9

To the best of my knowledge, an algebraic curve of degree $2$ is smooth if and only if it is irreducible. In other words, the only smooth conic sections are the non-degenerate ones.

Does this hold for any higher degree algebraic curves?

I know that being reducible implies that the curve is not smooth, which is the contrapositive of smooth implying irreducible. This should follow from the product/Leibniz rule of differentiation, because if an algebraic curve equals $p(x_1,\dots,x_n)q(x_1,\dots,x_n)$, then all of its partial derivatives are of the form $\frac{\partial}{\partial x_i}p(x)q(x_1,\dots,x_n) +p(x_1,\dots,x_n) \frac{\partial}{\partial x_i}q(x)$, and thus the curve will have a singular point at every point of intersection of the zero sets of $p$ and $q$, which by Bezout's theorem should be non-empty at least in projective space.

(Here I am using the definition of smooth as having no singular points at all, singular points being where the gradient vanishes and thus we can not define a tangent line.)

So I guess my question reduces to:

Does an algebraic curve being irreducible imply that it is smooth?

Chill2Macht
  • 20,920
  • 3
    Well, be careful — you seem to be mostly thinking about the plane even though there's an $n$. In higher ambient dimensions curves don't have to meet. Two skew lines in $\mathbb P^3$ have degree $2$ (check!) and that's a very smooth scheme. – Hoot Sep 06 '16 at 19:16
  • 1
    By the way, one should definitely study the cuspidal and nodal plane cubics. These never stop coming up. – Hoot Sep 06 '16 at 19:25
  • @Hoot Does Bezout's theorem only hold for the projective plane? I am new to algebraic geometry and actually haven't even gotten to the proof of it yet, so I don't remember all of the assumptions. It would make some sense intuitively if curves not in the same plane did not have to intersect in higher dimensions. – Chill2Macht Sep 06 '16 at 21:01
  • 1
    Bezout's theorem holds in complete generality but you have to generalize every part of it, not just the dimension of the ambient space, and you have to work harder to get the result. A definitive statement should be in section 8.4 of Fulton's Intersection Theory. I don't know if I'm a jerk for suggesting that book but I can't think of anything else right now. – Hoot Sep 06 '16 at 21:31

1 Answers1

8

As soon as the degree is larger than two, there are non-smooth irreducible curves of degree $d$. Take for instance $z^{d} - t w^{d-1}=0$ in the projective plane, in the homogeneous coordinates $[z:w:t]$. It has a cusp at $[0:0:1]$.

Addendum: in fact, there is a whole (difficult) field in algebraic geometry whose goal is to understand singularities of irreducible algebraic sets. Topics include classification and finding desingularizations (roughly speaking, finding a way to "fix" the singularity). Those questions can also be studied over different fields.

Albert
  • 9,170
  • I appreciate the simple counterexample, especially because it works for all $d>2$. Do you know if that field in algebraic geometry which you mentioned has a name, and if so, what the name is? I am curious to read up a little about it, since right now I am studying only smooth cubic curves, and the authors don't seem to give the assumption of smoothness as much attention as it might deserve. – Chill2Macht Sep 06 '16 at 20:57
  • 1
    Singularity is everywhere in algebraic geometry — it's hard to really pin it down in one place. I haven't looked at it in a while but Brieskorn and Knörrer have a book Plane Algebraic Curves that's pretty gentle. – Hoot Sep 06 '16 at 21:37
  • 1
    @William singularity theory ;) – Albert Sep 07 '16 at 06:27