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 ...
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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 $\...
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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)/\...
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15 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 ...
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15 votes
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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 ...
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  • 17.8k
14 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 $H^2(\text{Gal}(\bar K/K),\bar K^*)=\mathbb Q/\mathbb Z$...
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14 votes
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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}\...
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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 ...
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13 votes
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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 $X$ ...
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 $\...
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  • 4,400
11 votes
Accepted

Variant of Hilbert 90 for Galois extensions

This is an application of Artin's Lemma to reduce to Hilbert 90. Namely, let $K$ be any field at all, and $G$ any finite subgroup of ${\rm{Aut}}(K)$ (as noted in this comments, such finiteness holds ...
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  • 1,573
11 votes
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"Forms" of quadrics

It is not too difficult to see that any automorphism of a smooth quadric hypersurface $$X : Q(x) = 0,$$ over a field $k$ must be a projective automorphism (see for instance the argument I give in ...
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11 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{\...
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  • 17k
11 votes
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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....
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  • 116k
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 $\...
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  • 116k
11 votes
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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, ...
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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(...
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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 ...
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10 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 ...
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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 ...
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10 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{...
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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 ...
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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). ...
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9 votes
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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 $k'$ ...
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9 votes
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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 ...
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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 ...
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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 $\det(A) = -1$.
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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 ...
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