Let $f: S \to B$ be a fibration from an integral surface $S$ to integral curve $B$.
Here I use following definitions:
A surface (resp. curve) is a $2$-dim (resp. $1$-dim) proper $k$ scheme over fixed field $k$.
Fibration has two properties:
-
$O_B = f_*O_S$
-
all fibers of $f$ are geometrically connected
Now I wan't to see why the property 1. is "stable" with respect to the generic fiber; namely if we benote by $\eta \in B$ the unique generic point of $B$ and $S_{\eta}:=f^{-1}(\eta)$ it's fiber and by $g: S_{\eta} \to \eta$ the corresponding induced map then I wan't to know why
$$O_{k(\eta)}= g_* O_{S_{\eta}}$$
holds?
Some remarks:
Note that as scheme $\eta= Spec(k(\eta))$.
Futhermore if we denote the canonical immersions $i_B: \eta \to B$ and $i_S: S_{\eta} \to S$.
My attempts:
Firstly, how is the structure sheaf of $S_{\eta}$ is concretely defined. My intuitive choice would be $O_{S_{\eta}}= O_S \otimes_k k(\eta)$. Is this correct?
Now the problem (in order to verify $O_{k(\eta)}= g_* O_{S_{\eta}}$):
Let $U$ be open in $\eta$ so wlog $U= \eta$.
I have to show that $g_* O_{S_{\eta}}(\eta)=k(\eta)$(= $O_{k(\eta)}(\eta)$)
Then I proceed using definitions and that $i_B \circ g = f \circ i_S$ by construction:
$$g_* O_{S_{\eta}}(\eta)= g_* O_{S_{\eta}}(i_b^{-1}(\eta))= {i_B} _* g_*O_{S_{\eta}}(\eta)=f_* {i_S} _* O_{S_{\eta}}(\eta)= {i_S} _* O_{S_{\eta}}(S_{\eta})= {i_S} _* O_S \otimes_k k(\eta)(S_{\eta})= O_S(S) \otimes_k k(\eta)$$
And exactly this is the problem: I don't see why $O_S(S) \otimes_k k(\eta)= k(\eta)$.
Does anybody see where is the error in my reasonings?
Best Answer
This is much easier than you're making it. As $\eta$ is a point, it suffices to show that $\mathcal{O}_{S_\eta}(S_\eta)=k(\eta)$. As proper and fibers geometrically connected is preserved under fiber product, the map $g:S_\eta\to \eta$ is proper, so all global sections of $\mathcal{O}_{S_\eta}$ are constant in the fiber direction on each connected component. This means that the global sections are a vector space of dimension equal to the number of geometric connected components, but $S_\eta$ is geometrically connected. Thus $\mathcal{O}_{S_\eta}(S_\eta)=k(\eta)$ and the proof is finished.