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Let $C$ be the projective closure of $Z(f) \subset \mathbf{A}^2$ where $f$ is an irreducible polynomial of degree 4 in $x$ and degree 2 in $y$, so $C = Z(f^*) \subset \mathbf{P}^2$ where $f^*$ is the homogenization of $f$. For a particular example, let $f = x^4y - 2x^3y^2 + 6x^3y - 6x^2y^2 - 4x^3 + 12x^2y - 4xy^2 - 12x^2 + 12xy - 8x + 4y$. (I have a family of these, all of a particular form.)

Using MAGMA I find that this curve has a nonsingular model of genus 1:

> R<x,y,z> := ProjectiveSpace(Rationals(),2);
> C:= Curve(R, x^4*y - 2*x^3*y^2 + 6*x^3*y*z - 4*x^3*z^2 - 6*x^2*y^2*z + 12*x^2*y*z^2 - 12*x^2*z^3 - 4*x*y^2*z^2 + 12*x*y*z^3 - 8*x*z^4 + 4*y*z^4);
> Degree(C);
5
> Genus(C);
1
> P0:= C![-2,0,1];
> E, phi:=EllipticCurve(C,P0);
> E;
Elliptic Curve defined by y^2 - 4*x*y = x^3 - 11*x^2 + 12*x over Rational Field

How would one go about showing this by hand? I suspect that the algorithms used by MAGMA are not practical for this. Can we prove that the degree of the defining homogeneous polynomial of the nonsingular model is 3? In that case we would be done after an application of the genus-degree formula $g = (d-1)(d-2)/2$.

Note that I'm not asking for a construction of the nonsingular model, just a proof of the fact that it has genus 1. This may or may not be helpful.

Many thanks in advance.

Note: The singular points of my example $C$ that I know of are $[\sqrt{2} : \sqrt{2} : 1]$ and $[-\sqrt{2} : -\sqrt{2} : 1]$, and I believe they are ordinary singularities (multiplicity 2, two distinct tangents).

Note 2: Actually, $[0 : 1 : 0]$ is also a singular point, but I don't know what kind. If it's ordinary with multiplicity 3, then we are done by the generalized genus-degree formula discussed in the comments below. But how do I prove that these are all the singularities?

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2  
First of all you can know that the nonsingular model exists by using the normalization theorem, or something of the like. To find the genus, you have to look at the singularities of the curve, and then apply a generalized degree formula. For example, if you have the curve $y^2=x^2(x+1)$, then it has a node at $(0,0)$. This node subtracts 1 from your original degree formula, so the genus of the normalization of the curve is $(3-1)(3-2)/2-1=0$. For the general formula, look for example at "Singular Points of Plane Curves" by C. T. C. Wall, Corollary 7.1.3. –  rfauffar Jul 3 '13 at 15:29
    
@RobertAuffarth Thanks. A generalized genus-degree formula that takes singularities into account sounds like a good way to go. However, I haven't been able to successfully apply one to my example. –  Ricardo Buring Jul 5 '13 at 16:35
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Once you understand the singularities of your curve, you should basically be done, see for example the end of en.wikipedia.org/wiki/Genus%E2%80%93degree_formula it shouldn't be too bad to prove those are the only singularities, nor to prove they are of order 2 –  John Stalfos Jul 5 '13 at 20:42
    
@JohnStalfos With the notation of that article shouldn't I have $d = 5$ (the degree of $C$, which equals the degree of $f^*$), and $r = 2$ for each of the two singularities? Then I get $g = 6 - 1 - 1 = 4$, which is wrong. I'm also not sure how to prove those singularities are the only ones. –  Ricardo Buring Jul 6 '13 at 11:08
    
Actually there is at least one more singularity, namely $[0:1:0]$. I've edited my question to reflect this. –  Ricardo Buring Jul 6 '13 at 15:56

1 Answer 1

up vote 2 down vote accepted
+50

I don't know how Magma procedes, but in your particular example, it is easy. Namely you want to compute the genus of the projective smooth curve with function field equal to $k(x,y)$, where $k$ is the ground field (say of characteristic different from $2$) and $x,y$ are linked by the relation $f(x,y)=0$.

Write $f$ as $a_0(x)y^2+a_1(x)y+a_2$ with $a_i(x)\in k[x]$. Then you have $$ (2a_0y+a_1)^2+ ({4a_0a_2-a_1^2})=0. $$ This is the equation of a hyperelliptic. Let $4a_0a_2-a_1^2=g^2h$ with $g, h\in k[x]$ and $h(x)$ without square factor, then a non-singular equation of the function field is $$z^2+h(x)=0.$$ And it is known that the genus is $[(d-1)/2]$ if $d=\deg h(x)$. In you particular example, I found $h(x)=-(x^4 + 12x^3 + 32x^2 + 24x + 4)$ and the function field has genus $1$. It is a bad idea to write a projective equation and solve singularities.

To get an elliptic equation, you have to make a change of variables and send a rational point (say $x=0$, $z=2$) to $\infty$. I didn't check.

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Thanks, this seems like a better approach. How does it follow that $z^2 + h(x) = 0$ is a non-singular equation of the function field? Where can I find a proof that the genus is $[(d-1)/2]$ where $d = \deg h(x)$? Can we compute $\deg h(x)$ more generally? Say, in terms of the $a_i(x)$? I'm also not sure which change of variables to make in order to get an elliptic equation. –  Ricardo Buring Jul 12 '13 at 12:25
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Because you have something like $y_1^2+g^2h=0$, where $y_1=2a_0y+a_1$. So $(y_1/g)^2+h=0$ and put $z=y_1/g$. It is clear that $k(x,y)=k(x,z)$. The affine curve defined by the equation $z^2+h(x)$ is non-singular by Jacobian criterion. So this is the equation of a piece of the non-singular projective curve corresponding to $k(x,y)$. The computation of the genus of a hyperelliptic is standard, try any textbook (probably in Hartshorne as an exercise). I don't understand what do you mean by "more generally", and you ask too many questions in a comment:). –  user18119 Jul 13 '13 at 21:47

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