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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?

evinda
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2 Answers2

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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.

Klaramun
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  • So, for $n=1$ it is like that:

    $$F(X,Y,Z)=X+Y+Z$$

    $$\frac{\partial{F}}{\partial{X}}=1$$

    $$\frac{\partial{F}}{\partial{Y}}=1$$

    $$\frac{\partial{F}}{\partial{Z}}=1$$

    So, we cannot find a $P$ such that $\frac{\partial{F}}{\partial{X}}(P)=\frac{\partial{F}}{\partial{Y}}(P)=\frac{\partial{F}}{ \partial{Z} }(P)=0$, right?

    $$$$ For $n>1$, is it like that?

    $$\frac{\partial{F}}{\partial{X}}=nX^{n-1}$$

    $$\frac{\partial{F}}{\partial{Y}}=nY^{n-1}$$

    $$\frac{\partial{F}}{\partial{Z}}=nZ^{n-1}$$

     – evinda Dec 08 '14 at 20:28
  • Is it like that?

    $$\frac{\partial{F}}{\partial{X}}(P)=0 \Rightarrow nX^{n-1}=0 \Rightarrow X=0$$

    $$\frac{\partial{F}}{\partial{Y}}(P)=0 \Rightarrow nY^{n-1}=0 \Rightarrow Y=0$$

    $$\frac{\partial{F}}{\partial{Z}}(P)=0 \Rightarrow nZ^{n-1}=0 \Rightarrow Z=0$$

    If so, could you explain me why $[0,0,0]$ does not live in the projective space?

     – evinda Dec 08 '14 at 20:33
  • Yes, this is so, and by definition of the projective space $[0,0,0]$ is not in there, do you know the definition of it? – Klaramun Dec 08 '14 at 21:56
  • Klaramun Could you maybe explain me the definition? – evinda Dec 09 '14 at 15:31
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    Yes. Let $V$ be a $k$-vector space of dimension $n+1$. We define the projective space of $V$ (or simply the projectivization of $V$) $\mathbb{P}(V) = \mathbb{P}_V^n$ as \begin{equation} \mathbb{P}_V^n = V \backslash { 0 } / \sim \end{equation} where the equivalence relation $\sim$ is given by \begin{equation} v \sim w \text{ if and only if } v = \lambda w \text{ for some } \lambda \in k^ \end{equation*} That is, $\mathbb{P}_V^n$ is simply the set of $1-$dimensional subvector spaces of $V$. – Klaramun Dec 09 '14 at 16:29
  • Klaramun So, how could we justify it formally that there are no singular points? – evinda Dec 09 '14 at 16:42
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    As some other people have noticed, it depends on the field you are working with. If it has characteristic $0$, then the argument I gave to you is completely formal (recall $[0,0,0]$ is not in the projective space simply by definition, we are excluding the $0$ vector from $V$). If the field has characteristic $p > 0$, then the same argument can work only if $p$ does not divide $n$ (see the answer below). – Klaramun Dec 09 '14 at 17:06
  • We are working in $K=\mathbb{Q},\mathbb{R}, \mathbb{C}$. – evinda Dec 09 '14 at 17:20
  • So here you have it, – Klaramun Dec 09 '14 at 17:38
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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.

Georges Elencwajg
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