# Lecture 8: Riemann-Roch Spaces (Part 2) Recall the proposition we ended with last time: :::{.proposition title="?" ref="prop:deg_bounded_above"} There exists a $\delta = \delta(K/k) \in \ZZ$ such that for all $A\in \Div K$, we have \[ \deg A - \ell(A) \leq \delta .\] ::: :::{.exercise title="?"} This proposition is enough to show the existence of rational functions whose polar divisor has as its support any finite subset $S \subset \Sigma(K/k)$. ::: Most of the lecture will be the proof of this statement. ## Proof of Upper Bound Rewriting \cref{lemma:divisor_order_to_subspaces} yields \[ A_2, A_2 \in \Div K,\, A_1 \leq A_2 \implies \deg A_1 - \ell(A_1) \leq \deg A_2 - \ell(A_2) .\] We now proceed to prove \cref{prop:deg_bounded_above} in several steps. ### Step 1 Choose an $x\in K\sm k$ and set $B\da (x)_-$. :::{.claim} There exists a $C\geq 0$ such that for all $n\geq 0$, \[ \ell(nB + C) \geq (n+1) \deg B .\] ::: So we give ourselves a certain effective divisor: the divisor of poles of an arbitrary nonconstant element. We can then get a preliminary asymptotic lower bound, not on the same Riemann-Roch space, but on a new one after augmenting the space by some fixed effective divisor $C$. :::{.proof title="?"} Since $K/k(x)$ has finite degree, let $u_1,\cdots, u_d$ be a basis for $K$ consisting of finitely many rational functions. Note that $d = [K: k(x)]$, and is also equal to $\deg B$ since $B$ was a divisor of poles. Noting that the divisor groups are free commutative groups, so taking any finite number of elements in $\bigoplus \ZZ$, we can find an element that is less than or equal to all of them. Thus we can choose a $C\geq 0$ such that \[ (u_i) \geq -C && \forall 1\leq i \leq d .\] Since the $u_i$ are $k(x)\dash$linearly independent in $K$, the functions $\ts{x^i u_j \st\, 0\leq i \leq n,\,\, 1\leq j \leq d}$ are $k\dash$linearly independent, since any $k\dash$linear relation would immediately yield a $k(x)\dash$linear relation among the $u_i$. :::{.exercise title="?"} If $f_i\in \mathcal{L}(D_i)$, so the poles of $f$ are no worse than $D_i$, then the poles of $f_1 f_2$ are bounded by $D_1 + D_2$ and thus $f_1 f_2 \in \mathcal{L}(D_1 + D_2)$. ::: Now we can note that there are $(n+1)d = \deg B$ many elements here, and moreover, these all lie in $\mathcal{L}(nB + C)$ since each $(u_j) \geq -C$ and $(x) \geq - B$ and $i\leq n$. From this we can conclude \[ \ell(nB+c) \geq (n+1) d = (n+1) \deg B .\] ::: ### Step 2 We'll now show that throwing in the fixed divisor $C$ can't increase the Riemann-Roch space that much, and in fact \[ \ell(nB + C) \leq \ell(nB) + \deg C ,\] and so we get a bound \[ \ell(nB) &\geq \ell(nB + C) - \deg C \\ &\geq (n+1) \deg B - \deg C \\ &= \deg(nB) + \qty{ [K:k(x)] - \deg C} \\ &\da \deg(nB) \pm \gamma ,\] which shows that \[ \forall n\geq 0,\, \, \deg(nB) - \ell(nB) \leq \gamma \label{eq:step_2_bound} .\] A problem here is that $\gamma$ depends upon everything that we've done so far, and this inequality only holds for multiples of a fixed divisor (an infinite ray emanating from $B$). ### Step 3 :::{.claim} For all $A\in \Div K$, there exist $A_1, D \in \Div K$ and $n\geq 0$ such that $A \leq A_1$, $A_1 \sim D$, and $D\leq nB$. I.e. although it can't literally be true that $A \leq nB$, it will be up to linear equivalence. ::: To see this, set $A_1 \da \max(A, 0)$. Using the bound from \cref{eq:step_2_bound}, for $n\gg 0$ we have \[ \ell(nB - A_1) &\geq \ell(nB) - \deg A_1 \\ &\geq \deg(n B) - \gamma - \deg A_1 \\ &> 0 ,\] and so there exists a $z\in \mathcal{L}(nB - A_1)\nonzero$, a nontrivial element in the linear system. :::{.remark} The first inequality is an application of our lemma because $A_1$ is effective, which was the point of this maneuver. I.e., in order to get from $nB - A_1$ to $nB$, we added $A_1$, which can only increase the dimension of the space by at most $\deg A_1$. Finally, in the last inequality, we use the fact that $B$ has positive degree since it's a divisor of poles of a nonconstant rational function, and the remaining terms don't depend on $n$, so we can make $\deg(nB)$ arbitrarily large. ::: So now set $D \da A_1 - (z)$, then $A_1 \sim D$ and since it's in the linear system, \[ (z) \geq - (nB - A_1) = A_1 - nB \] so $-(z) \leq nB - A_1$ and by adding $A_1$ to both sides, we obtain \[ 0 = A_1 - (z) \leq nB .\] What have we shown? For any divisor $D$, we can make it less than $nB$ for some $n$, up to linear equivalence. ### Step 4 Finally, for $A\in \Div K$, choose $A_1, D$ as in the previous step, so $A\leq A_1 \sim D \leq nB$. Then \[ \deg A - \ell(A) &\leq \deg(A_1) - \ell(A) && \text{using } A\leq A_1\\ &= \deg(D) - \ell(D) && \text{changing within linear equivalence class}\\\ &\leq \deg(nB) - \ell(nB) \\ &\leq \gamma .\] $\qed$ ## Genus :::{.definition title="Genus (Important!)"} The **genus** of $K/k$ is defined as \[ g \da \max_{A\in \Div K}(\deg(A) - \ell(A) + 1) .\] This exists by the \cref{prop:deg_bounded_above}, since this set is bounded above. ::: :::{.exercise title="?"} Show that $g\geq 0$ always and \[ g(k(t)/k) = 0 .\] ::: :::{.remark} Note that if the $+1$ is mostly a correction factor to match up with the topological genus of $\PP^1_{/\CC}$. That the genus is non-negative should come from the lower bound we had from before. It turns out that over $k=\CC$, this genus will agree on the nose with the topological genus of the corresponding compact Riemann surface. ::: :::{.theorem title="Riemann's Inequality" def="thm:riemann_inequality"} If $K/k$ is a function field of genus $g$, a. For all $A\in \Div K$, \[ \ell(A) \geq \deg(A) + 1 - g .\] b. There exists a $c = c(K) \in \ZZ$ such that for all $A \in \Div K$, \[ \deg(A) \geq c \implies \ell(A) = \deg(A) - g + 1 .\] ::: :::{.remark} This says that the dimension of the linear system is very close to the degree of the corresponding divisor, and is only off by a constant factor $g$. Part (a) is literally just a rearrangement of the definition of the genus. Part (b) says that if you assume $A$ has sufficiently large degree, this upper bound becomes an equality. ::: :::{.proof title="of b"} By the definition of $g$, since it is a maximum there exists an $A_0$ such that \[ g = \deg(A_0) - \ell(A_0) + 1 .\] Set $c\da \deg(A_0) + g$. Then if $\deg(A) \geq c$, we have \[ \ell(A - A_0) &\geq \deg(A-A_0) - g + 1 \\ &\geq c - \deg(A_0) - g + 1 \\ &= 1 ,\] so there exists a $z\in \mathcal{L}(A- A_0)\nonzero$ since the dimension is at least 1. Now set $A' \da A + (z)$, and note that $A' \geq A_0$. Thus \[ \deg(A) - \ell(A) &= \deg(A') - \ell(A') \\ &\geq \deg(A_0) - \ell(A_0) && \text{by the lemma}\\ &=g-1 .\] By maximality of the genus, we have $\deg(A) - \ell(A) \leq g-1$, which forces equality ::: Next up: how to we make this inequality into an equality? It turns out that there is some different divisor $D'$ and we can subtract off $\ell(D')$, and that will be the Riemann-Roch theorem.