Show $\operatorname{Pic}(X)=0$? Exercise $14.2$.Q Vakil’s notes

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I'm reading Vakil's notes and I'm struggling with the exercise $14.2$.Q. I've been able to prove everything except $\operatorname{Pic}(X)=0$ with
$$
X=\operatorname{Spec}\frac{k[x,y,z]}{(xy-z^2)}.
$$

I tried to find an affine covering where all the rings are UFDs, but without results. Also I tried to compute $\operatorname{H^1}(X,\mathcal{O}_X^{\times})$ using the Cech cohomology, but also here I don't know how to set my covering.

Some (maybe usefull) facts: this ring is normal and the class group of $X$ is $\mathbb{Z}/2\mathbb{Z}$.

Thank you in advance to those who can answer or suggest something.

Best Answer

This is example 6.11.3 in Hartshorne. In 6.11.2 he argues that by the proof of Prop 6.11, Cartier divisors correspond to certain Weil divisors: if a Cartier divisor is given by $(U_i, f_i),$ then one associates a Weil divisor $\sum\limits_Y v_Y(f_i) [Y]$ to it. Weil divisors one gets this way are locally principal, meaning their restrictions on some open covering are principal.

Also note that by 6.15, $\operatorname{Pic}X \simeq \operatorname{CaCl}X.$

Now, the generator of $\operatorname{Cl}X \simeq \mathbb{Z}/2\mathbb{Z} $ is given by $V(z,y)$, which is not locally principal because $p:=(z,y)$ is not locally principal in the local ring of the point $(0,0,0) =: [m]$. Indeed, $m/m^2$ is a vector space spanned by $x, y$ and $z$, and $p \otimes \frac{k[x,y,z]}{(xy-z^2)}_m$ is its subspace which contains $y$ and $z$, so $p$ can't be principal in $\frac{k[x,y,z]}{(xy-z^2)}_m$.

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