part(b): I can prove the kernel is $(y^2-xy-x^3)$ and identify the image, but what kind of intuition should I get from the variety or to explain what? Thank you in advance.
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
The geometric intuition here is that (a) describes a curve with a cusp at the origin - see the picture at the top of this wikipedia page, which happens to use precisely the same curve to illustrate what a cusp is - whereas (b) describes a deformation of that curve so that the cusp is removed and replaced with a simple self-intersection at the origin.
Let's think of $t$ as time, and imagine a particle whose position at time $t$ is $(t^2-t,t^3-t)$. The particle starts at the origin when $t=0$, takes a detour in the all-negative quadrant between $t=0$ and $t=1$, before returning to the origin at $t=1$ (and flying off to infinity along the positive quadrant at all other times).
We can interpolate between cases (a) and (b) by modifying the motion of the particle such that its position at time $t$ is $(t^2-\epsilon t, t^3-\epsilon^2 t)$. When $0<\epsilon<1$, this particle returns to the origin at time $t=\epsilon$, and if you perform a similar computation as in case (b) you will find that the image is the set of polynomials $p(t)$ satisfying $p(0)=p(\epsilon)$. Finally, taking the limit as $\epsilon\to 0$ we see that this condition becomes $p'(0)=0$, or in physical terms the particle has speed $0$ at the origin - it takes an "infinitesimal rest" while changing its direction.
