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How to compute the genus of $ \{X^4+Y^4+Z^4=0\} \cap \{X^3+Y^3+(Z-tW)^3=0\} \subset \mathbb{P}^3$?

We know that the genus of $ \{X^4+Y^4+Z^4=0\} \subset \mathbb{P}^3$ is 3 because the degree is 4. Now, I want to know the genus of the intersection as a curve. For that I have to use the adjunction formula and the fact that $K_{\mathbb{P}^3}=O(-4)$.

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    $\begingroup$ WHat do you man by genus in the second paragraph? That is not a curve... $\endgroup$ – Mariano Suárez-Álvarez Apr 6 '11 at 18:36
  • $\begingroup$ Added algebraic-geometry tag $\endgroup$ – Jyrki Lahtonen Jun 17 '11 at 5:24
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If $X$ is the the intersection in $\mathbb{P}^3$ of two hypersurfaces of degrees $d_1$, $d_2$, respectively, and $\mathrm{dim} \ X = 1$, then the genus of $X$ is: $$ g = \frac{d_1^2d_2 + d_1 d_2^2}{2} - 2 d_1 d_2 + 1. $$ So, in your case $d_1 = 4$ and $d_2 = 3$, therefore $g = 19$.

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Alas, I don't know how to use $K_{\mathbf{P}^3}$ here, so this solution may not be of use to you.

Assuming that $t\neq0$, and that your base field $k$ is algebraically closed with char $k > 3$, then (writing $U=Z-tW$) the function field of this variety is $k(y,z,u)$, with $y=Y/X$, $z=Z/X$, $u=U/X$, $z^4+y^4+1=0$, $u^3+y^3+1=0$. This is a tower of Kummer extensions, so ramification is easy to check. All the ramification of $k(y,z)/k(y)$ occurs at the 4 (finite) places with $y^4+1=0$. We have $e=4$ for all these points, so the genus $g'=g(k(y,z))$ can be solved from $$ 2g'-2=4(2\cdot0-2)+4\cdot(4-1)\Rightarrow g'=3, $$ as expected.

Similarly all the ramification in $k(y,z,u)/k(y,z)$ takes place at those points, where $y^3+1=0$. This equation has 3 solutions in $k$ and none of them are also solutions of $y^4+1=0$. Therefore the 3 ramified places of $k(y,u)/k(u)$ become 12 ramified places of $k(y,z,u)/k(y,z)$. For all these we have $e=3$. Therefore the genus $g''$ of $k(y,z,u)$ is $$ 2g''-2=3(2\cdot 3-2)+12(3-1)\Rightarrow g''=19. $$

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