# Quartic curve - what is the genus ?

Hi,

I am studying the following quartic curve:

$f(x,y) = c_1x^2 + c_2x^4 + c_3x^2y + c_4x^2y^2 + c_5y^2 + c_ 6y^3 + c_7y^4$

where $c_i$ are constant (in fact they are expressions in terms of other constants). Starting to learn a bit about curves, I found that a necessary condition for a point $(x_0, y_0)$ be singular (a double point) is that

$$F(x_0, y_0) = 0,\qquad F_x (x_0, y_0) = 0,\qquad F_y (x_0, y_0) = 0$$

and that the second derivatives calculated at that point are not all equal to zero. Solving these three equations (trial and error) I got two solutions:

$$(x_0, y_0) = (0,0),\qquad (x_0, y_0) = (0, -2 c_5/c_6)$$

The second solution is a solution due to the fact that the coefficients $c_i$ are interrelated. For both points the second derivatives are not equal to zero.

Therefore, this curve has apparently has two double points, both with multiplicity equal to 2. Thus, this curve would have genus = 1, if there are no more singular points.

My questions are:

1) What I said above accurate?

2) is there any simple way to test if there is more singular points?

3) if no more singular points, how to parameterize a quartic curve like that? (I tried to transform this curve in an elliptic one, making $x^2 = z$, but i'm not sure if this is correct.)

Thank you very much

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If the curve is singular, there is more than one notion of genus for it. – Mikhail Bondarko Nov 19 '10 at 12:32
In order to have an ordinary double point (= a node), you also need to have distinct tangents at the point, which means the Hessian matrix at that point is invertible. – François Brunault Nov 19 '10 at 13:01
If indeed your curve has genus 1, because it has two simple double points as you claim (I have not checked that), then you can transform it into a smooth cubic in the following way. Take a quadratic Cremona transformation based on three points on the curve, two of them being the singular points. The result must be a smooth cubic. – Chris Wuthrich Nov 19 '10 at 13:37
I think you should look also at the points "at infinity", by homogenizing your equation, to see if there are more singular points there, since this is an affine equation. – roy smith Nov 19 '10 at 16:29
I don't think your second "singular" point is on the curve for general values of the coefficients. It looks like it's generically a curve of genus two. It is clearly elliptic if $c_2=0$. – Felipe Voloch Nov 19 '10 at 16:49

The proper setting for the question, as pointed out in a comment, is the projective plane $P^2$, so I'm going to add a variable $z$ and make everything homogeneous. Also, since you say nothing about the coefficients, I will work over the complex numbers.
Consider the linear system of plane quartics spanned by $z^2x^2$, $x^4$, $zx^2y$, $x^2y^2$, $z^2y^2$, $zy^3$ and $y^4$. The only base point of this system is the point $P=[1,0,0]$. It is easy to see that every curve of the system is singular at $P$ (it is true for all the generators) and that there is at least one curve (e.g. $z^2(x^2+y^2)=0$) that has an ordinary double point at $P$. Hence by Bertini's theorem the general curve of the system has an ordinary double point at $P$ and is smooth elsewhere. It is easy to show directly that such a curve cannot be reducible, so by the genus formula it has geometric genus 2.