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?
Best Answer
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.