# Euler characteristic of the zero set of $z_1^d+z_2^d+z_3^d=0$ in $CP^2$

I'm doing this qualifying exam problem from my university and got stuck with this one:

Let $$Z$$ be the zero set of the polynomial $$z_1^d+z_2^d+z_3^d=0$$ in $$CP^2$$, here $$z_1,z_2,z_3$$ are complex numbers; what is the Euler characteristic of $$Z$$?

I really have no clue what to do with this. I know that the Euler characteristic is the same as the self intersection number for a compact oriented manifold. However, I'm not sure how to deform the zero set to be transversal. Can anyone get me started please? Thank you! (This question might be related to algebraic geometry, however algebraic geometry is not in our syllabus.)

Edit: the available results from my syllabus are those in Lee ,Hatcher, and Guillemin and Pollack only.

• A non-singular curve in the projective plane defined by a degree $d$ equation has genus $g=(d-1)(d-2)/2$ should get you started. No sure that result is available at a qual, though. You could try and use Hurwitz genus formula instead (if available). Transversal may be the way to go. Commented Mar 13, 2021 at 8:15
• If you're expecting some help at the level of your qualifying exams, it would be a good idea to for you to mention (at least broadly) what results are available and what topics are usually covered. Commented Mar 13, 2021 at 9:28
• d lines have a "total" Euler characteristic of 2d and intersect in d(d-1)/2 points; smoothing each reduces the naive Euler characteristic count by 2 (as in from 4 to 2 in the case of d=2), leaving us with 2d-d^2+d=3d-d^2 (and genus g=(-e+2)/2=(d^2-3d+2)/2=(d-1)(d-2)/2).
– Max
Commented Mar 13, 2021 at 17:58