# Monday, September 14

:::{.exercise}
\hfill

a. Show that for all divisors $D$, $\mathcal{L}(D)$ is a $k\dash$vector space.
b. If $D\sim D'$, then $\mathcal{L}(D) \sim \mathcal{L}(D')$.
:::

:::{.lemma}
Let $A\leq  B$ in $\Div(K)$.
Then

a. $\mathcal{L}(A) \subset_k \mathcal{L}(B)$.
b. We have $\dim \mathcal{L}(B) / \mathcal{L}(A) \leq \deg(B) - \deg(A) = \deg(B-A)$.
:::


:::{.corollary title="?"}
For $D\in \Div(K)$,

a. If $\deg D< 0$ then $\mathcal{L}(0) = (0)$.
b. If $\deg D \geq 0$ then $\dim \mathcal{L}(D) \leq \deg D + 1$.
:::

:::{.proof title="of corollary"}
\hfill

a. A divisor of negative degree is not linearly equivalent to an effective divisor.
b. WLOG $D$ is effective, so write $D = P_1 + \cdots + P_r$.

  Using the fact that $\dim \mathcal{L}(0) = 1$, apply the previous lemma $r$ times.
:::

:::{.proof title="of lemma"}
**Part a**:

Easily reduce to the case $B = A + P$ for some place $P$ not necessarily of degree 1.


**Part b**:

> Try to work this out when the place is degree 1.

Choose $t\in K$ with $v_p(t) = v_p(B) = v_p(A) + 1$.
For $f\in \mathcal{L}(B)$ we have 
\[  
v_p(f) \geq -v_p(B) = -v_p(t) \implies ft\in R_p
.\]

We can thus define a $k\dash$linear map into the residue field $k(p)$:
\[  
\psi: \mathcal{L}(B) &\to k(P) = R_p / \mfm_p \\
f &\mapsto ft \mod \mfm_p
.\]

We can compute the kernel, 
\[  
\ker \psi = \ts{  f\in \mathcal{L}(B) \st v_p(t) \geq -v_p(t) + 1 = -v_p(A) } = \mathcal{L}(A)
.\]
thus $\dim \mathcal{L}(B) / \mathcal{L}(A) \leq [k(P) : k]$ since this map descends to an injection.

:::

:::{.remark}
For $P \in \Sigma(K/k)$ with residue field $k_p$ and $[k_p: k] = d$ and $K_p$ is the completion of $K$ with respect to $\abs{\wait}_P$, then $K_P \cong k_p((t))$.

This comes from the structure theory of complete discretely valued fields.
:::

:::{.exercise title="Important"}
If $D\in \Div k(t)$, show that
\[  
\mathcal{L}(D) = 
\begin{cases}
\deg(D) + 1 & \deg(D) \geq 0 \\
0           & \deg(D) < 0
\end{cases}
.\]
:::

The *Riemann-Roch Problem* asks for good upper and lower bounds on $\mathcal{L}(D)$ and especially on $\mathcal{L}(nD)$ as a function of $n$.
The last corollary gives an upper bounds on $\mathcal{L}(D)$ in terms of $\deg(D)$:
\[  
\deg(D) \geq 0 &\implies \deg(D) - \mathcal{L}(D) \geq -1
.\]

:::{.proposition title="?"}
There exists some constant $\gamma = \gamma(K/k) \in \ZZ$ such that for all $A\in \Div(K)$ we have
\[  
\deg A - \mcl(A) \leq \gamma
.\]

:::