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25
votes
What is known about the cohomological dimension of algebraic number fields?
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 $\B …
15
votes
2
answers
841
views
Elements of arbitrary large order in the first Galois cohomology of an elliptic curve
Let $E$ be an elliptic curve over $k=\mathbb{Q}$. Consider $H^1(k,E)$.
In this answer Daniel Loughran writes: "I'm pretty sure that this cohomology group has elements of arbitrarily large order". I …
14
votes
3
answers
1k
views
Infiniteness of the Galois cohomology over a number field with coefficients in a finite Galo...
Let $k$ be a number field and $M$ be a nonzero finite discrete $\mathrm{Gal}(\bar k/k)$-module. Is it true that $H^1(k,M)$ is infinite?
This would complete the answer of Daniel Loughran. There is a c …
11
votes
1
answer
318
views
Galois cohomology class of a reductive group not coming from a torus
Let $G$ be a (connected) reductive group over a perfect field $k$, and let $\xi\in H^1(k,G)$ be a cohomology class.
By a theorem of Steinberg (Serre, Galois cohomology, Appendix 1 to Chapter III, Theo …
10
votes
Accepted
Galois cohomology class of a reductive group not coming from a torus
$\newcommand{\la}{\langle}\newcommand{\ra}{\rangle}$The following example is due to Vladimir Chernousov (private communication).
Let $K={\Bbb Q}(x,y,x',y')$, where $x,y,x',y'$ are variables.
Consider …
10
votes
1
answer
537
views
A Galois extension over $\mathbb{Q}$ with Galois group $A_4$ and with cyclic decomposition g...
Does there exist a Galois extension $L/\mathbb{Q}$ with Galois group $A_4$ (the alternating group on four letters) such that all the decomposition groups are cyclic?
This question is motivated by the …
9
votes
1
answer
1k
views
Nonabelian $H^2$ and Galois descent
I would like to know whether the following metatheorem on nonabelian $H^2$ has been ever stated and/or proved.
Let $k$ be a perfect field and $k^s$ its fixed separable closure.
Let $X^s$ be a variety …
9
votes
1
answer
788
views
Forms of ${\rm SL}(2)$
I know all real forms of ${\rm SL}(2,{\Bbb C}$). They are ${\rm SL}(2,{\Bbb R})$ and ${\rm SU}(2)$.
Moreover, ${\rm SL}(2,{\Bbb R})$ is isomorphic to ${\rm SU}(1,1)$. Thus I can say that all real fo …
9
votes
1
answer
369
views
For which subgroups the transfer map kills a given element of a group?
$\newcommand{\ab}{{\rm ab}}
\newcommand{\ord}{{\rm ord}}
$Let $G$ be a finite or profinite group. Consider the abelianized group
$$G^\ab=G/G'$$
where $G'$ is the commutator subgroup of $G$.
Let $H\sub …
8
votes
Accepted
Group cohomology question, trivial Galois action on discrete Galois module means we can say ...
If the $G_K$-action on $M$ is trivial, then
$$H^1(K,M)=\mathrm{Hom}(G_K,M),$$
and by Chebotarev's density theorem
$$ F^1(K,M)=0.$$
For details see Lemma 1.1(i) of Sansuc, J.-J. Groupe de Brauer et ari …
7
votes
1
answer
440
views
Imperfect Tate (cup product) pairing in Galois cohomology?
Let $k$ be a field of characteristic 0 with a fixed algebraic closure $\bar k$
and absolute Galois group $\Gamma={\rm Gal}(\bar k/k)$.
Let $M$ be a finite $\Gamma$-module, that is, a finite abelian g …
7
votes
1
answer
332
views
Explicit cocycles for the first Galois cohomology of a $p$-adic torus
Let $K$ be a $p$-adic field (a finite extension of the field of $p$-adic numbers ${\mathbb Q}_p$).
Let $T$ be a $K$-torus with character group $X={\sf X}^*(T)$ and cocharacter group $Y={\sf X}_*(T)=X^ …
7
votes
Accepted
Real forms of complex reductive groups
I answer Question 1. It is just a calculation.
Instead of a real torus, say ${\bf T}$, I consider a pair $(T,\sigma)$,
where $T$ is a complex torus and $\sigma\colon T\to T$ is an anti-holomorphic inv …
7
votes
1
answer
237
views
Hasse principle for $H^2$ of a maximal torus of a split simply connected group?
Let $k$ be a number field, and let $G$ be a split simply connected algebraic group over $k$.
Let $\Omega_k$ denote the set of places of $k$.
Let $T$ be a maximal torus of $G$ (defined over $k$). Cons …
7
votes
2
answers
902
views
Is this exact sequence known?
$\newcommand{\Tors}{{\rm Tors}}
\newcommand{\tf}{{\rm\, t.f.}}
\newcommand{\Gt}{{\Gamma\!,\,\Tors}}
\newcommand{\Gtf}{{\Gamma\!,\tf}}
\newcommand{\Q}{{\mathbb Q}}
\newcommand{\Z}{{\mathbb Z}}
\newcomm …