The Stacks project

Proof. Let $f = \pi \circ f'$ be the factorization of Lemma 37.53.1. Note that besides the conclusions of Lemma 37.53.1 we also have that $f'$ is separated (Schemes, Lemma 26.21.13) and finite type (Morphisms, Lemma 29.15.8). Hence $f'$ is proper. By Cohomology of Schemes, Proposition 30.19.1 we see that $f_*\mathcal{O}_ X$ is a coherent $\mathcal{O}_ S$-module. Hence we see that $\pi $ is finite, i.e., (2) holds.

This proves all but the most interesting assertion, namely that all the fibres of $f'$ are geometrically connected. It is clear from the discussion above that we may replace $S$ by $S'$, and we may therefore assume that $S$ is Noetherian, affine, $f : X \to S$ is proper, and $f_*\mathcal{O}_ X = \mathcal{O}_ S$. Let $s \in S$ be a point of $S$. We have to show that $X_ s$ is geometrically connected. By Lemma 37.53.3 we see that it suffices to show $X_ u$ is connected for every étale neighbourhood $(U, u) \to (S, s)$. We may assume $U$ is affine. Thus $U$ is Noetherian (Morphisms, Lemma 29.15.6), the base change $f_ U : X_ U \to U$ is proper (Morphisms, Lemma 29.41.5), and that also $(f_ U)_*\mathcal{O}_{X_ U} = \mathcal{O}_ U$ (Cohomology of Schemes, Lemma 30.5.2). Hence after replacing $(f : X \to S, s)$ by the base change $(f_ U : X_ U \to U, u)$ it suffices to prove that the fibre $X_ s$ is connected when $f_*\mathcal{O}_ X = \mathcal{O}_ S$. We can deduce this from Derived Categories of Schemes, Lemma 36.32.7 (by looking at idempotents in the structure sheaf of $X_ s$) but we will also give a direct argument below.

Namely, we apply the theorem on formal functions, more precisely Cohomology of Schemes, Lemma 30.20.7. It tells us that

where $X_ n$ is the $n$th infinitesimal neighbourhood of $X_ s$. Since the underlying topological space of $X_ n$ is equal to that of $X_ s$ we see that if $X_ s = T_1 \amalg T_2$ is a disjoint union of nonempty open and closed subschemes, 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 $\mathcal{O}^\wedge _{S, s}$ is a local ring. Thus the assumption was wrong, i.e., $X_ s$ is connected, and we win. $\square$