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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 $\Bbb{Q}$.

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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.

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  • $\begingroup$ Maybe it's worth remarking that the infinite cohomological dimension is a local problem at the prime 2 only. For $K$ any number field and $p$ an odd prime, the $p$-cohomological dimension (i.e. the sup of the degrees of nonzero cohomology among all finite $Gal(\overline{K} / K)$-modules whose cardinality is a power of $p$) is always 2. $\endgroup$ Jan 19, 2020 at 8:37
  • $\begingroup$ @DavidLoeffler: Thank you! I will edit my answer. $\endgroup$ Jan 19, 2020 at 10:20

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