I have the projective curve in $\mathbb{P}^2$ given by \begin{align} F(X,Y,Z)=Y^2 Z^2-X^4-Y^4. \end{align} I want to calculate the genus of the curve. My approach would be to calculate the partial derivatives and use the multiplicities $r_i$ of the singular points in the formula \begin{align} g(F)=\frac{(n-1)(n-2)}{2}-\sum_i \frac{(r_i-1) r_i}{2}. \end{align} I get $DF(X,Y,Z)=(4X^3,2YZ^2-4Z^3,2Y^2Z)$, and I get the singular point polynomials $X^3, Z^2(Y-2Z),Y^2Z$. Is the next step to calculate $r_i$ by the intersection number between the curve $F$ and the singular point polynomials? Or do I need to dehomogenize $F$ to get the tangents to intersection between the tangents? Or do I need another approach for the $r_i$?

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    $\begingroup$ there is only one singular point [0:0:1] with multiplicity 2. Just taking the common zero point of DF and F. $\endgroup$ – Chen Jiang Nov 19 '16 at 4:07
  • $\begingroup$ I believe you made a mistake when computing the differential $\endgroup$ – Xuqiang QIN Nov 19 '16 at 6:28

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