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Let $k$ be a field and $X$ a proper $k$-scheme. It is a theorem of Murre and Oort that the Picard functor is representable by a $k$-group scheme $\operatorname{Pic}_{X/k}$ which is locally of finite type. Thus we can talk about $\operatorname{Pic}_{X/k}^0$, the connected component of the origin. It is a characteristic subgroup scheme.

My question is: Is the reduced closed subscheme $(\operatorname{Pic}_{X/k}^0)_{\operatorname{red}}$ a subgroup scheme of $\operatorname{Pic}^0_{X/k}$? What if $X$ is normal, or even smooth?

It is a general fact about group schemes $G$ which are locally of finite type over a perfect field, that the reduced closed subscheme $G_{\operatorname{red}}$ is in fact a subgroup scheme of $G$. This can fail if the base field is imperfect. So my question is really a question about imperfect base fields. I was looking through the 'standard' references, but couldn't find a discussion of this question.

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    $\begingroup$ Ugh, that's what happens when you don't include Grothendieck in 'standard references'. My question is answered positively in FGA 236-17, Cor. 3.2 in the case that X is normal. $\endgroup$
    – Lars
    Commented Oct 21, 2014 at 16:02
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    $\begingroup$ What Grothendieck has written there is not quite what he meant; if you look at the proof (which refers back to an earlier result) then you'll see that "normal" should have been "geometrically normal" in the statement (due to the method of proof via comparison with the geometric fiber). That typo was missed in the errata published a bit later. Grothendieck leaves it to the reader to check that a proper connected commutative group scheme over a field $k$ has a Zariski-dense set of $k_s$-points, which is indeed a nice exercise. $\endgroup$
    – user27920
    Commented Oct 22, 2014 at 6:08
  • $\begingroup$ Yes, I noticed that incongruence (normal vs. geom. normal) after I postet the above comment. $\endgroup$
    – Lars
    Commented Oct 22, 2014 at 7:50

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