I've just started with algebraic geometry, so i apologize in advance if my question is too easy to show.

Given is a curve $\Gamma $ in $\mathbb{P}^{2}(\mathbb{F_{q^{m}}})$ defined by $X^{q}Z+XZ^{q}-Y^{q+1}=0$. Find the number of the rational points and the genus $g$ of this curve.

Well, i see that the curve is already homogenized and the degree of it is $d=q+1$. I know that the genus can be at most $g=\frac{(d-1)(d-2)}{2}$ according to the general formula of genus. This is only in case i have a smooth curve. Here i see immediately that the points $(1,0,0)$ and $(0,0,1)$ are singular and they reduce the genus. How can i find the other singular points and the number of points on the curve? Can somebody help me with this question?

Thanks in advance!

  • 3
    $\begingroup$ Your curve is smooth! $\endgroup$ – Georges Elencwajg Dec 7 '14 at 22:18

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