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Does anyone maybe have a reference to the proof of the following result by Tate?

Let $\Gamma$ be the absolute Galois group of the rationals. Then the second cohomology group (for trivial $\Gamma$-action) H$^2(\Gamma, \mathbb{Q}/\mathbb{Z})$ is trivial.

Unfortunately I couldn't find it online or in the library. Any help would be greatly appreciated!

Kind regards!

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    $\begingroup$ What is $H^2(\Gamma ,\mathbb{Q},\mathbb{Z})$? $\endgroup$
    – abx
    Commented Aug 30, 2016 at 10:24
  • $\begingroup$ Thanks, abx, I'm am so sorry for this mess! I'll edit right away $\endgroup$
    – JH_94
    Commented Aug 30, 2016 at 10:46
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    $\begingroup$ Look at the long exact sequence $H^2(\Gamma, \mathbb{Q}) \rightarrow H^2(\Gamma, \mathbb{Q}/\mathbb{Z}) \rightarrow H^3(\Gamma,\mathbb{Z})$. This should give you want you want, since $H^3=0$. $\endgroup$
    – Venkataramana
    Commented Aug 30, 2016 at 11:13
  • $\begingroup$ Okay, thank you very much Venkataramana! I'm sorry but I can't vote up yet.. $\endgroup$
    – JH_94
    Commented Aug 30, 2016 at 11:19
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    $\begingroup$ @Venkataramana: It is also worthwhile to note that $H^2(\Gamma,\mathbb{Q})=0$, since $\Gamma$ is profinite and $\mathbb{Q}$ is uniquely divisible. $\endgroup$
    – GH from MO
    Commented Aug 30, 2016 at 15:39

1 Answer 1

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By the Galois cohomology long exact sequence, this is isomorphic to $\operatorname{H}^3(\Gamma,\mathbb{Z})$, and the vanishing of this is Chapter I, Corollary 4.17 in Milne's Arithmetic Duality Theorems.

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    $\begingroup$ Actually, the proof in Milne's book reduces $H^3(\Gamma,\mathbb{Z})=0$ to $H^2(\Gamma,\mathbb{Q}/\mathbb{Z})=0$, which in turn is reduced to $H^3(\mathbb{R},\mathbb{Z})=0$ by Theorem 4.10. $\endgroup$
    – GH from MO
    Commented Aug 30, 2016 at 16:09
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    $\begingroup$ A nice alternative reference for the proof that ${\rm{H}}^2(k, \mathbf{Q}/\mathbf{Z})=0$ for number fields $k$ is somewhere in Serre's survey paper "Weight-1 modular forms and Galois representations" (or some such title as that), in the Durham conference proceedings. (The result is expressed in the more exotic-looking form ${\rm{H}}^2(k, \mathbf{C}^{\times})=1$, but $\mathbf{C}^{\times}/\mu_{\infty}$ is uniquely divisible and $\mu_{\infty}=\mathbf{Q}/\mathbf{Z}$, so it is the same assertion.) $\endgroup$
    – nfdc23
    Commented Aug 31, 2016 at 8:11

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