I have to solve this exercise but I have really no clue even how to start with it:
Identify the topological space underlying the cubic $Y^2Z=X^2(X-Z)$ in $\mathbb{PR}^2$. How does it fit with the shape of $Y^2Z=X(X-Z)(X-\lambda Z)$ in $\mathbb{R}^2$ ($\lambda\in\mathbb{R}-\{0,1\}$)?
I would be really thankful if anybody could help me, thanks in advance!
On
a) The topological space underlying the cubic $Y^2Z=X^2(X-Z)$ in $\mathbb{PR}^2$ is disconnected .
One (closed) connected component is the singleton set consisting of just the origin {[0:0:1]}.
The other component (the rest) is homeomorphic to a circle.
b) Every curve $Y^2Z=X(X-Z)(X-\lambda Z)$ with $\mathbb{R}^2$ ($\lambda\in\mathbb{R}-\{0,1\}$) is disconnected too, and is homeomorphic to the disjoint union of two circles.
Think of $\Bbb R\Bbb P^2$ as $\Bbb R^2$ with a copy of $\Bbb R\Bbb P^1$ at infinity. Setting $Z=1$, you get the affine "curve" $y^2=x^2(x-1)$, consists of the origin and a smooth curve (for $x\ge 1$). It meets the "line at infinity" in the one point $[0,1,0]$. Thus, it is homeomorphic to the union of a point and a circle. For extra credit (:)), can you show that the curve has an inflection point at infinity (and meets the line at infinity tangentially)?
I mistakenly was thinking of the curve $Y^2 Z = X^2(X+Z)$; setting $Z=1$, you get the affine nodal cubic $y^2=x^2(x+1)$. When you include the point at infinity, you should be able to convince yourself that this is homeomorphic to a figure eight curve (i.e., the wedge of two circles). It looks like this:
In the case of the smooth cubic, the standard affine picture has two components. To what standard one-manifold do you think the projective curve is homeomorphic?