Htykuut

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?

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