23
votes
Accepted
Proof of Prop 1.1 in Wiles' "Modular elliptic curves and Fermat's last theorem"
Well, although there is a typo (Wiles forgot to close his parenthesis, and wrote $H^2(G,\mu_{p^r}\to H^2(G,\mu_{p^s})$ in his proof), his claim is correct. Let, as ibid. $F$ be the finite extension of ...
- 5,600
22
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 $\...
- 13.4k
16
votes
Galois cohomologies of an elliptic curve
When thinking of cohomology as describing a defect to a functor being exact, it has to be expected that the first few $H^i$ appear more often. But there are of course higher coholomology groups and ...
- 7,986
16
votes
Accepted
Third Galois cohomology group
The group $H^3(K,\bar{K}^\times)$ naturally arises when trying to calculate the Brauer group of a variety. Explicitly, the Hochschild-Serre sequence yields the exact sequence
$$0 \to \mathrm{Br}_1(X)/\...
- 20.6k
15
votes
Accepted
A Galois extension over $\mathbb{Q}$ with Galois group $A_4$ and with cyclic decomposition groups
Yes. Note that Daniel Loughran's comment to David Speyer's answer to this question states that for any solvable group $G$, there is a Galois extension $L/\mathbb{Q}$ with all decomposition groups ...
- 19.9k
15
votes
Galois cohomologies of an elliptic curve
$H^2$ appears quite prominently in the duality theory of elliptic curves. For a local field such as a finite extension of $\mathbb Q_p$, one has $H^2(\text{Gal}(\bar K/K),\bar K^*)=\mathbb Q/\mathbb Z$...
- 45.3k
15
votes
Accepted
Forms of ${\rm SL}(2)$
$\DeclareMathOperator\SL{SL}\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\Br{Br}\DeclareMathOperator\U{U}\DeclareMathOperator\disc{disc}\DeclareMathOperator\Nm{Nm}\DeclareMathOperator\diag{diag}\...
- 35.8k
14
votes
Accepted
Etale cohomology with coefficients in $\mathbb{Q}$
The following is surely expressing whatever is in the core non-formal aspect of Joe Berner's answer (which is above my pay grade); it is offered as an alternative version of the same ideas.
Let $X$ ...
13
votes
First Galois cohomology of Weil restriction of $\mathbb{G}_m$
One can do much better: it is not necessary to assume $L/K$ is Galois (merely separable is sufficient). And in fact one can formulate the result in a manner which works beyond that of fields, working ...
13
votes
Hasse principle for rational times square
$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\KK{\mathbb{K}}\def\LL{\mathbb{L}}\def\Gal{\mathrm{Gal}}\def\FF{\mathbb{F}}$Here are some examples of $\mathbb{K}$ for which this Hasse principle holds, and ...
- 150k
12
votes
Consequences of Shafarevich conjecture
The Shafarevich conjecture belongs to the broader program of Inverse Galois theory, and in that context it is just another step in that particular approach to understanding $\mathrm{Gal}(\overline{\...
- 17.4k
12
votes
Accepted
Brauer groups and field extensions
No: the conic $C:X^2+Y^2+1=0$ splits over the field $L=\mathbb{Q}(x)[y]/(x^2+y^2+1)$, since $(X,Y)=(x,y)$ is an $L$-point of $C$. However $L$ has no subfields algebraic over $\mathbb{Q}$ other than $\...
- 4,665
12
votes
Accepted
The Mumford-Tate conjecture
Yes.
Under the Hodge conjecture, the Hodge cycles are the algebraic cycles, so the $\mathbb Q_\ell$-linear combinations of Hodge cycles are the $\mathbb Q_\ell$-linear combinations of algebraic cycles....
- 131k
11
votes
Accepted
Is it true that $ H^{2r} ( X , \, \mathbb{Q}_{ \ell } (r) ) \simeq H^{2r} ( \overline{X} , \, \mathbb{Q}_{ \ell } (r) )^G $?
This is false for a general field $k$. It is true for some special fields, like finite fields.
Counterexample: Take $k = \mathbb C((t))$, $E$ an elliptic curve over $\mathbb C$ base-changed to $\...
- 131k
11
votes
Accepted
Embedding torsors of elliptic curves into projective space
Suppose that $C \subset X$ is a smooth projective curve of genus $1$ embedded in a Brauer-Severi surface over a field $k$. We have $C^2 = 9$ since this holds after passing to the algebraic closure, ...
- 20.6k
11
votes
Proof of $V\cong \overline{K} \otimes_{K} V_K$ using $H^1(G_{\overline{K}/K},\operatorname{GL}_n(K))＝0$
Here's a sketch of the proof. I encourage you to fill in the details yourself. The definition of $V_K$ is $V_K=H^0(G_{\overline K/K},V)$. The key part of the proof is to show that $V$ has a $\overline{...
- 45.3k
11
votes
Accepted
For which subgroups the transfer map kills a given element of a group?
The answer to Q1 is yes, the order of $a$ might be smaller than the gcd: Let $G=\langle x,y\mid x^8=y^2=1,x^y=x^3\rangle$ be the semidihedral group of order $16$. Let $a=[x]\in G_{\text{ab}}=C_2\times ...
- 1,604
10
votes
Infiniteness of the Galois cohomology over a number field with coefficients in a finite Galois module
An easy proof:
Let $L/K$ be a Galois extension with Galois group $G$ such that $M$ becomes trivial over $L$. Thus we can also think of $M$ as a $G$-module.
The kernel of $H^1(K,M[p]) \rightarrow H^1(...
- 101
10
votes
Accepted
Do $PGL_n$-torsors induce elements of the Brauer group
Your first statement is not quite true. A $\mathrm{PGL}_2$-torsor does indeed give an element of order $2$ in the Brauer group, but there can be elements of order $2$ in the Brauer group of a general ...
- 4,155
10
votes
Embedding torsors of elliptic curves into projective space
The answer to the first question is no. Let $k=\mathbb{R}$ be the real numbers. Then every cubic in $\mathbf{P}^2_k$ has a rational point. Take the genus $1$ curve $C$ in $\mathbf{P}^3_k$ obtained by ...
- 15.1k
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 ...
9
votes
Accepted
Selmer Group versus Selmer Variety
You should read the Bloch--Kato paper in the Grothendieck Festschrift. This was, I believe, the first paper to consider Selmer groups of Galois representations defined by local conditions coming from ...
- 35.8k
9
votes
Accepted
Nonabelian $H^2$ and Galois descent
Let me elaborate more on the remark above. Let $k$ be a perfect field. Let $\mathrm{Field}_k$ denote the category of finite extensions of $k$, i.e., the objects of $\mathrm{Field}_k$ are fields $k'$ ...
- 9,186
9
votes
Accepted
A cup product in Galois cohomology of Elliptic curve
One can use the exact sequence
$$ 0 \to E(K)/mE(K) \to H^1(K,E[m]) \to H^1(K,E)[m] \to 0 $$
to define a pairing
$$ E(K)/mE(K) \times H^1(K,E)[m] \to H^2(K,\mu_m) $$
by taking $(Q,\xi)$ to $\phi(Q'\cup ...
- 45.3k
9
votes
Accepted
Elements of arbitrary large order in the first Galois cohomology of an elliptic curve
Here is the kind of method I had in mind.
We have the elliptic curve Kummer sequence
$$0 \to E[n] \to E \to E \to 0,$$
Here I denote by $E[n]$ the $n$-torsion group scheme of $E$. Applying Galois ...
- 20.6k
9
votes
Accepted
Infiniteness of the Galois cohomology over a number field with coefficients in a finite Galois module
In fact, $H^1(K,M)$ is infinite not only to the family of number fields, but also to the more general family of Hilbertian fields (those fields which satisfies the Hilbert irreducibility theorem). ...
- 3,282
9
votes
Galois cohomology of finite fields
Consider the short exact sequence
$$0 \to \mathbb Z \to \mathbb Q \to \mathbb Q/\mathbb Z \to 0.$$
Note that $\operatorname{Ext}^i_\mathcal C(N,\mathbb Q) = 0$ for all $i$: it is torsion since $N$ is ...
- 26.5k
9
votes
Accepted
If $A \in \text{End}(\bigwedge^k \mathbb{R}^d)$ equals $\bigwedge^k B$ for some complex matrix $B$, does it have a real source?
Counterexample: $d=3$, $k=2$, $A = -I$, $B = iI$. There is no real $M$
because $\det \bigl(\bigwedge^2 M \bigr) = \det^2 M$ and $\det(A) = -1$.
- 76.8k
9
votes
Accepted
Is a 8-dimensional quadratic form recognized by its Lie algebra, modulo equivalence and scalar multiplication?
Yes: Proposition C.3.14 in Brian Conrad's article Reductive group schemes is that $SO(q)$ determines $q$ up to similarity for all $q$ of dimension $> 2$. (This was pointed out by @user74230 in a ...
- 1,966
Only top scored, non community-wiki answers of a minimum length are eligible