# Special fiber is geometrically connected if the generic fiber is under properness assumption?

I am a beginner in algebraic geometry, and want to understand the following proof towards criterion of Néron–Ogg–Shafarevich for abelian varieties:

Why can Zariski's connectedness theorem imply fiber is geometric connected? I think this requires $$f_*O_X=O_{Y}$$ and follows by Stein decomposition. Is there another proof or generalization for this lemma?

• You should always cite the documents that you use (namely, this one). – Watson Nov 27 '18 at 7:36
• @Watson thank you! The picture contains the whole part of the proof, and I am only confused at the last sentence about connectedness. But anyway, it's good to add a reference. – sawdada Nov 27 '18 at 7:41

Let $$Y = Spec(R)$$. Let us consider the map induced by $$f$$

$$f : \mathcal{O}_{Y} \rightarrow f_*\mathcal{O}_X$$

Since $$X$$ is proper over $$R$$, we get that $$M := f_*\mathcal{O}_X$$ is a finite module over $$R$$. Note that $$M$$ is also a reduced $$R-$$algebra. Consider the natural base change map $$\varphi^0(y)$$ for $$y \in Y$$ not necessarily closed point

$$\varphi^0(y) : R^0f_*\mathcal{O}_X = f_*\mathcal{O}_X \otimes k(y)\rightarrow H^0(X_y, \mathcal{O}_{X_y})$$

where $$X_y$$ is the fiber over the point $$y$$. For $$y = Spec(k)$$, the map $$Spec(k) \rightarrow Spec(R)$$ is a flat map and hence we know this map to be an isomorphism by flat base change theorem. Also note that $$X_y$$ is geometrically connected by hypothesis and hence $$H^0(X_k, \mathcal{O}_{X_k}) = k$$. Thus we get that the $$R-$$module $$M$$ satisifes $$M \otimes k \cong k$$. More geometrically, this says $$Spec(M) \otimes Spec(k) \cong Spec(k)$$. That is the map $$Spec(M) \rightarrow Spec(R)$$ is a normalization map since both have same function fields. Since $$R$$ being a dvr is already a normal ring, hence $$Spec(M) \xrightarrow{\sim} Spec(R)$$ is an isomorphism. That is $$M \cong R$$. Thus one has

$$\mathcal{O}_Y \rightarrow f_*\mathcal{O}_X$$

is an isomorphism. That is what was required.

All this is done in the following lemma. https://stacks.math.columbia.edu/tag/0AY8

• Thank you, that's helpful. – sawdada Nov 27 '18 at 15:48