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The algebraic curve $Y^2 - X^3 = 0$ has a singularity at the origin because, while the curve is irreducible over the ring of polynomials, the curve is not irreducible when viewed in the set of differentiator functions. We have a factorization as $(Y-\sqrt{X^3})(Y+\sqrt{X^3})$, so the curve is the union of two differentiable curves with intersect badly at the origin. Similarily, $Y^2 - X^2 - X^3$ defines another algebraic curve with a singularity at the origin, because the polynomial is reducible over the ring of differentiable functions to $(Y-X\sqrt{X + 1})(Y + X\sqrt{X+1})$, and these differentiable curves define two curves with different tangents at the origin.

My question is whether there is a general characterization of singularity theory over algebraic curves over the real and complex numbers, in the sense that every singularity results from the fact that the algebraic curve contains two differentiable curves nonsingular at the point. I tried working with the ring of germs of functions locally differentiable around the point, and I found that if an element is reducible in this ring, it must be singular at the point. But I had difficulties proving that if a function is irreducible it must be nonsingular at a point. What kind of results exist in the literature which expand upon this idea of singularities?

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  • $\begingroup$ I think this is more or less the idea behind the geometric version of the Puiseux theorem. Basically, one can split any curves singularities into "branches", and these branches can be used for solving the singularity. Googling this should lead to relevant informations I hope. $\endgroup$ – user171326 Jun 8 '17 at 15:50
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The cusp $\mathcal{C}$ given by $Y^2 - X^3 = 0$ is an example of what you seek.

In both of your examples, what you are doing geometrically is constructing a two-fold cover of the curve, and it is this two-fold cover that reduces to the union of two curves.

However, the two examples are qualitatively quite different. Assuming these are curves over $\mathbb{R}$, then when you look at the Euclidean topology:

  • In the case of the cusp, any neighborhood of $(\sqrt{X}, Y) = (0,0)$ on $Y - \sqrt{X^3}$ contains an entire neighborhood of $(X,Y) = (0,0)$ on $Y^2 - X^3$
  • In the case of the node, sufficiently small neighborhoods of $(\sqrt{X+1}, Y) = (0,0)$ on $(Y - X\sqrt{X+1})$ do not contain an entire neighborhood of $(X,Y) = (0,0)$.

Algebraically, it's more convenient to just consider one of the new curves in isolation, by introducing a new variable $Z$.

You are considering the curve $Y - Z^3 = 0$, mapping onto the cusp by $(X,Y) = (Z^2, Y)$, and the curve $Y - (Z^2 - 1) Z = 0$ mapping onto the node by $(X,Y) = (Z^2 - 1, Y)$.

In the case of the cusp, the singular point lifts to a single point of the cover; geometrically it's more like 'straightening out a kink'. But in the case of a node, the singular point lifts to two different points of the cover, representing the fact you've split the single point into two different points.

Another way to express the difference algebraically is that, as polynomials over the ring of power series $F[[X]]$, $Y^2 - X^2 - X^3$ factors into two irreducible factors, but $Y^2 - X^3$ is still irreducible.

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  • $\begingroup$ @JacobDenson: I've completely rewritten my post, so if you saw it earlier, it's worth looking again. $\endgroup$ – Hurkyl Jun 8 '17 at 16:07

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