Sections of morphisms of schemes up to a finite morphism

Let $f:X\longrightarrow S$ be a flat projective morphism of regular integral noetherian schemes such that that the generic fibre $X_\eta\longrightarrow K(S)$ is a smooth projective connected curve over $K(S)$.

Assumptions. For simplicity, let us assume that $S$ is one-dimensional, i.e., $S$ is a connected Dedekind scheme and that $K(S)$ is perfect. (I don't think we need these assumptions though.)

Let $\pi:S^\prime \longrightarrow S$ be a finite morphism of Dedekind schemes and suppose that there is a morphism $P:S^\prime\longrightarrow X$ of $S$-schemes, i.e., we have that $\pi = f\circ P$.

Question. Does the morphism $f:X\longrightarrow S$ factor through a flat projective morphism $X\longrightarrow S^\prime$ ?

Of course, this is not going to be true in general. So here's a better question.

Less precise question. If this is not true, can we ''explain'' when $f$ factors through $\pi$ and when not?

Unprecise question. Does a situation like the one above arise ''often''? Any references that might help?

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If $f \colon X \to S$ has connected fibers, and $S' \to S$ has not, $X \to S$ cannot factor over $S' \to S$ or do I miss the point here? –  Holger Partsch Mar 20 '11 at 18:12

Consider the Stein factorization $X\to S_1\to S$ of $f$ (thus $S_1=\mathrm{Spec}(f_*\mathcal{O}_X)$). Then $S_1\to S$ is finite, and since $X$ is normal, I think $S_1$ must be the normalization of $S$ in $K(X)$. Then $f$ factors through $S'$ if and only if $S_1\to S$ does, which in turn is equivalent to Karl's condition $K(S')\subset K(X)$.

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Geometrically, you can consider the fiber product $X' = X\times_S S'$. Then the map $X \to S$ factors through $S' \to S$ if and only if the map $X' \to X$ (which is a finite morphism obtained from $S' \to S$ by a base change) has a section.

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Based on the above comment of Holger Partsch, you can't factor a lot of stuff. You can do silly things like:

A necessary condition for factoring $f$ through a given $S'$ is that you can embed: $$K(S) \subseteq K(S') \subseteq K(X).$$

If you are willing to blow-up $X$ to resolve indeterminacies, then this might be the only obstruction in the geometric setting anyway (assuming $S$ and $S'$ are normal).

On the other hand THIS question, and the numerous excellent answers, seems like it might be very relevant depending on your context.

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