If we have some homogeneous polynomial $f(x, y)$ (for example $f(x, y) = x^3y + x^2y^2 - 17y^4$) than it's solution set is a union of straight lines (or a single point/null set). I've seen arguments for this, but I was wondering if this 'geometric' argument could be made formal.
Suppose we have a polynomial $p(x, y).$ We can look at its low degree part (all terms of highest degree) to get another polynomial $q(x, y)$ such that the graph of $q(x, y) = 0$ models the behavior of $p(x, y)$ in a neighborhood of $0.$ Hence, it's graph needs to be a union of straight lines (or degenerate point/null set) since the algebraic curve, by the implicit function theorem, describes a bunch of functions that are differentiable at $0$ and thus have linear approximations at $0$ and thus look like lines locally; hence the graph of $q(x, y) = 0$ should be a bunch of lines. As every homogeneous polynomial is clearly the low degree part of some curve, this property is true of all homogeneous polynomials.
(This argument is much more dubious...) Similarly, we could approach this problem from the opposite perspective, looking at the high degree part of the polynomial. Then, place the polynomial in the projective plane and, like before, find its derivatives 'at infinity' and therefore approximate it locally as a bunch of lines.
Can either of these two arguments be made rigorous? Is there even a reasonable way to talk about the derivative 'at infinity' in a projective plane?
It's easier: if $f(x,y)$ is homogeneous of degree $d$, then $f(\lambda x,\lambda y)=\lambda^d f(x,y)$. So if $f(x,y)=0$ for a point $(x,y)\neq (0,0)$, then $f$ also vanishes on all the points on the line through the origin passing through $(x,y)$.