What is known about the cohomological dimension of algebraic number fields? What is the cohomological dimension of algebraic number fields like $\Bbb{Q}$, $\Bbb{Q}[i]$, $\Bbb{Q}[\sqrt{3}]$ or similar? I'm interested in computing the cohomological dimension of $\Bbb{A}^1_k$ with $k$ a finite extension of $\Bbb{Q}$.
 A: By definition,  an algebraic number field is  a finite extension 
of the field of rational numbers $\Bbb Q$.
An algebraic number field $K$ is called totally imaginary 
if it has no embeddings into $\Bbb R$.
For example, the field $\Bbb Q(i)$ is totally imaginary, 
while  $\Bbb Q$ and $\Bbb Q(\sqrt{3})$ are not.
We fix an algebraic closure $\overline K$ of $K$, and we write $\Gamma_K={\rm Gal}(\overline K/K)$ for the absolute Galois group of $K$. 
Let $p$ be a prime number. By definition, the $p$-cohomological dimension ${\rm cd}_p(K)$ is  the supremum of the degrees of nonzero cohomology  over all finite $\Gamma_K$-modules $M$ whose cardinality is a power of $p$.
The cohomological dimension ${\rm cd}(K)$ is the supremum of ${\rm cd}_p(K)$
over all prime numbers $p$.

Theorem 1 (well-known) The cohomological dimension ${\rm cd}(K)$ of an algebraic number field $K$ is $2$ if $K$ is totally imaginary, and $\infty$ otherwise.

Following a suggestion of David Loeffler, I state and prove two other theorems, from which Theorem 1 follows.

Theorem 2. The 2-cohomological dimension ${\rm cd}_2(K)$ of an algebraic number field $K$ is $2$ if $K$ is totally imaginary, and $\infty$ otherwise.

Proof. Assume that $K$ is not totally imaginary.  Let $\Omega_{\Bbb R}$ denote the set of real embeddings of $K$; then  $\Omega_{\Bbb R}$ is nonempty. 
We take $M=\mu_2$. For any odd natural number $n\ge 3$ we have
$$ H^{n}(K,\mu_2)\cong\bigoplus_{v\in\Omega_{\Bbb R}} H^n(K_v,\mu_2)\cong
\bigoplus_{v\in\Omega_{\Bbb R}} 
H^1(\Gamma_{\Bbb R},\mu_2)
\cong\bigoplus_{v\in\Omega_{\Bbb R}}\Bbb Z/2\Bbb Z\neq 0.$$
(For the first isomorphism, see J. S. Milne, Arithmetic Duality Theorems, Theorem I.4.10(c). For the second isomorphism, see Atiyah and Wall, Cohomology of Groups, IV.8, Theorem 5, in: Cassels and Fröhlich (eds.), Algebraic Number Theory.)
Thus ${\rm cd}_2(K)=\infty$, as required.
Now assume that $K$ is totally imaginary. Then for any $n\ge 3$ and any finite $\Gamma_K$-module $M$ (in particular, for any $\Gamma_K$-module $M$ whose cardinality is a power of 2)
we have 
$$H^n(K,M)\cong\bigoplus_{v\in\Omega_{\Bbb R}}H^n(K_v,M)=0,$$
because $\Omega_{\Bbb R}=\varnothing$. (For the  isomorphism, again see Milne, Theorem I.4.10(c).) Thus ${\rm cd}_2(K)\le 2$.
Let $\Omega_f$ denote the set of finite places of $K$. We take $M=\mu_2$. We have
$$H^2(K, \mu_2)\cong\left\{(a_v)\in\bigoplus_{v\in\Omega_f}\Bbb Z/2\Bbb Z\ 
\mid\ \sum_{v\in \Omega_f} a_v=0\right\}, $$
and hence, $H^2(K,\mu_2)\ne 0$. 
Thus ${\rm cd}_2(K)=2$, as required.

Theorem 3. For any odd prime $p$,
   the $p$-cohomological dimension ${\rm cd}_p(K)$ of any algebraic number field $K$ equals  $2$.

Proof. For any $n\ge 3$ and any finite $\Gamma_K$-module $M$ whose cardinality is a power of $p$, we have 
$$H^n(K,M)\cong\bigoplus_{v\in\Omega_{\Bbb R}}H^n(K_v,M)\cong
\bigoplus_{v\in\Omega_{\Bbb R}} H^1(\Gamma_{\Bbb R},M)=0,$$
where $H^1(\Gamma_{\Bbb R},M)=0$ because the group $\Gamma_{\Bbb R}$
is of order 2 while $M$ is of odd cardinality (this is an easy exercise; see also Atiyah and Wall, IV.6, Corollary 1 of Proposition 8). Thus ${\rm cd}_p(K)\le 2$.
Let $\Omega_f$ denote the set of finite places of $K$. We take $M=\mu_p$. We have
$$H^2(K, \mu_p)\cong\left\{(a_v)\in\bigoplus_{v\in\Omega_f}\Bbb Z/p\Bbb Z\ 
\mid\ \sum_{v\in \Omega_f} a_v=0\right\}, $$
and hence, $H^2(K,\mu_p)\ne 0$. 
Thus ${\rm cd}_p(K)=2$, as required.
