2=Z_1^2, \quad Z_0Z_3=Z_1Z_2,\quad Z_1Z_3=Z_2^2$$

Make a change of variable of the image of $\pi_p$: $$y_0=Z_0-Z_3, \quad y_1=Z_1, \quad y_2=Z_2$$

Plug these into the defining equations of the twisted cubic: $$y_0y_2+Z_3y_2=y_1^2\\ (y_0+Z_3)Z_3=y_1y_2\\ y_1Z_3=y_2^2$$

The last equation gives $$Z_3=\frac{y_2^2}{y_1}$$

Plug this into the first two will give you the defining equation of the projection.

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    $\begingroup$ Why is $\pi_p(Z_0:Z_1:Z_2:Z_3)=(Z_0-Z_3:Z_1:Z_2)$? How would this change if, for instance, $p=(0:0:1:0)$? $\endgroup$ – Barbara Nov 23 '17 at 13:46
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    $\begingroup$ @Barbara because the line joining a generic point $(t^3:st^2:s^2t:s^3)$ and the point $(1:0:0:1)$ is $(a+bt^3:bt^2s:bts^2:a+bs^3)$, then the intersection with $Z_3=0$ is given when $a=-bs^3$, so $\pi_p$ sends $(t^3:st^2,s^2t:s^3)\mapsto (t^3-s^3:st^2:s^2t)$, i.e., $(Z_0-Z_3:Z_1:Z_2)$. For any other point is the same idea. $\endgroup$ – e.turatti Nov 23 '17 at 15:43

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