If $L/K$ is a Galois extension with group $G$ then we can consider $H^*(G;L^\times)$. This is useful in algebraic number theory, and there are many results about it.
Now let $L/K$ be a finite separable extension that need not be Galois. The tensor products $L\otimes_K\dotsb\otimes_KL$ form a cosimplicial ring $R^\bullet$, the unit groups form a cosimplicial abelian group $(R^\bullet)^\times$, this gives a cochain complex in the usual way, and we can write $HU^*(L/K)$ for the resulting cohomology groups. In the Galois case it is not hard to show that $HU^*(L/K)=H^*(G;L^\times)$ (starting from the standard isomorphism $\phi\colon L\otimes_KL\to\text{Map}(G,L)$ given by $\phi(a\otimes b)(\sigma)=a\,\sigma(b)$). The cosimplicial approach makes various functorial properties much more obvious than the group cohomology approach, however. This is all surely well-known to many people.
Is there a convenient reference that shows which parts of class field theory etc remain true in this setting? I know how to prove that $HU^1(L/K)=0$, for example. Given $K\leq L\leq M$ it is not hard to produce a sequence $$ 0 \to HU^2(L/K) \to HU^2(M/K) \to HU^2(M/L), $$ and to prove that the composite is zero. In the case where everything is Galois it is a standard fact that the sequence is exact. I could probably generalise the proof but it would be nice if it was already written somewhere. For a degree $n$ Galois extension of local fields it is known that $HU^2(L/K)\simeq\mathbb{Z}/n$. I don't know whether that should still be true in the non-Galois case. Some facts can be proved by reduction to the Galois case but it would feel more satisfying to prove them directly.
