21
$\begingroup$

I need to calculate the genus of the Fermat Curve, and I'd like to be reviewed on what I have done so far; I'm not secure of my argumentation.

Such curve is defined as the zero locus $$X=\{[x:y:z]\in\mathbb{P}^2:x^d+y^d+z^d=0\}$$ It is defined by the homogeneous polynomial $F(x,y,z)=x^d+y^d+z^d$ which is obviously non-singular, therefore $X$ is a smooth projective plane curve. Now, I consider the projection $$ \begin{array}{cccc} \pi:&X&\longrightarrow&\mathbb{P}^1\\ &[x:y:z]&\longmapsto&[x:y] \end{array} $$ which is a well-defined holomorphic function. I need to calculate $\pi$'s degree, and for this I must calculate the sum of the multiplicities of the set $\{p:p\in\pi^{-1}(y)\}$, for any $y$ in $\pi(X)$; since the degree is constant, any such $y$ will do.

We note that, from Miranda's Riemann Surfaces and Algebraic Curves' Lemma 4.6 at page 46, that $\pi$ has a branch point at $p\in X$ if and only if $\frac{\partial F}{\partial z}(p)=0$

Having this in mind, to calculate the degree, I chose $[1:0]\in\mathbb{P}^1$; and for this one, $$\pi^{-1}([1:0])=\{[1:0:z]:1+z^d=0\}$$ Now I can see clearly that $|\pi^{-1}([1:0])|=d$ (the d-th roots of $-1$), and each one has multiplicity 1 (because they are not ramification points, as the above mentioned lemma states); therefore the degree of $\pi$ is $d$.

Now from the Lemma again, the ramification points of $\pi$ are those that $z=0$, that is, the points $\{[x:y:0]\in\mathbb{P}^2:x^d+y^d=1\}$ In homogeneous coordinates those points are of the form $[1:y:0]$, where $y$ is a d-th roots of $-1$ and from this we conclude that $\pi$ has $d$ points of ramification, and those are $$ \left[ 1:e^{\dfrac{2\pi i k}{d}}:0\right] $$

To calculate the multiplicity of each one of these points I used the following argument, which I'm not sure if it is correct: for a point $[x:y]\in\mathbb{P}^1$ such that $x^d+y^d=0$, $\pi^{-1}([x:y])=\{[x:y:0]\}$ and therefore $|\pi^{-1}([x:y])|=1$. Since the degree of $\pi$ is $d$, $d$ is also the multiplicity of every point of ramification of $\pi$.

What follows noy is applying Hurwitz's formula (and noting that $g(\mathbb{P}^1)=0$ and a very easy calculation to conclude that $g(x)=\dfrac{(d-1)(d-2)}{2}$.

Now I must apologize for the long, long text, but I will be much grateful for any correction made in my argumentation!

$\endgroup$
  • 9
    $\begingroup$ Why the downvote? There's research effort, I believe it is quite clear and also very useful question. $\endgroup$ – Marra Apr 26 '13 at 21:47
5
$\begingroup$

When you discuss ramification points of $\pi$, the equation should be $x^d + y^d = 0$, not $x^d + y^d = 1$.

Also, whenever you wrote about $d$th roots of unity, you needed the $d$th roots of $-1$.

The rest seems right. Yes, the number of preimages of the ramification points is all that you need in order to apply Hurwitz's formula, it is not necessary to find the degree of each preimage.

$\endgroup$
  • $\begingroup$ When I meant $x^d+y^d=1$ it was a mistake of mine, I really meant $x^d+y^d=0$. I have edited the mistake about the roots of $-1$. Also, didn't you mean that I don't have to find the multiplicity (instead of degree) of every preimage of the ramification point? Thanks for the answer! I spent a lot of time and effort on this! :) $\endgroup$ – Marra Apr 26 '13 at 22:22
2
$\begingroup$

Another approach to calculate the multiplicities is to use Lemma 4.4 (page 45). We calculate the multiplicities of the points $[\alpha:1:0]$ and show they are $d$.

First we need a local representation of $F$ near $[\alpha:1:0]$. In the patch where $y\neq 0$, this corresponds to $(\alpha, 0)$ in $\mathbb C^2$ (with coordinates $x, z$), cut out by $x^d+1+z^d=0$. The $z$ partial vanishes at $(\alpha, 0)$ so by implicit function theorem, the chart near $(\alpha, 0)$ is projection to the $z$ axis, and there is a holomorphic function $g(z)$ with the inverse to projection is $z \mapsto (g(z),z)$. The map $F$ on the patch $y\neq 0$ sends $(x, z)$ to $x$. So the local representation of $F$ is $z \mapsto g(z)$.

Now we need to calculate the order of vanishing of $g'$ at $0$. We have $g(z)^d+1+z^d=0$ so after differentiating and manipulating, $$ g(z)^{d-1} \cdot g'(z) = -z^{d-1}. $$ Notice $g'(0)=0$ because $g(0)=\alpha$. It follows the order of vanishing of $g'$ at $0$ is $d-1$. Hence the multiplicity is $d$.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.