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This is a follow-up (of sorts) to this question.

Let $f : X \to T$ be a proper morphism of schemes. Then the notion of a relative ample (or $f$-ample) line bundle can be defined in several equivalent ways; Lazarsfeld's book gives the clearest discussion I have found. The condition that $L$ is $f$-ample is equivalent to each of the following:

  1. $L\big\vert_{f^{-1}(t)}$ is ample on $f^{-1}(t)$ for all $t \in T$.

  2. If $D$ is the divisor class corresponding to $L$, then $D^{\dim V}\cdot V > 0$ for each subvariety of $X$ which maps to a point in $T$.

  3. Given an ample line bundle $A$ on $T$, $L\otimes f^*A^{\otimes m}$ is ample on $X$ for sufficiently large positive $m$.

My question is: are these statements still equivalent if we only know, a priori, that $X$ is an algebraic space? I suspect they do, and that the proofs in Lazarsfeld basically go through the same, but an independent reference would be welcome if one exists.

Edit: I'm not sure whether it matters, but I'm only really interested in the case where $X$ is smooth (i.e. a Moishezon manifold).

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    $\begingroup$ Do you mean 1 and 2 are always equivalent, and 3 provides another characterization when such an $A$ exists? Anyway, try using the theorem of formal functions for proper maps of algebraic spaces to deduce whatever you want from the proper infinitesimal fibers (which certainly are schemes in the presence of a fibral ample line bundle). That should let you bootstrap once you settle the equivalence of 1 and 2 when $T=$Spec($k$) for fields $k$. $\endgroup$
    – Marguax
    Sep 19, 2013 at 11:07
  • $\begingroup$ @Marguax: Yes, the third condition requires $T$ to be projective of course; in this case, it is equivalent. Also, thanks for the suggestion! $\endgroup$
    – Rhys Davies
    Sep 19, 2013 at 11:17

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