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Let $K$ be a field. What is the Picard group of the ring $R := \bar K \otimes_K \bar K$?

I have seen the question Picard group of product of spaces, but it only deals with varieties over an algebraically closed field.

If $K$ is perfect, then $R$ is a direct product of copies of $\bar K$ indexed by the absolute Galois group of $K$. I guess in that case that $\operatorname{Pic}(R)=0$, but I am not sure. Moreover, in general, $R$ may not be reduced.

KReiser
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Alphonse
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  • The seesaw theorem could help (https://mathoverflow.net/questions/81611/picard-groups-of-fiber-products) – Alphonse Dec 16 '22 at 14:52
  • I don’t think the seesaw is the correct viewpoint here. I think a simple limit argument suffices, ie that the Picard group is the colimit of the Picard groups of $\overline{K} \otimes L$ over finite $L/K$, which are all zero by KReiser’s deleted answer. – Aphelli Dec 16 '22 at 14:59
  • Good catch, @Aphelli! – KReiser Dec 16 '22 at 15:06

1 Answers1

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Indeed, $\operatorname{Pic} R=0$. This follows from a few facts:

  1. If $S=\lim S_i$ is a limit of a directed system of qcqs schemes $S_i$ with affine transition maps, then $\operatorname{Pic} S = \operatorname{colim} \operatorname{Pic} S_i$ (ref)
  2. Every finite-dimensional algebra over a field is a finite direct product of Artinian local rings (ref 1 + ref 2)
  3. The Picard group of a finite direct product of rings is the product of the Picard groups of the factors (ref)
  4. Line bundles over affine schemes correspond to projective modules (ref)
  5. Any projective module over a local ring is free (ref)

By 1, we may reduce to considering the case of $\overline{K}\otimes_K L$ for finite $L/K$, each of which are a finite-dimensional algebra over $\overline{K}$. By 2, such an algebra splits as a finite direct product of Artinian local rings. Then by 3 we can reduce to considering each of the factors. Next, by 4, we're looking to classify all rank one projective modules over each of these local rings, but by 5 they're all free, and so the Picard group vanishes. (Thanks to Aphelli for pointing out part 1.)

KReiser
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  • Thanks for the prompt answer. But here $R$ is not finite-dimensional over $K$. I should see what happens for Picard groups of infinite products. – Alphonse Dec 16 '22 at 14:53
  • Oops, my bad - I'll delete this while I fix it – KReiser Dec 16 '22 at 14:56
  • Thanks, it sounds good. Thanks for all the references, I am going to look at those, and accept the answer once I check everything. – Alphonse Dec 16 '22 at 15:24