Let $X$ be a projective integral curve over algebraically closed $k$ and let $f: Y \to X$ be the normalization of $X$. Let $p \in X$ a closed point $q \in Y$ be a closed point contained in the preimage or fiber $f^{-1}(p)$ of $p$. Denote by $D_q= [q]$ the associated Weil divisor (a $0$-cycle of $Y$) to $q$.
I have to show that there exist some other $0$-cycle $Z$ which is disjoint from the fiber: $Z \cap f^{-1}(p)= \emptyset$ of $p$ rational equivalent to $[q]$: recall rational equivalent ( $[q]\sim Z$) means that $[q]=Z +[div(s)]$ for some rational function $s$ on $Y$.
In the hint is adviced to use Riemann-Roch (for curves). I have no idea how to exploit RR here. RR gives
$$\dim _{k}H^{0}(Y,{\mathcal {O}}_{Y}(D))-\dim _{k}H^{0}(Y,\omega_Y \otimes {\mathcal {O}}_{X}(-D))=\operatorname{deg}(D)+1-g_Y$$
for arbitrary divisor $D$ on $Y$, dualizing sheaf $\omega_Y$ und genus $g_Y=\dim_k H^1(Y,\mathcal {O}_Y)$.
Now I'm looking for a $0$-cycle $Z$ with property $D:= Z – [q]$ and $H^{0}(Y,{\mathcal {O}}_{Y}(D)) \neq 0$ contains a section $s$ with $[div(s)]=D$. I not understand why RR provides the existence of this $s$. Could somebody help me to understand how this RR hint has be to applied here? beside RR, how the normality of $Y$ comes into play?
Best Answer
Riemann Roch tells you that for divisors of sufficiently large degree, the global sections of $O_Y(D)$ have dimension $\deg(D)-g_Y+1$, and this lets you build sections with desired zero sets. In your case, consider a large positive divisor $D$ of support disjoint from your point $P$, and consider the sections of $O_Y(D)$ and $O_Y(D+P)$, viewing these as subsheaves of the fraction field $K(Y)$. As subsheaves of $K(Y)$ these are given by $$L(D)(U)=\{f\in K(X)|\text{div}(f)+D|_U\geq 0\}$$
By our sufficiently large hypothesis, there exists a global section of $L(D+P)$ that doesnt arise from $L(D)$, which after translating through the definitions gives a function $f$ in $K(X)$ which has an order $1$ pole at $P$, and all other poles contained in the support $D$, and zero set also disjoint from $P$. So by construction we have the divisor $Z=P+\text{div}(f)$ disjoint from $P$.
The normality here is critical, in that we needed our curve to be nonsingular to use Riemann Roch, and normal=nonsingular in dimension $1$, so by normalising, we resolved the singularities of $X$.