This question is probably naive, but I was wondering: what exactly is a self-intersection in the graph of a polynomial's zero set?
For some parametric curve described by $x(t),y(t)$, a self-intersection is plainly the point $(x,y)$ for which, for some $w\neq t$, $x(w)=x(t)=x$ and $y(w)=y(t)=y$, with some trivial assumptions about the curve itself.
Yet for curves that are not parametrized, I have a hard time pinning down exactly what goes on at a point where the graph intersects itself. The point itself is the same point here, there, and everywhere else; no funny $w$ and $t$ business.
My question: Is there a definition of a self-intersection of the zero set of a polynomial, and what is remarkable about these points compared to others? Also, are there different kinds of intersection points? Because clearly, what we see here:
Is not the same as what we see here:
Even though both graphs are the zero sets of one polynomial.
The cubic $C:y^2=x^2(x+1)$ has a self-intersection at $(0,0)$.
To see it algebraically let $k=\Bbb{R}$ or any field of characteristic not $2$, the ring of polynomials functions $C\to k$ is $$k[C]=k[x,y]/(y^2-x^2(x+1))$$ the subring regular at $(0,0)$ of the field of rational functions is $$k[C]_{(x-0,y-0)} = \{ \frac{u}{v},u,v\in K[C],v(0,0)\ne 0\}$$ $m=(x-0,y-0)$ is its only maximal ideal and its $m$-adic completion is $$\varprojlim_{n\to\infty} k[C]_{(x-0,y-0)}/m^n=k[[x,y]]/(y^2-x^2(x+1))$$ $$=k[[x,y]]/((y-x(1+x)^{1/2})(y+x(1+x)^{1/2}))$$ $$\cong k[[x,y]]/(y-x(x+1)^{1/2}))\times k[[x,y]]/(y+x(1+x)^{1/2}))$$ where $$(1+x)^{1/2}=\sum_{l=0}^\infty {1/2\choose l} x^l \in k[[x]]$$ That the completion at $(0,0)$ is the product of two integral domains indicates a self-intersection of two smooth pieces of curves.
Then search about https://en.wikipedia.org/wiki/Intersection_number and https://en.wikipedia.org/wiki/Intersection_theory