Third Galois cohomology group

Question

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?

Answer 1

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.

One of the only examples I know of is due to Umenatsu- On the Brauer group of diagonal cubic surfaces.

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.