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I have some difficulties to prove that the image of the function $f:\,\mathbb P^1\longrightarrow\mathbb P^3$ such that $$(u,v)\longmapsto (u^3,\,u^2v,\,uv^2,\,v^3)$$ is the algebraic projective set $$V(XT-YZ,\, Y^2-XZ,\,Z^2-YT)$$ Clearly I have problem to prove that the algebraic projective set is contained in the image of $f$. In particular "solving brutally" the polynomial system I'm losing my mind in calculations, so I hope that there is a simpler method.

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Every point of $V=V(XT-YZ,\, Y^2-XZ,\,Z^2-YT)$ is in at least one of the four standard copies of $\mathbb A^3$affine spaces covering $\mathbb P^3$. So check them successively. I'll do the $T=1$ part.

When $T=1$, $V$ is given by $$x=yz, \: y^2=xz,\: z^2=y$$ But then a point $(x,y,z)=[x:y:z:1] \in V $ satisfying these equations is simply the image under $f$ of $[u:v]=[z:1]\in \mathbb P^1 $, since the first and third displayed equations for $V$ immediately imply that $x=yz=z^2.z=z^3$

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    $\begingroup$ In fact, for this particular calculation, you only need to check the charts where $T \neq 0$ and where $X \neq 0$, since $X = T = 0 \Longrightarrow Y = Z = 0$ in $V$. $\endgroup$ – Michael Joyce May 12 '12 at 23:42
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    $\begingroup$ An excellent observation, @Michael: +1. And then for $[1:y:z:t]\in V$ , we have $f([1:y])=[1:y:y^2:y^3]=[1:y:z:t]$ $\endgroup$ – Georges Elencwajg May 13 '12 at 0:14
  • $\begingroup$ @Galoisfan: Welcome to mathstackexchange! You are encouraged to 'accept' an answer that addresses your question, which Georges' certainly does. $\endgroup$ – Michael Joyce May 13 '12 at 0:59

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