Let $\pi:\mathbb{P}^3 \to V(x_2) \cong \mathbb{P}^2$ the linear projection with center $P =(0:1:0:0)$. Find the equation for the image of $C=\{(s^3:s^2t:st^2:t^3)|~(s:t) \in \mathbb{P}^1 \}$ under $\pi$.

I am currently trying to solve this problem.
What I know is that $\pi(C)=\{(s^3:0:st^2:t^3)|~(s:t) \in \mathbb{P}^3 \}$. However I am uncertain of what is meant by 'Find the equation for the image of $C$'. My guess is that I need to find a polynomial $f \in k[x_1,x_2,x_3,x_4]$ such that $a\in C \Leftrightarrow f(a)=0$. Is that correct? And how do I find that?
Any help would be appreciated!

  • $\begingroup$ First, $(s,t)\in \mathbb{P}^1$ not 3 as you write. $\pi(C)$ is a curve in $\mathbb{P}^2$, so you need to find an irreducible homogeneous polynomial $f(x_1,x_2,x_3)$ such that $f(s^3,st^2,t^3)=0$ for all $(s:t)$. $\endgroup$ – Mohan Jan 13 at 19:07
  • $\begingroup$ so $f=x_1x_3^2-x_2^3$ would do the job right? Is that really the whole exercise, it seems too simple $\endgroup$ – get rekt m8 Jan 13 at 19:17
  • $\begingroup$ @Mohan could you tell me what I would have to do if $P$ was a different point? I was told I need a change of coordinates, but i am not sure how that works $\endgroup$ – get rekt m8 Jan 13 at 19:57
  • $\begingroup$ What does $V(x_2)$ mean? $\endgroup$ – amd Jan 13 at 20:32
  • $\begingroup$ @amd the subset of $\mathbb{P}^3$ where the second coordinate is $0$. So $z \in V(x_2)$ has the form $(a:0:c:d)$ where not every coordinate is zero. $\endgroup$ – get rekt m8 Jan 13 at 20:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.