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During my work I came across the group $H^3(\mathrm{Gal}(L/K),L^\times)=H^3(L/K,L^\times)$ for certain (infinite) Galois extensions $L/K$ (for an arbitrary field $K$) and I wondered if there is an arithmetic/ natural interpretation of $H^3(L/K,L^\times)$ (resp. $H^3(K^{sep}/K,(K^{sep})^\times)$) and/or corresponding inflation maps $H^3(L/K,L^\times)\to H^3(L'/K,(L')^\times)$. I was thinking of an analog of the relative Brauer group and the fact that inflation is an inclusion.

I did some research online and found an entry on nlab (https://ncatlab.org/nlab/show/line+n-bundle) that seems to be promising although I don't understand much of what is written there. Some rough idea of what is going on would be very helpful.

I also found the following MO post but I don't think it's helpful in this particular situation, as it appears "at the end" of an exact sequence and doesn't seem to allow much information to be extracted: Third Galois cohomology group

My background is more in group/ number theory, but I have a little knowledge of étale cohomology and algebraic geometry if that helps.

Any help or suggestion for a reference would be greatly appreciated!

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  • $\begingroup$ Arithmetic Duality Theorems by Milne may have some relevant material $\endgroup$
    – Bma
    Commented May 1 at 7:43
  • $\begingroup$ @Bma Thank you I'll have a look $\endgroup$
    – Firebolt2222
    Commented May 2 at 12:17
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    $\begingroup$ At least if you have the corresponding roots of unity in our ground field, then you can construct elements of this Galois cohomology group using the cup product $H^1(k,\mu_n) \times H^1(k,\mu_n) \times H^1(k,\mu_n) \to H^3(k,\mu_n)$, which can be related to your cohomology using Kummer theory. Whether such an element represents a non-trivial cohomology class is a tricky question. $\endgroup$
    – Daniel Loughran
    Commented May 7 at 19:25
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    $\begingroup$ Elements of this form appear in the paper "Surfaces defined by pairs of polynomials" by Skorobogatov and Gvirtz-Chen. it seems in general to get something non-trivial you need to be using transcendental elements (indeed this cohomology group is trivial over number fields). $\endgroup$
    – Daniel Loughran
    Commented May 7 at 19:26
  • $\begingroup$ @DanielLoughran Thank you very much for the reference! This seems very helpful. I'll have a deeper look into it. Though I have to admit that I'm not that strong in algebraic geometry. $\endgroup$
    – Firebolt2222
    Commented May 16 at 17:42

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