The Stacks project

Proof. Let $f = \pi \circ f'$ be the factorization of Lemma 76.36.1. Note that besides the conclusions of Lemma 76.36.1 we also have that $f'$ is separated (Morphisms of Spaces, Lemma 67.4.10) and finite type (Morphisms of Spaces, Lemma 67.23.6). Hence $f'$ is proper. By Cohomology of Spaces, Lemma 69.20.2 we see that $f_*\mathcal{O}_ X$ is a coherent $\mathcal{O}_ Y$-module. Hence we see that $\pi$ is finite, i.e., (2) holds.

This proves all but the most interesting assertion, namely that the geometric fibres of $f'$ are connected. It is clear from the discussion above that we may replace $Y$ by $Y'$. Then $Y$ is locally Noetherian, $f : X \to Y$ is proper, and $f_*\mathcal{O}_ X = \mathcal{O}_ Y$. Let $\overline{y}$ be a geometric point of $Y$. At this point we apply the theorem on formal functions, more precisely Cohomology of Spaces, Lemma 69.22.7. It tells us that

where $X_ n = \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, \overline{y}}/\mathfrak m_{\overline{y}}^ n) \times _ Y X$. Note that $X_1 = X_{\overline{y}} \to X_ n$ is a (finite order) thickening and hence the underlying topological space of $X_ n$ is equal to that of $X_{\overline{y}}$. Thus, if $X_{\overline{y}} = T_1 \amalg T_2$ is a disjoint union of nonempty open and closed subspaces, then similarly $X_ n = T_{1, n} \amalg T_{2, n}$ for all $n$. And this in turn means $H^0(X_ n, \mathcal{O}_{X_ n})$ contains a nontrivial idempotent $e_{1, n}$, namely the function which is identically $1$ on $T_{1, n}$ and identically $0$ on $T_{2, n}$. It is clear that $e_{1, n + 1}$ restricts to $e_{1, n}$ on $X_ n$. Hence $e_1 = \mathop{\mathrm{lim}}\nolimits e_{1, n}$ is a nontrivial idempotent of the limit. This contradicts the fact that $\mathcal{O}^\wedge _{Y, \overline{y}}$ is a local ring. Thus the assumption was wrong, i.e., $X_{\overline{y}}$ is connected as desired. $\square$