# Third Galois cohomology group

It is well known that when $$K$$ is a local or global field the Galois cohomology group $$H^{3}(K,K_{\text{sep}}^{\times})=0$$ where $$K_{\text{sep}}$$ denotes the separable closure of $$K$$. Could someone give an example of a field $$K$$ where $$H^{3}(K,K_{\text{sep}}^{\times}) \neq 0$$ and why it is non-zero in this case?

• This question about the Teichmüller class might interest you mathoverflow.net/q/143031/41291 Commented Jun 1, 2021 at 10:03
• My example, whilst I think is interesting, is probably not the most natural way to approach the question "what is $H^3(K,\bar{K}^\times)$?" For example, does anyone know how to calculate this cohomology group when $K$ is a purely transcendental extension of $\mathbb{Q}$ or $\bar{\mathbb{Q}}$? Is it always non-trivial in this case? Commented Jun 1, 2021 at 15:54
• @DanielLoughran When $K$ is of transcendental degree 1 over a separably closed subfield, isn’t this group known to be zero (standard Galois cohomology) ?
– A.B.
Commented Jun 1, 2021 at 19:23
• @მამუკაჯიბლაძე Thanks for the reference!
– H U
Commented Jun 1, 2021 at 19:25
• @A.B.: Ah yes the vanishing in this case follows from Tsen's Theorem. Still, it would be interesting to know what happens for higher transcendence degree or non-closed field of constants. Commented Jun 2, 2021 at 8:45

The group $$H^3(K,\bar{K}^\times)$$ naturally arises when trying to calculate the Brauer group of a variety. Explicitly, the Hochschild-Serre sequence yields the exact sequence $$0 \to \mathrm{Br}_1(X)/\mathrm{Br}(K) \to H^1(K,\mathrm{Pic}(X_{\bar{K}})) \to H^3(K,\bar{K}^\times)$$ for a projective variety $$X$$ over a perfect field $$K$$, where $$\mathrm{Br}_1(X) = \ker(\mathrm{Br}(X) \to \mathrm{Br}(X_{\bar{K}}))$$ is the algebraic part of the Brauer group of $$X$$.
Over number fields the vanishing you mention allows one to calculate the Brauer group using $$H^1(K,\mathrm{Pic}(X_{\bar{K}}))$$, which is often easier to understand. But there are examples where this doesn't happen where $$K$$ is a function field.
This concerns the variety $$X: \quad x^3 + by^3 + cz^3 + dt^3 = 0$$ over the function field $$K=k(b,c,d)$$ where $$k$$ is any field containing a third root of unity. Then in the above cited paper it is shown that $$\mathrm{Br}(X)/\mathrm{Br}(K) = 0$$ but $$H^1(K,\mathrm{Pic}(X_{\bar{K}})) \cong \mathbb{Z}/3\mathbb{Z}$$. Thus $$H^3(K,\bar{K}^\times)$$ is non-trivial, and gives a geometric interpretation of this through the failure of elements of $$H^1(K,\mathrm{Pic}(X_{\bar{K}})$$ to come from Brauer group elements.