I am thinking about how norms $ν : L ⭢ ℤ$ on higher local fields could induce long exact sequences in different cohomologies.
$ₚ^{\text{sep}}((t)),ℚₚ^{\text{sep}}$, and $ℂ$ are a local fields. What is the easiest way to show that their Brauer group vanishes?
$ₚ^{\text{sep}}((t₁,...,t_{n+1})),ℚₚ^{\text{sep}}((t₁,...,t_{n}))$, and $ℂ((t₁,...,t_{n}))$ are higher local fields. What is the easiest way to find their étale cohomology?
I was hoping that they would produce the cohomology ring of 0.
I'll answer the first question:
If a field is separably closed, then the Brauer group vanishes, because the absolute Galois group is trivial, this covers $\Bbb Q_p^{\mathrm{sep}}$ and $\Bbb C$.
To see that $\overline{\Bbb F_p}((t))$ has vanishing Brauer group, we can use local class field theory. Note that we have $\overline{\Bbb F_p}((t))=\bigcup_{n \geq 1}\Bbb F_p((t))(\zeta_{p^n-1})$. Therefore, we get $$\mathrm{Br}(\overline{\Bbb F_p}((t)))=\varinjlim_{n \geq 1}\mathrm{Br}(\Bbb F_p((t))(\zeta_{p^n-1}))$$ Where the direct limit is taken over inflation maps. But each $\Bbb F_p((t))(\zeta_{p^n-1})$ is a local field, so local class field theory tells us what those Brauer groups and the inflation map look like: More generally, for an extension of non-archimedean local fields $L/K$; there's a commutative diagram:
$\require{AMScd} \begin{CD} \mathrm{Br}(K) @>{\cong}>> \Bbb Q/\Bbb Z\\ @V{\mathrm{Inf}}VV @VV{\cdot [L:K]}V\\ \mathrm{Br}(L) @>>{\cong}> \Bbb Q/\Bbb Z \end{CD}$
Where the horizontal maps are the local invariant maps.
This actually implies in our situation that the direct limit is zero: it suffices that for all $n$ and all $\alpha \in \mathrm{Br}(\Bbb F_p((t))(\zeta_{p^n-1})$, there is some $m$ such that $\alpha$ is in the kernel of the inflation map $\mathrm{Br}(\Bbb F_p((t))(\zeta_{p^n-1})\to \Bbb F_p((t))(\zeta_{p^{nm}-1})$. But using the commutative diagram above, we can just take $m$ to be the order of $\alpha$, because the inflation map corresponds after applying invariant isomorphisms to multiplication by $[\Bbb F_p((t))(\zeta_{p^{nm}-1}):\Bbb F_p((t))(\zeta_{p^n-1})]=m$.
Basically the same argument shows that $\Bbb Q_p^{ur}$ has trivial Brauer group.