Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
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
add comment

3 Answers

up vote 3 down vote accepted

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 $K(S')\subset K(X)$.

share|improve this answer
add comment

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.