# Wednesday, September 20

:::{.remark}
Recall: $D = \sum a_i D_i$ is Cartier iff $\exists: \phi: \abs{ \Sigma}\to \RR$ a piecewise linear function with $\phi(u_i) = a_i$ where $D_i$ is associated to ray $i$ with primitive generator $\rho_i$.
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:::{.example title="?"}
Consider the cone on $\rho_1 = (1, 0)$ and $\rho_2 = (1, 2)$, which yields a singular toric variety $X_ \Sigma=\AA^2/\ZZ/2$ with $\Cl(X_ \Sigma) = \ZZ/2$.
Are $D_1, D_2$ Cartier? What is $\Pic(X_ \Sigma)$?

Write $D_1 = a_1D_1 + a_2 D_2$ with $a_1 = 1, a_2 = 0$, and similarly $a_1 = 0, a_2=1$ for $D_2$.
So we need $\phi(1, 0) = -1$ and $\phi(1, 2) = 0$ with $\phi(N) \subseteq \ZZ$.
Write $\phi = \inp{\wait}{?}$ and take $? = (-1, t)$ for any $t$ to get the former, and for the latter take $t=1/2$.
Thus set $\phi(x, y) = -x + {1\over 2}y$ for $D_1$; this is defined on $\sigma_1$.
This is not an integral map though, since $(0, 1) \mapsto {1\over 2}$, so there are no such integral functions.

Similarly, for $D_2$, we want $\phi(1, 0) = 0$ and $\phi(1, 2) = -1$.
Require $\phi = \inp{\wait}{(a, b)}$ and note $(1, 0)\cdot(a, b) = 0 \implies a=0$ and $(1, 2) \cdot (a, b) = -1\implies 1a+2b = 1 \implies b = -{1\over 2}$.
For the same reason as above, this is not Cartier.
Note that $2D_2$ is Cartier, since $\phi(x, y) = -x + y$.

For affine toric varieties, all Cartier divisors are principal since there is just one cone and any piecewise linear function with one piece (on the one cone) is simply linear.
So $\Pic(X_ \Sigma) = 0$.
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## Quotients in algebraic geometry

:::{.remark}
Note we can write $\PP^n = \CC^{n+1}\smts{0}/\cstar$; more generally any toric variety can be written as $X_ \Sigma \sm Z \gitquot G$ for some subset $Z$ and some group $G$.
Assuming $\gens{\Sigma(1)}_\RR = N_\RR$, we have a SES 
\[
M \injects \bigoplus _{\rho\in \Sigma(1)} \ZZ D_{\rho} \surjects \Cl(X_ \Sigma) 
,\]
and applying the contravariant $\zmod(\wait, \cstar)$ we obtain
\[
G \da \Hom_\ZZ(\Cl(X_ \Sigma), \cstar) \injects (\ctar)^{\size \Sigma(1)} \surjects (\cstar)^n = T_N
\]
which is exact since $\cstar$ is divisible.
Thus $T_N \cong (\cstar)^N / G$ for $N\da \size \Sigma(1)$.
Note that in the case of $\PP^n$, we have $\Hom_\ZZ(\Cl(\PP^n), \cstar) \cong \Hom_\ZZ(\ZZ, \cstar) \cong \cstar$.
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:::{.remark}
How to find equations for $Z \subseteq \CC^N$ for $N = \size \Sigma(1)$: e.g. for $\PP^2$, we have $\CC^3$ with coordinates $x_1,x_2,x_3$.
For each cone $\sigma$, we get one equation $\prod_{\rho \not\in \sigma} x_{ \rho}$; intersect all of these equations.
So on the fan for $\PP^2$, $\sigma_1\leadsto x_3=0, \sigma_2 \leadsto x_1=0, \sigma_3 \leadsto x_2 = 0$.
Thus $Z = \ts{0} \subseteq \CC^3$ for $\PP^2$ in this case.
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