stein factorisation with connected fibres

Question

Let $X\rightarrow Y\rightarrow Z$ be a stein factorisation. If we know the the fibres of the composite morphism is connected, then wouldn't it imply that $Y=Z$?

All spaces above are integral varieties.

Answer 1

The main results of Stein factorisation are the following:

Lemma 1. Let $f \colon X \to Y$ be a proper morphism such that $f_* \mathcal O_X = \mathcal O_Y$. Then $f$ has geometrically connected fibres. $\square$

Lemma 2. Let $\phi \colon X \to Z$ be a finitely presented proper morphism. Then $\phi$ factors as $$X \stackrel f\to Y \stackrel g\to Z,$$ where $f_* \mathcal O_X = \mathcal O_Y$, and $g$ is finite. $\square$

Corollary. In Lemma 2, the map $f$ has geometrically connected fibres. $\square$

Remark. It is very tempting to ask to what extent the converse of Lemma 1 holds: if $\phi \colon X \to Z$ is a proper morphism with geometrically connected fibres, then does $\phi_* \mathcal O_X = \mathcal O_Z$ hold? It turns out that this is false in general. The best thing we can prove is that in the Stein factorisation $X \to Y \to Z$, the map $g$ is both finite and has geometrically connected fibres. This implies that it is radicial.

However, not every radicial map is an isomorphism. For example, $Z$ can be a cuspidal curve and $X$ its normalisation. Or $\phi \colon \mathbb P^1_{\mathbb F_p} \to \mathbb P^1_{\mathbb F_p}$ can be the Frobenius morphism, which is finite with geometrically connected fibres (but not geometrically reduced fibres), but induces a nontrivial extension of function fields!

The best positive result one can prove is the following:

Lemma. Suppose $\phi \colon X \to Z$ is a dominant proper morphism of $k$-varieties with geometrically connected fibres, and $Z$ is geometrically normal. If $\operatorname{char} k = 0$, then $\phi_* \mathcal O_X = \mathcal O_Z$.

Proof. Since $\phi_*$ commutes with flat base change (hence so does the formation of the Stein factorisation), we may assume $k$ is algebraically closed. If $X \to Y \to Z$ is the Stein factorisation, then $g \colon Y \to Z$ is a finite radicial morphism. Then the function field extension $K(Z) \to K(Y)$ is purely inseparable, hence an isomorphism. Since $Z$ is normal, this implies that $Y \to Z$ is an isomorphism. $\square$

Remark. The example $\operatorname{Frob} \colon \mathbb P^1_{\mathbb F_p} \to \mathbb P^1_{\mathbb F_p}$ shows that the result is false in characteristic $p > 0$.