There are plenty of sources discussing dual abelian varieties, but I'm looking for a reference that discusses the construction and properties of the dual abelian scheme. I'm willing to accept general theory that the Picard scheme of an abelian scheme exists, but why is it an abelian scheme?
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3 Answers
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You can also look at chapter 6 of Mumford's "Geometric Invariant Theory", together with part of Kleiman's article in "FGA Explained" (section 9.6) for the construction of $Pic_{X/S}^\tau$ from $Pic_{X/S}$.
answered Jan 15, 2013 at 18:01
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$\begingroup$ In terms of addressing properties of the dual abelian scheme (when it exists, which it always does by Faltings-Chai), it is also appropriate to mention Oda's thesis. $\endgroup$ Commented Jan 15, 2013 at 18:14
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Alternately, see Faltings, Chai, "Degeneration of Abelian Varieties" Chapter I, especially Theorem 1.9
The general idea is: it can be shown that the Picard functor of a scheme $X/S$ is represented by a group algebraic space over $S$. If $X$ is an abelian scheme, then it can be shown that $Pic^0(X/S)$ is an abelian algebraic space. A Theorem of Raynaud shows that any such algebraic space is automatically a scheme. I think I recall that BLR only proves this fact for certain types of schemes $S$.
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$\begingroup$ BLR don't prove it for any class of schemes, they cite other references. Also, please be careful about hypotheses: ${\rm{Pic}}_{X/S}$ is only an algebraic space under some hypotheses on $f:X \rightarrow S$ such as being proper fppf and $f_{\ast}O_X = O_S$ universally (e.g., reduced and connected geometric fibers). $\endgroup$ Commented Jan 15, 2013 at 18:10
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$\begingroup$ You are of course correct. I'm being far too glib. $\endgroup$ Commented Jan 15, 2013 at 19:15
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See Bosch, Luetkebohmert, Raynaud, "Néron Models", chap. 8, 8.4, p. 234.
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$\begingroup$ This "reference" consists entirely of a punt to GIT (= Bellovin's answer). $\endgroup$ Commented Jan 15, 2013 at 18:05
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$\begingroup$ @pranavk. I agree (althoug BLR gives some information about the dual abelian scheme). Even so, it is an entry point into the literature about this subject. I think that your comment is somewhat offensive. $\endgroup$ Commented Jan 16, 2013 at 9:43