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Let $S$ be the spectrum of an excellent discrete valuation ring with field of fractions $K$ and $C$ be a proper integral regular curve over $K$.

Assume, $C$ admits a proper regular flat model $\mathscr{C} \to S$. Let $\mathcal{L}_{\eta}$ a line bundle on $C$.

Q: Is there any general reason why $\mathcal{L}_{\eta}$ should always be always extendable to a line bundle $\mathcal{L}$ over the model $\mathscr{C} $ based crucialy on it's assumed regularity?

Source: It seems that this principle was used in arXiv:2110.00545 in the proof of Theorem 4.3 (p 41/42)

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    $\begingroup$ Extend to any coherent sheaf and then take reflexivization (double dual). A reflexive sheaf on a two-dimensional regular scheme is locally free, so you obtain a line bundle this way. $\endgroup$ Commented Oct 13, 2023 at 9:35
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    $\begingroup$ Why don't we just take the associated divisor $D$ of $\mathcal{L}_{\eta}$ on $C$ and the closure of this divisor on $\mathcal{C}$? This is a Weil divisor and by regularity it is Cartier. So we get a line bundle on $\mathcal{C}$. $\endgroup$ Commented Oct 13, 2023 at 13:32
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    $\begingroup$ @AriyanJavanpeykar: yes, that looks to be extendable to even more general situations beyond dimension two setting - namely when we dealing with proper flat fibrations $X \to S$ with regular $X$ and $S$ Dedekind. Then one can under relative mind conditions show that the closure in $X$ of a Weil divisor of generic fiber is flat over $S$ and conseqeuntly is a Weil divisor in $X$ and apply again this Cartier- Weil identification $\endgroup$ Commented Oct 13, 2023 at 14:02
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    $\begingroup$ As Ariyan says you can take the closure of a divisor, which works as long as the model is locally factorial. For surfaces you get more: you can extend vector bundles, not only line bundles. $\endgroup$ Commented Oct 13, 2023 at 14:52
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    $\begingroup$ @PiotrAchinger: yes Ariyan Javanpeykar's argument working with Weil divisors I understand so far. The question is "how far" goes the argumentation via the techniques using the reflexive sheaf? Say we have a reflexive sheaf on an integral scheme $X$. As you said the if $X$ 2-dim + regular (... even loc factorial suffice), then it is locally free. Now how far can the assumptions on $X$ be "maximally" weakened (regarding dimension and regularity) in order to have still this implication that "reflexive" implies "locally free"? $\endgroup$ Commented Oct 13, 2023 at 16:06

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