Does the projective Fermat curve have singular points?

Question

I want to check if the projective Fermat curve,

$$X^n+Y^n+Z^n=0, n \geq 1$$

has singular points.

Could you give me a hint how we could do this? Do we have to check maybe if the curve is irreducible? If so, how could we do this?

Answer 1

If you let $F(X,Y,Z) = X^n+Y^n+Z^n$, the singular points of the curve $C = \{ F = 0\}$ are defined by having all its derivatives equal to $0$, that is, $P \in C$ is a singular point if and only if \begin{equation*} \frac{\partial F}{\partial X}(P) = \frac{\partial F}{\partial Y}(P) = \frac{\partial F}{\partial Z}(P) = 0 \end{equation*}

The case $n=1$ is trivial (no such points). Let $n > 1$, check that the only possible point is $[0,0,0]$, but this point does not live in the projective space.

Hence this curve has no singular points.

Answer 2

If the curve $C$ is defined over a field $k$ , then $C$ is smooth unless the characteristic of the field is a prime number $p$ and that characteristic divides $n$, namely $n=p\nu$ for some $0\neq n\in \mathbb N$.

a) In this latter case the curve is non reduced, since its equation is $(x^\nu+y^\nu+z^\nu)^p=0$, and thus certainly not smooth.
b) If on the other hand $char (k)=0$ or $char(k)=p\gt0$ but $p$ does not divide $n$, then $C$ is smooth and thus irreducible.