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Note: This question was asked on MSE first, but got zero reactions. So I deleted it there, and am now posting it here.

I am looking for a reference or explanation of the fact that is used in Mumford's first paper on Enriques classification of surfaces in char p (page 328, Step IV and V), as well as in Badescu's book "Algebraic Surfaces" (Prop. 13.4 & Prop. 13.5).

The version of the statement in Step IV is:

Let $X$ be a minimal surface, and $D$ an effective divisor on X. If we know that $\text{Pic}^0X$ is an abelian variety (remember that in char p it might not be reduced), and that $\text{Pic}^0X \rightarrow \text{Pic}^0D$ is surjective, then we can conclude that $\text{Pic}^0D$ is also abelian. Why can we conclude that?

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    $\begingroup$ The Picard scheme of a proper 1-dimensional scheme is always smooth since the infinitesimal criterion is a consequence of the vanishing of coherent ${\rm{H}}^2$ on such "curves" (see 8.4/2 in the book Neron Models). So the target ${\rm{Pic}}^0_{D/k}$ is a smooth connected group variety, and inherits properness by surjectivity, whence it is an abelian variety. $\endgroup$
    – nfdc23
    Commented Nov 11, 2017 at 5:07
  • $\begingroup$ @nfdc23 thank you for your swift comment! I saw this (I believe) in Mumford's Lecture notes on curves on algebraic surfaces, lecture 27, as indicated in both the paper and Badescu's book. I currently forget the exact formulation, and don't have access to either M's lectures or the book you mentioned until monday. Do you happen to have the exact statement? $\endgroup$
    – rollover
    Commented Nov 11, 2017 at 12:51
  • $\begingroup$ If $Pic^0$ of any proper 1-dim. scheme is smooth, what do you make of the formulation "we see that $Pic^0X \rightarrow Pic^0D$ is surjective, and because $Pic^0X$ is an Abelian variety, we conclude that $Pic^0D$ is also an Abelian variety"? If $Pic^0D$ is smooth and therefore Abelian anyways, then this seems quite redundant.. $\endgroup$
    – rollover
    Commented Nov 11, 2017 at 12:52
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    $\begingroup$ The identity component of the Picard scheme of a projective scheme $Z$ over a field $k$ is generally not proper, even for a singular integral curve (e.g., for $Z$ the nodal cubic it is $\mathbf{G}_m$); the real content is that ${\rm{Pic}}^0_{D/k}$ is proper (given that a-priori it is smooth since $\dim D = 1$), and this is what we get from the surjectivity of the map from ${\rm{Pic}}^0_{X/k}$ that is already known to be proper. $\endgroup$
    – nfdc23
    Commented Nov 11, 2017 at 17:24
  • $\begingroup$ @nfdc23 Thank you! It all makes sense now. :) Feel free to post this as an answer so I can accept it. $\endgroup$
    – rollover
    Commented Nov 12, 2017 at 3:36

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