# How to obtain the genus of the Riemann surface corresponding to an algebraic curve

I am trying to read about the genus of an algebraic curve. I have been told that there is a connection between topological genus and genus defined for an algebraic curve. Since an algebraic curve gives rise to a $2$ dimensional compact manifold(I am plotting the curve in $\mathbb{CP}^2$) I understand that it has to be connected sums of torus.

But I cannot get the genus this way. I am aware that one can find the genus by applying Quadratic Transformation but I am interested in knowing if we can find the genus by blowing up i.e by blowing up I can get a nonsingular curve which is birational to the given curve which is not necessarily a plane curve. Is there some formula to calculate genus from here?

Also say I have a projective curve then can I dehomogenise it , then blow up to make a non singular plane curve and finally homogenise it and find the genus.

• for nonsingular algebraic curves, topological genus = algebraic genus – oxeimon Jul 1 '15 at 4:35
• @oxeimon Is it true even if it is not a plane curve? – happymath Jul 1 '15 at 4:37
• – oxeimon Jul 1 '15 at 4:39
• @oxeimon I tried to calculate the genus of $y^2=(x-a_1)(x-a_2)(..)(x-a_4)$ It has a singular point at $[0,1,0]$ after blowing up i got $1=x^2(1-a_1y)(...)(1-a_4y)$ now from this it seems to be $5*4/2=10$. but I think the answer is 1 – happymath Jul 1 '15 at 4:45
• The problem is that blowing up a plane curve gives you a curve that no longer lies in a plane, and the degree calculation that you show is no longer valid. Indeed, the genus of the nonsingular model for that plane curve is $1$. – John Brevik Jul 1 '15 at 8:56