Proving the image of a map $v$ in $\mathbb{P}$ is the zero locus of $f:=x_{0}x_{2}^2-x_{1}^3-x_{0}x_{1}^2$.

Question

Consider the map $v:\mathbb{P}\longrightarrow \mathbb{P}^2$ suct that $v(x_{0},x_{1})=(x_{0}^3,x_{0}x_{1}^2-x_{0}^3,x_{1}^3-x_{0}^2x_{1})$. How to prove that $v(\mathbb{P})$ is the zero locus of $f:=x_{0}x_{2}^2-x_{1}^3-x_{0}x_{1}^2$. The first inclusion is clear, that is $v(\mathbb{P})\subset Z(f)$. Also I know that if $[x_{0}:x_{1}:x_{2}]\in Z(f)$ and $x_{1}= 0$, then $[x_{0}:x_{1}:x_{2}]=[1:0:0]$ or $[0:0:1]$. And if $x_{1}\neq 0$, then we can suppose it one and obtain the following: $y_{0}y_{2}^2-y_{0}=1$, how can I complete to finish this discussion and obtain the second inclusion?

Answer 1

First of all, I think it is better to change the name of the variables, and we can restrict to the affine space $x_0\ne 0$ as you said. So we want to show that the map $f:\mathbb{A}^1\to \mathbb{A}^2$ given by $f(t)=(t^2-1,t^3-t)$ has image the curve $C$ given by the equation $y^2=x^3+x^2$ (where I use $(x,y)$ as coordinates of $\mathbb A^2$). It is easy to show the image of $f$ is inside $C$. Now, to see that it is equal to the image, you can use that in the image of $f$ it happens that $$\frac{y}{x}=\frac{t^3-t}{t^2-1}=t.$$ So, take a point $(x,y)\in C$, and define $t(x,y)=\frac{y}{x}$ if $x\ne 0$. Then one easily show form the equation $y^2=x^3+x^2$ that $$x=\left(\frac{y}{x}\right)^2-1$$ and that $$y=\left(\frac{y}{x}\right)^3-\frac yx.$$ so $(x,y)=f(t(x,y))$. It remains to consider the case $x=0$, so $y=0$, but then $f(1)=(0,0)=f(-1)$.