# A kind of Stein factorization for non-proper morphisms

Let $S$ and $T$ be a noetherian connected normal scheme, say. Let $K$ (resp. $L$) be the function field of $S$ (resp. $T$). Assume that $char(L)=0$. Let $f: S\to T$ be a smooth surjective morphism. Is it true that $f$ factors as $$S\to T'\to T$$ where $S\to T'$ is smooth and surjective with geometrically connected generic fibre and $T'\to T$ is a finite morphism?

(A candidate for $T'$ could be the following scheme: Let $F$ be the algebraic closure of $L$ in $K$. Let $T'$ be the normalization of $T$ in $F$. Then the morphism $f$ factors through $T'$, but I do not know how to prove that the resulting morphism $S\to T'$ is smooth and surjective.)

My main interest is the case where $L$ is a number field and $T$ is a dense open subscheme of $Spec(O_L)$.

• I think you should try some examples on your own. For instance, what happens if $T$ is $\text{Spec}(\mathbb{Z})$ and $S$ is the maximal open subscheme of $\text{Spec}(\mathbb{Z}[x]/\langle x^3 + x + 1 \rangle)$ that is smooth over $T$? – Jason Starr Apr 29 '14 at 19:53
• Then I can take $T'=S$? – Sebastian Petersen Apr 30 '14 at 5:53
• "Then I can take $T'=S$?" No, you cannot. The induced morphism from $S$ to $T$ is not a finite morphism, and it does not have geometrically connected generic fiber. So you cannot take $T'$ equals $T$, and you also cannot take $T'$ equals $S$. – Jason Starr Apr 30 '14 at 12:01
• OK, now I get it. Maybe I should require $S\to T'$ dominant instead of surjective? I am thinking for a moment... – Sebastian Petersen Apr 30 '14 at 13:29