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Let $\pi: X \to S$ be a morphism of schemes. I will say $\pi$ is "pseudoconnected" if $\mathcal{O}_S \to \pi_* \mathcal{O}_X$ is an isomorphism (this is not standard language).

  1. If $\pi$ is proper with connected fibers, can we deduce that $\pi$ is pseudoconnected? I think this follows from Zariski's main theorem (the version that says a proper morphism is a pseudoconnected morphism followed by a finite morphism) but I am squeamish because Hartshorne doesn't explicitly say this (even for projective morphisms).

  2. What if instead $\pi$ is proper with geometrically connected fibers?

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1 Answer 1

up vote 5 down vote accepted
  1. Non. Let $X\to S$ correspond to a finite extension of fields $L/K$ of degree $>1$. Then the unique fiber is connected, but $O_S=K\to \pi_*O_X=L$ is not an isomorphism.

  2. Again non. Take $L/K$ purely inseparable in the above example.

(A sufficient condition when $S$ is reduced is $\pi$ proper with geometrically reduced and geometrically connected fibers.)

Edit A more geometric example. Let $S$ be a curve over $\mathbb C$ with a cusp and normalization $\rho : S'\to S$. Let $X=\mathbb P^1 \times S'$ with the natural (projective) morphism $\pi$ to $S$. Then $\pi_*(O_X)=\rho_*(O_{S'})\ne O_S$, while the fibers are geometrically connected.

The above sufficient condition is too strong. It implies that for any base change $T\to S$, the direct image of $O_{X\times_S T}$ is equal to $O_T$ (in other words, the isomorphism $O_S\to \pi_*O_X$ holds universally). A weaker condition is the following: suppose that $S$ is Noetherian, integral and normal, that the generic fiber of $X\to S$ is geometrically integral, and $X$ is integral. Then $O_S\to \pi_*O_X$ is an isomorphism. This is because $\mathrm{Spec}(\pi_*O_X)\to S$ is finite and birational.

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Thank you for the edit! I felt silly re-asking the question asking for a geometric example, so it was nice that you anticipated my needs! –  user29743 May 14 '12 at 14:07

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