1
$\begingroup$

I'm trying to compute the genus of the normalization of the curve: $y^5=x(x-1)(x-2)$ Now I calculate the ramification points of the projection x: they are $(0,0),(1,0),(2,0)$ and they are of ramification order 5, and possibly ramification points at the infinite point $(1:0:0)$.

Now the curve is singular at $(1:0:0)$, and after a long blow up computation I found another ramified point of degree 5. The Riemann Hurwitz formula then gives that the genus is 4.

EDIT: As suggested below the genus-degree formula has a correction term for singularities. The curve has one singularity at infinity: $(1:0:0)$. The equation for a neighborhood around this point is given by $y^5-z^2x^2-3z^3x-2z^4$. Following the procedure for calculating the multiplicty of a singular point outlined in the link below I find that it is equal to 2. Hence the genus of the curve is $g(C)=(d-1)d/2-r(r-1)/2=(5-1)(4-1)/2-2(2-1)/2=6-1=5$.

Again however the genus does not agree with the computation with ramification points. Moreover even if the ramification at infinity is different this can only decrease the genus of the curve to be smaller than 4.

Link: How does one calculate genus of an algebraic curve?

$\endgroup$
  • 3
    $\begingroup$ The degree-genus formula only holds for smooth curves, right? $\endgroup$ – user4571 Dec 13 '15 at 23:47
  • 1
    $\begingroup$ Without doing the computation, my suspicion is that the R-H formula computation is correct, and that once you account for the singularities in the degree-genus formula (there are modifications to handle these that lower the result) you will get the same answer. $\endgroup$ – user4571 Dec 13 '15 at 23:50
  • $\begingroup$ Patrick is correct; for (possibly) singular curves the degree-genus formula is only an inequality. For instance, a curve of degree $d$ with a single ordinary node will have genus one less than a smooth curve of degree $d$. In general, the correction term is a quantity called the delta invariant of the singularity. $\endgroup$ – Tabes Bridges Dec 14 '15 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.