Mixing
How to find the equation of an image under a central projection
$begingroup$
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 $ain C Leftrightarrow f(a)=0$. Is that correct? And how do I find that?
Any help would be appreciated!
algebraic-geometry algebraic-curves projective-geometry projective-space projection
asked Jan 13 at 18:46
get rekt m8get rekt m8
807
$endgroup$
add a comment |
$begingroup$
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 $ain C Leftrightarrow f(a)=0$. Is that correct? And how do I find that?
Any help would be appreciated!
algebraic-geometry algebraic-curves projective-geometry projective-space projection
asked Jan 13 at 18:46
get rekt m8get rekt m8
807
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $ain C Leftrightarrow f(a)=0$. Is that correct? And how do I find that?
Any help would be appreciated!
algebraic-geometry algebraic-curves projective-geometry projective-space projection
asked Jan 13 at 18:46
get rekt m8get rekt m8
807
$endgroup$
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 $ain C Leftrightarrow f(a)=0$. Is that correct? And how do I find that?
Any help would be appreciated!
algebraic-geometry algebraic-curves projective-geometry projective-space projection
algebraic-geometry algebraic-curves projective-geometry projective-space projection
asked Jan 13 at 18:46
get rekt m8get rekt m8
807
asked Jan 13 at 18:46
get rekt m8get rekt m8
807
edited Jan 13 at 19:11
get rekt m8
asked Jan 13 at 18:46
get rekt m8get rekt m8
807
asked Jan 13 at 18:46
get rekt m8get rekt m8
807
asked Jan 13 at 18:46
get rekt m8get rekt m8
807
807
$begingroup$
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
add a comment |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@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
$begingroup$
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
$begingroup$
@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
add a comment |
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$begingroup$
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