Klein quartic (divisor)

Question

help me please with this problem (it's taken from book and I don't know how to prove)

Klein quartic is the curve $ X\subset \mathbb{P}^{2}$ with the homogeneous equation:

$x^{3}y+y^{3}z+z^{3}x=0$

Prove that the divisor of the linear form $z$ on $X$ is $(z)=Q_{1}+3Q_{2}$ where $Q_{1}=(1:0:0)$ and $Q_{2}=(0:1:0)$

Thanks!

Book

Answer 1

I couldn't find that book in our library so I need to guess the meaning of the divisor of a form on a curve. I assume that it means that each point in the intersection of the projective curves $$ X^3Y+Y^3Z+Z^3X=0\qquad\text{and}\qquad Z=0 $$ is multiplied by some kind of intersection multiplicity. I further guess that this multiplicity is defined by calculating the (exponential) valuation of the defining function of the latter curve at a point of intersection (or, more precisely, in the local ring).

So here we can see $Q_1=(1:0:0)$ as the origin $u=v=0$, where $u=Y/X$ and $v=Z/X$ are the affine coordinates on this piece. This affine piece of the Klein quartic is defined by the equation $P(u,v)=u+u^3v+v^3=0$. Also, in this piece $X=1$, so $Z=v$ (this is admittedly shady - hopefully an algebraic geometer can comment eventually). The partial derivative $\partial P/\partial u=1+3u^2v+3v^2$ does not vanish at $Q_1$, so we can use $v-v(Q_1)=v-0=v$ as a local parameter. So if we denote by $\nu_1$ the discrete valuation at $Q_1$, we have $\nu_1(v)=1$. Therefore the point $Q_1$ appears with multiplicity 1.

With $Q_2$ we need to go to the affine piece $Y=1$, and use affine coordinates $r=X/Y$ and $s=Z/Y$. The equation of the quartic then becomes $R(r,s)=r^3+s+s^3r=0$. As above we see that $r$ is a local variable at $Q_2$. If $\nu_2$ is the corresponding valuation, we thus have $\nu_2(r)=1$. The form $Z$ agrees with the function $s$ on this affine plane, so we need to compute $\nu_2(s)$. Because $s$ vanishes at $Q_2$, we know that $\nu_2(s)\ge 1$. The equation of the curve gives us the relation $$ 3=\nu_2(r^3)=\nu_2(s+s^3r). $$ Here $\nu_2(s^3r)=3\cdot\nu_2(s)+\nu_2(r)\ge 4$, so by the non-archimedean triangle inequality we must have $\nu_2(s)=3$.

All this suggests that the divisor you are looking for really is, indeed, $1\cdot Q_1+3\cdot Q_2$. Unfortunately it is impossible for me to check that this calculation is a valid answer in view of what has been covered in your book up to this point.