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



Galois invariant line bundle and base change
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.


