6
$\begingroup$

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!

$\endgroup$
5
  • 4
    $\begingroup$ What is $H^2(\Gamma ,\mathbb{Q},\mathbb{Z})$? $\endgroup$
    – abx
    Aug 30 '16 at 10:24
  • $\begingroup$ Thanks, abx, I'm am so sorry for this mess! I'll edit right away $\endgroup$
    – JH_94
    Aug 30 '16 at 10:46
  • 3
    $\begingroup$ 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 $H^3=0$. $\endgroup$ Aug 30 '16 at 11:13
  • $\begingroup$ Okay, thank you very much Venkataramana! I'm sorry but I can't vote up yet.. $\endgroup$
    – JH_94
    Aug 30 '16 at 11:19
  • 1
    $\begingroup$ @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
    Aug 30 '16 at 15:39
6
$\begingroup$

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.

$\endgroup$
2
  • 4
    $\begingroup$ 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
    Aug 30 '16 at 16:09
  • 1
    $\begingroup$ 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
    Aug 31 '16 at 8:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.