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Let $X = V(xy-zw) \subset \mathbb{P}^3$ (the variables are $x,y,z,w$). I know from various sources that $Pic(X) \cong \mathbb{Z} \oplus \mathbb{Z}$, where generators are the two lines $l_x = V(x,w) \subset X$ and $l_y = V(y,w)$. (see for example, Hartshorne, II, Example 6.6.1,2).

I can show that $Pic(X)$ is generated by $\{l_x,l_y\}$, but I'm having trouble at showing that the kernel of $\mathbb Z l_x \oplus \mathbb Z l_y \to Pic(X)$ is trivial. It seems to me that to show this it is equivalent to show that for any $f \in K(X)$, the associated divisor $(f)$ has coefficient zero at prime divisors $l_x$ and $l_y$. In turn, this amounts to showing that $v_{l_x}(f) = 0 = v_{l_y}(f)$.

However, I cannot understand why this would be true. More precisely, it seems that if I put $f = x/y$, then $f$ is a non-zero rational function on $X$, and $v_{l_x}(f) \geq 1$ since $f$ is zero on the whole $D(y)\cap l_x$, whence in the maximal ideal of $\mathcal O _{l_x}$.

Thanks for your help. Also I don't know to what extent this is a trivial/stupid question, so please be indulgent if that would be the case..

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No, you don't need to show that, for every $f\in K(X)$, $\operatorname{div}(f)$ has $0$ coefficient at $l_x$ and $l_y$. What you need to show is that, if $\operatorname{div}(f)\in\mathbb{Z}l_x\oplus\mathbb{Z}l_y$, then $\operatorname{div}(f)=0$. It is slightly different.

In fact, $\operatorname{div}(x/y)=V(x,w)+V(x,z)\notin\mathbb{Z}l_x\oplus\mathbb{Z}l_y$.

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  • $\begingroup$ Indeed, you are right. Thank you so much! $\endgroup$
    – Olivier
    May 12, 2015 at 7:13

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