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Given a field $k$, how do I have to think about the Galois cohomology group $H^1(k,\mathbb{Q}/\mathbb{Z})$? I know this is the group of continuous homomorphisms from $\Gamma_k$ to $\mathbb{Q}/\mathbb{Z}$ but I'd like to see this in a more concrete way still...

EDIT: Also, if $K$ is an extension of $k$, which are the elements of $H^1(k,\mathbb{Q}/\mathbb{Z})$ which map to $0$ in $H^1(K,\mathbb{Q}/\mathbb{Z})$ via the restriction map? Is there a concrete way to look at these?

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1 Answer 1

up vote 6 down vote accepted

More concrete still? It depends on the field $k$. First, since $\mathbf{Q}/\mathbf{Z}$ is abelian, any continuous homomorphism from $\Gamma_k$ to $\mathbf{Q}/\mathbf{Z}$ factors through the topological abelianization $\Gamma_k^{ab}$ (maximal Hausdorff abelian quotient of $\Gamma_k$), so $H^1(k,\mathbf{Q}/\mathbf{Z})=\mathrm{Hom}_{cts}(\Gamma_k^{ab},\mathbf{Q}/\mathbf{Z})$ is the Pontryagin dual of $\Gamma_k^{ab}$, and is therefore a discrete torsion abelian group. But this can be pretty complicated. If $k$ is a finite field, then $\Gamma_k^{ab}=\Gamma_k=\hat{\mathbf{Z}}$ with the Frobenius automorphism $x\mapsto x^{\# k}$ generating a dense subgroup of $\Gamma_k$. In this case $\mathrm{Hom}_{cts}(\hat{\mathbf{Z}},\mathbf{Q}/\mathbf{Z})=\mathbf{Q}/\mathbf{Z}$ via the map sending a homomorphism to its image on Frobenius. If $k$ is a local or global field, then $\Gamma_k^{ab}$ is described by class field theory, but I'm not sure this gives a "concrete" description of its Pontryagin dual. When $k=\mathbf{Q}_p$ with $p$ odd the abelianization of $\Gamma_k$ is the profinite completion of $\mathbf{Q}_p^\times$, which is isomorphic to $\mathbf{Z}_p\times\mu_{p-1}\times\mathbf{Z}$, so, completing gives $\mathbf{Z}_p\times\mu_{p-1}\times\hat{\mathbf{Z}}$, and then the dual is isomorphic to $\mathbf{Q}_p/\mathbf{Z}_p\times\mu_{p-1}\times\mathbf{Q}/\mathbf{Z}$. This can be generalized to finite extensions of $\mathbf{Q}_p$ (you'll get extra copies of $\mathbf{Z}_p$). Probably another user smarter than I can elaborate on this in the case of global fields (or point out any oversights or mistakes on my part). (This isn't really an answer but was too long for a comment.)

EDIT (added to address the second question):

The restriction map $H^1(k,\mathbf{Q}/\mathbf{Z})\rightarrow H^1(K,\mathbf{Q}/\mathbf{Z})$ just sends a homomorphism $G_k\rightarrow\mathbf{Q}/\mathbf{Z}$ to its restriction to $G_K$, so a homomorphism is in the kernel if its restriction to the latter subgroup is zero. When $K$ is Galois, this kernel therefore consists of the continuous homomorphism $G_k\rightarrow\mathbf{Q}/\mathbf{Z}$ that factor through $\mathrm{Gal}(K/k)$, which is $\mathrm{Hom}_{cts}(\mathrm{Gal}(K/k),\mathbf{Q}/\mathbf{Z})=H^1(K/k,\mathbf{Q}/\mathbf{Z})$. This is the inflation-restriction sequence for $K/k$.

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I don't know about smarter users, but this is a pretty nice answer already. –  Georges Elencwajg Oct 24 '11 at 21:02
    
Thank you Georges. That means a lot coming from someone with as many great answers as you. –  Keenan Kidwell Oct 24 '11 at 21:11
    
What happens when (in your edit) $K/k$ is not Galois?? –  Evariste Oct 25 '11 at 10:18
    
It's the same. The kernel is still the set of homomorphisms from $G_k$ containing $G_K$ in their kernel. There just isn't any quotient of which to speak. –  Keenan Kidwell Oct 25 '11 at 11:33

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