# Friday, September 15

:::{.remark}
Recall last time: for $f\in \CC(X)^*$, we set $v_D(f) = n$ where $\ro{f}{U} = g^n h$ where $h$ is a regular function on $U$ that is not identically zero on $D$.
Idea: how much does $f$ vanish at a generic point of $D$.
Correction from last time: what is $v_D(f)$ for $D = V(x) \subseteq \AA^2$ and $f(x, y) = {y^n \over x^p + y^q}$? 
One may try $f(x) = x^{-p}{y^n \over 1 + y^qx^{-p}} \da x^{-p} h(x, y)$ and compute $v_D(f) = -p$, but this doesn't work because $\lim_{x\to 0}h(x, y) = 0$.
So we not only need $h$ to not identically zero, but also that its limit isn't zero in the case in which the function is not defined.
For rational functions, use that $v_D(f/g) = v_D(f) - v_D(g)$.
:::

:::{.example title="?"}
We have 

- $v_{V(x)}(x^2) = 2$,
- $v_{V(x)}(1 + x^2) = 0$,
- $v_{V(y)}(y-x^2) = 0$

:::

:::{.remark}
Recall that we discussed linear equivalence, $\Cl(X)$, and $\Pic(X)$ -- these can both be explicitly computed for toric varieties.
A divisor $D$ is principal if $D = \div(f)$ for some rational function $f$.
We'll consider only torus-invariant divisors. 
For any ray $\rho \in \Sigma$, there will exist a corresponding torus-invariant divisor $D_ \rho$, which is the closure of the orbit corresponding to $\rho$.
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:::{.example title="?"}
Consider $\Sigma_{\PP^1}$ with rays \( \rho_1, \rho_2 \), then \( D_{ \rho_1} = \ts{ 0 } , D_{\rho_2} = \ts{ \infty } \).
For \( \PP^2 \) with coordinates \( [x_0: x_1: x_2] \) and rays \( \rho_0, \rho_1, \rho_2 \), we'll have $D_{\rho_i} = V(x_i)$, so e.g. $D_{ \rho_0} = V(x_0) \cong \PP^1$ with coordinates $[x_1: x_2]$.
Note that $\PP^2 \sm \qty{V(x_0) \union V(x_1) \union V(x_2)} \cong (\cstar)^2$.
:::

:::{.remark}
On a toric variety, we have a distinguished set of rational functions given by characters of $(\cstar)^n \actson X_{ \Sigma}$.
For $m\in M$, these are defined by 
\[
v_m: (\cstar)^n \to \cstar &\injects \CC \\
(x_1,\cdots, x_n) &\mapsto \prod x_i^{m_i}
.\]
What is $\div(v_m)$? 
The divisors appearing in $\div(f)$ are only those where $f$ has zeros or poles.
Note that $v_m$ has no poles since it is a regular function.
The only divisors appearing in $\div(v_m)$ are in fact toric divisors $D_{\rho_i}$ along which $v_m$ has zeros.
We have
\[
\div(v_m) = \sum_i v_{D_ { \rho_i}} D_{\rho_i}
.\]

:::

:::{.lemma title="?"}
Let $u_{\rho_i}$ be the primitive integral generator of the ray $\rho_i$, then
\[
v_{D_{ \rho_i}}(v_m) = \inp{ m}{u_{ \rho_i} }
.\]
:::


:::{.example title="?"}
Consider $\PP^2$ with $\rho_i$ labeled in counterclockwise order in the fan.
So in particular $D_{\rho_0} = \PP^1_{[x_1: x_2]}$ where $x_0 \neq 0$.
Write $z^{1, 0} = x, z^{0, 1} = y$, and in coordinates on $\CC^2 \subseteq \PP^2$ where $x_2\neq 0$, we can define a rational map 
\[
\PP^2 &\rationalto \CC \qquad z^{1, 0} = x \\
[x_0: x_1: x_2] &\mapsto {x_0 \over x_2}
.\]
This is essentially the function $x$.
This has a zero along $x_0 = 0$ and a pole along $x_2 = 0$.
We now check $V_{D_{\rho_0}}(z^{1, 0}) = \inp{(1, 0)}{(1, 0)} = 1$, yielding a first order zero along $x=0$.
Similarly check $V_{\rho_2}(z^{1, 0}) = \inp{(1, 0)}{(-1, -1)} = -1$, yielding a first order pole along $x=0$.


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