I have this curve $C =$ { $[X,Y,Z] \in \mathbb{P^2C} | Y^3Z = X^4 - X^2Z^2$}.

I've found that it's singular in $[0,0,0]$. So, I have to blow up this point. I don't know if $[0,0,1]$ is a base point (How to find one ?), but I've considered the equation for the affine curve related which is $Y^3 = X^4 - X^2$. Then for one chart of the blow-up I put $Y = ut$ and $X = t$ and I obtain $u^2(ut^3 - u^2 +1)$. I take $ut^3 - u^2 + 1 = 0$ which is $ut^3 -u^2w^2 + w^4$ in the projective space (Is this the curve without singularities I obtain blowing up the singular point ?)

If it is correct I've found a Riemann surface related to the curve $C$, on this surface I consider the meromorphic function $f = \dfrac{X}{Y}$. Is it possible that f has a zero in $[0,0,1]$ and a pole in $[0,1,0]$ both of order one ?

I can consider the holomorphic function $G : C{\setminus}[0,0,0] \longrightarrow \mathbb{P^1C}$ which sends the poles in $\infty$. I have then in $mult_{[1,0,0]}G =1 = mult_{[0,1,0]}G$.

This Riemann surface has genus 3 by Plücker's formula. So I have to verify Hurwitz's formula $$2g(C{\setminus}[0,0,0]) -2 = degG(2g(\mathbb{P^1} - 2) + \sum_p (mult_pG - 1).$$ So, I don't know how to compute the degree of G and if G has any branch points.

Sorry for the long question but I'm a little bit confused. Thank you.

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    $\begingroup$ 1) (0,0,0) does not define a point in $\Bbb P^2$. 2) What do you mean by a base point ? A base point is usually associated to a linear system. $\endgroup$ – user171326 Jul 1 '17 at 14:00
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    $\begingroup$ Keep in mind that $(0,0,0)$ is a singular point of the affine cone over the projective curve, but that point is systematically removed in the definition of projective variety. $\endgroup$ – Tabes Bridges Jul 1 '17 at 16:09
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    $\begingroup$ Regarding what to blowup, eyeballing it looks to me like $(0,0,1)$ is indeed going to be singular (double check with partials though!). And yes, you can blowup an affine patch containing the singular point; blowing up is a local process. $\endgroup$ – Tabes Bridges Jul 1 '17 at 16:12
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    $\begingroup$ You did the blow up correctly. You don't need to take the projective closure of the curve. $ut^3 - u^2 + 1 = 0$ has gradient $(t^3 - 2u, 2ut^2)$ so the only point where this vanishes is $(0,0)$ which is not on the affine curve, i.e you curve is indeed smooth after one blow-up. I would say that your curve is of degree $4$ with one singularity so it's a rational curve, i.e $C$ is birational to $\Bbb P^1$. On the other hand, if $X/Y$ has one zero and one pole of order one and is defined everywhere it would gives an isomorphism $C \cong \Bbb P^1$. And indeed it's not defined at $[0:0:1]$. $\endgroup$ – user171326 Jul 1 '17 at 19:42
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    $\begingroup$ So everything looks correct except the computation of the genus (smooth curve of degree $d$ have genus $\frac{(d-1)(d-2)}{2}$ and singularity decrease the genus.) $\endgroup$ – user171326 Jul 1 '17 at 19:43

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