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Let $K$ be a number field and consider a finite Galois extension $L|K$. Moreover let $X$ be a projective, regular, integral variety over $K$. After a base change we obtain a morphism of varieties $f:X_L\to X$. Assume that $\mathcal L$ is a $\operatorname{Gal}(L|K)$ invariant line bundle on $X_L$; why is it true that there exists $n\in \mathbb N$ and a line bundle $M$ on $X$ such that $\mathcal L^n\cong f^\ast M$? Clearly the Galois invariance here plays a crucial role, but I cannot figure out how to use it.

Many Thanks in advance

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    $\begingroup$ There is a "local-global" spectral sequence mapping the Picard group of $X$ to the Galois-invariant subgroup of the Picard group of $X_L$ with cokernel contained in the Brauer group of $K$. Since every element of the Brauer group is torsion, it follows that every Galois-invariant element of the Picard group of $X_L$ descends after raising to an appropriately divisible tensor power. (There are also direct proofs, but the spectral sequence gives this and many other results as consequences.) $\endgroup$
    – Jason Starr
    Oct 31, 2021 at 23:30
  • $\begingroup$ manifold, did you read Jason Starr's comment? It completely answers your question. $\endgroup$
    – abx
    Nov 1, 2021 at 14:52
  • $\begingroup$ Yes, I would accept if it was an answer. $\endgroup$
    – manifold
    Nov 1, 2021 at 14:57
  • $\begingroup$ Yes, sorry. Noticed now $\endgroup$
    – manifold
    Nov 1, 2021 at 18:59
  • $\begingroup$ @manifold You have nothing to apologize for. I only posted my answer after the exchange between yourself and abx. It is bad form, but I often just write comments rather than full answers. $\endgroup$
    – Jason Starr
    Nov 1, 2021 at 20:37

1 Answer 1

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I am posting my comment as an answer. This result is discussed in many sources. I do not have Serre's "Galois cohomology" with me at this moment, but I am certain that it is discussed there. It should be discussed also in "Dix exposes sur le Groupe de Brauer".

In fact the reference where I first learned this is Igor Dolgachev's textbook on invariant theory, Section 2 of Chapter 7. He writes the long exact sequence of the spectral sequence in a slightly different setting of a group $G$ acting on $X_L$ through morphisms of $L$-schemes, rather than the "twisted" action of the Galois group of $L/K$ acting through morphisms of $K$-schemes. However, the long exact sequence of the short exact sequence is the same in both cases.

$$ 0\to \text{Pic}(X)\to \text{Pic}(X_L)^{\text{Aut}(L/K)} \to \text{Br}(K) $$

Since the Brauer group of $K$ is a torsion group, every $\text{Aut}(L/K)$-invariant element of $\text{Pic}(X_L)$ is in the image of $\text{Pic}(X)$ after replacing the invertible sheaf by all sufficiently divisible tensor powers.

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