Algebraic Curve
-
- Community Team
- Posts: 314
- Joined: Tue Nov 10, 2015 8:25 pm
Algebraic Curve
Prove that the set \( \displaystyle \left\{ (t^2,t^3+1) \in \mathbb{C}^{2} \, \big| \, t \in \mathbb{C} \right\} \) defines an (affine) algebraic curve.
-
- Community Team
- Posts: 426
- Joined: Mon Nov 09, 2015 1:52 pm
Re: Algebraic Curve
If \(\displaystyle{\left(t^2,t^3+1\right)\in\mathbb{C}^2}\) is a typical element of the given set, then
by setting \(\displaystyle{x=t^2\,,y=t^3+1}\), we get :
\(\displaystyle{y-1=t^3\implies (y-1)^2=t^6=x^3\iff x^3-y^2+2\,y-1=0}\) .
On the other hand, if \(\displaystyle{f(x,y)=x^3-y^2+2\,y-1\in\mathbb{C}[x,y]}\), then
\(\displaystyle{f(x,y)=0\iff x^3-y^2+2\,y-1=0\iff x^3=(y-1)^2}\). So, by considering the family
\(\displaystyle{y-1=t\,x\,,t\in\mathbb{C}}\), we get :
\(\displaystyle{f(x,y)=0\,\land y-1=t\,x\iff x=t^2\,,\land y=t^3+1}\) and if \(\displaystyle{x=0}\)
then \(\displaystyle{y=1}\) so :
\(\displaystyle{\left\{\left(t^2,t^3+1\right)\in\mathbb{C}^2:t\in\mathbb{C}\right\}=V(f(x,y))}\) .
by setting \(\displaystyle{x=t^2\,,y=t^3+1}\), we get :
\(\displaystyle{y-1=t^3\implies (y-1)^2=t^6=x^3\iff x^3-y^2+2\,y-1=0}\) .
On the other hand, if \(\displaystyle{f(x,y)=x^3-y^2+2\,y-1\in\mathbb{C}[x,y]}\), then
\(\displaystyle{f(x,y)=0\iff x^3-y^2+2\,y-1=0\iff x^3=(y-1)^2}\). So, by considering the family
\(\displaystyle{y-1=t\,x\,,t\in\mathbb{C}}\), we get :
\(\displaystyle{f(x,y)=0\,\land y-1=t\,x\iff x=t^2\,,\land y=t^3+1}\) and if \(\displaystyle{x=0}\)
then \(\displaystyle{y=1}\) so :
\(\displaystyle{\left\{\left(t^2,t^3+1\right)\in\mathbb{C}^2:t\in\mathbb{C}\right\}=V(f(x,y))}\) .
Create an account or sign in to join the discussion
You need to be a member in order to post a reply
Create an account
Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute
Sign in
Who is online
Users browsing this forum: No registered users and 21 guests