Certain Galois cohomology computation
$begingroup$
Let $L/mathbb{Q}$ be a Galois extension of degree $p$ and $E$ be an elliptic curve defined over $mathbb{Q}$. Let $p$ be a fixed prime (of good ordinary reduction if required). We use $L_infty, mathbb{Q}_infty$ as the notation for the cyclotomic $mathbb{Z}_p$-extension. Let $w|p$ be primes in $L$. We have $L_{infty,w}/ mathbb{Q}_{infty,p}$, a Galois extension and say the Galois group is denoted by $G$.
How does one one compute the Galois cohomology $H^i(G, E(L_{infty,w})_{p^infty})$ for $i=1,2$.
My understanding is that using a result of Lang and a deep result of Coates-Greenberg one can say $H^i(G, E(L_{infty,w}))=0$ for $i=1,2$. Consider the following short exact sequence
$$
0rightarrow mathcal{F}(mathfrak{m}_w) rightarrow E(L_{infty,w}) rightarrow tilde{E}(ell_w) rightarrow 0.
$$
Here $mathcal{F}(mathfrak{m}_w)$ is the corresponding formal group and $ell_w$ the residue field. $tilde{E}$ is reduction modulo $mathfrak{m}_w$. The result of Lang says $H^i(G, tilde{E}(ell_w))=0$ for $i=1,2$ and that of Coates-Greenberg says $H^i(G,mathcal{F}(mathfrak{m}_w))=0$. This implies $H^i(G, E(L_{infty,w}))=0$
number-theory homology-cohomology elliptic-curves galois-cohomology
asked Jan 21 at 17:37
debanjanadebanjana
407211
$endgroup$
add a comment |
$begingroup$
Let $L/mathbb{Q}$ be a Galois extension of degree $p$ and $E$ be an elliptic curve defined over $mathbb{Q}$. Let $p$ be a fixed prime (of good ordinary reduction if required). We use $L_infty, mathbb{Q}_infty$ as the notation for the cyclotomic $mathbb{Z}_p$-extension. Let $w|p$ be primes in $L$. We have $L_{infty,w}/ mathbb{Q}_{infty,p}$, a Galois extension and say the Galois group is denoted by $G$.
How does one one compute the Galois cohomology $H^i(G, E(L_{infty,w})_{p^infty})$ for $i=1,2$.
My understanding is that using a result of Lang and a deep result of Coates-Greenberg one can say $H^i(G, E(L_{infty,w}))=0$ for $i=1,2$. Consider the following short exact sequence
$$
0rightarrow mathcal{F}(mathfrak{m}_w) rightarrow E(L_{infty,w}) rightarrow tilde{E}(ell_w) rightarrow 0.
$$
Here $mathcal{F}(mathfrak{m}_w)$ is the corresponding formal group and $ell_w$ the residue field. $tilde{E}$ is reduction modulo $mathfrak{m}_w$. The result of Lang says $H^i(G, tilde{E}(ell_w))=0$ for $i=1,2$ and that of Coates-Greenberg says $H^i(G,mathcal{F}(mathfrak{m}_w))=0$. This implies $H^i(G, E(L_{infty,w}))=0$
number-theory homology-cohomology elliptic-curves galois-cohomology
asked Jan 21 at 17:37
debanjanadebanjana
407211
$endgroup$
$begingroup$
$mathbb{Q}_infty$ is the unique cyclotomic $mathbb{Z}_p$ extension, ie the $mathbb{Z}_p$ extension of $mathbb{Q}$ contained in $mathbb{Q}(mu_{p^infty})$.
$endgroup$
– debanjana
Jan 21 at 20:05
1
$begingroup$
So you meant $mathbb{Q}_infty=mathbb{Q}(mu_{p^infty})^{langle sigma rangle}$ where $sigma(mu_{p^k}) = mu_{p^k}^{zeta_{p-1}bmod p^k}$ and $zeta_{p-1}$ is the primitive root in $p$-adic integers. Then what results ?
$endgroup$
– reuns
Jan 21 at 20:15
$begingroup$
The result I have in mind of Lang is from the paper "Algebraic groups over finite fields" and the statement I am interested in says (something to the effect) that a connected commutative algebraic group over a finite field is cohomologically trivial. I am paraphrasing what I need from a paper of Mazur, where he quotes this theorem.
$endgroup$
– debanjana
Jan 21 at 20:34
add a comment |
$begingroup$
Let $L/mathbb{Q}$ be a Galois extension of degree $p$ and $E$ be an elliptic curve defined over $mathbb{Q}$. Let $p$ be a fixed prime (of good ordinary reduction if required). We use $L_infty, mathbb{Q}_infty$ as the notation for the cyclotomic $mathbb{Z}_p$-extension. Let $w|p$ be primes in $L$. We have $L_{infty,w}/ mathbb{Q}_{infty,p}$, a Galois extension and say the Galois group is denoted by $G$.
How does one one compute the Galois cohomology $H^i(G, E(L_{infty,w})_{p^infty})$ for $i=1,2$.
My understanding is that using a result of Lang and a deep result of Coates-Greenberg one can say $H^i(G, E(L_{infty,w}))=0$ for $i=1,2$. Consider the following short exact sequence
$$
0rightarrow mathcal{F}(mathfrak{m}_w) rightarrow E(L_{infty,w}) rightarrow tilde{E}(ell_w) rightarrow 0.
$$
Here $mathcal{F}(mathfrak{m}_w)$ is the corresponding formal group and $ell_w$ the residue field. $tilde{E}$ is reduction modulo $mathfrak{m}_w$. The result of Lang says $H^i(G, tilde{E}(ell_w))=0$ for $i=1,2$ and that of Coates-Greenberg says $H^i(G,mathcal{F}(mathfrak{m}_w))=0$. This implies $H^i(G, E(L_{infty,w}))=0$
number-theory homology-cohomology elliptic-curves galois-cohomology
asked Jan 21 at 17:37
debanjanadebanjana
407211
$endgroup$
Let $L/mathbb{Q}$ be a Galois extension of degree $p$ and $E$ be an elliptic curve defined over $mathbb{Q}$. Let $p$ be a fixed prime (of good ordinary reduction if required). We use $L_infty, mathbb{Q}_infty$ as the notation for the cyclotomic $mathbb{Z}_p$-extension. Let $w|p$ be primes in $L$. We have $L_{infty,w}/ mathbb{Q}_{infty,p}$, a Galois extension and say the Galois group is denoted by $G$.
How does one one compute the Galois cohomology $H^i(G, E(L_{infty,w})_{p^infty})$ for $i=1,2$.
My understanding is that using a result of Lang and a deep result of Coates-Greenberg one can say $H^i(G, E(L_{infty,w}))=0$ for $i=1,2$. Consider the following short exact sequence
$$
0rightarrow mathcal{F}(mathfrak{m}_w) rightarrow E(L_{infty,w}) rightarrow tilde{E}(ell_w) rightarrow 0.
$$
Here $mathcal{F}(mathfrak{m}_w)$ is the corresponding formal group and $ell_w$ the residue field. $tilde{E}$ is reduction modulo $mathfrak{m}_w$. The result of Lang says $H^i(G, tilde{E}(ell_w))=0$ for $i=1,2$ and that of Coates-Greenberg says $H^i(G,mathcal{F}(mathfrak{m}_w))=0$. This implies $H^i(G, E(L_{infty,w}))=0$
number-theory homology-cohomology elliptic-curves galois-cohomology
number-theory homology-cohomology elliptic-curves galois-cohomology
asked Jan 21 at 17:37
debanjanadebanjana
407211
asked Jan 21 at 17:37
debanjanadebanjana
407211
edited Jan 21 at 20:02
debanjana
asked Jan 21 at 17:37
debanjanadebanjana
407211
asked Jan 21 at 17:37
debanjanadebanjana
407211
asked Jan 21 at 17:37
debanjanadebanjana
407211
407211
$begingroup$
$mathbb{Q}_infty$ is the unique cyclotomic $mathbb{Z}_p$ extension, ie the $mathbb{Z}_p$ extension of $mathbb{Q}$ contained in $mathbb{Q}(mu_{p^infty})$.
$endgroup$
– debanjana
Jan 21 at 20:05
1
$begingroup$
So you meant $mathbb{Q}_infty=mathbb{Q}(mu_{p^infty})^{langle sigma rangle}$ where $sigma(mu_{p^k}) = mu_{p^k}^{zeta_{p-1}bmod p^k}$ and $zeta_{p-1}$ is the primitive root in $p$-adic integers. Then what results ?
$endgroup$
– reuns
Jan 21 at 20:15
$begingroup$
The result I have in mind of Lang is from the paper "Algebraic groups over finite fields" and the statement I am interested in says (something to the effect) that a connected commutative algebraic group over a finite field is cohomologically trivial. I am paraphrasing what I need from a paper of Mazur, where he quotes this theorem.
$endgroup$
– debanjana
Jan 21 at 20:34
add a comment |
$begingroup$
$mathbb{Q}_infty$ is the unique cyclotomic $mathbb{Z}_p$ extension, ie the $mathbb{Z}_p$ extension of $mathbb{Q}$ contained in $mathbb{Q}(mu_{p^infty})$.
$endgroup$
– debanjana
Jan 21 at 20:05
1
$begingroup$
So you meant $mathbb{Q}_infty=mathbb{Q}(mu_{p^infty})^{langle sigma rangle}$ where $sigma(mu_{p^k}) = mu_{p^k}^{zeta_{p-1}bmod p^k}$ and $zeta_{p-1}$ is the primitive root in $p$-adic integers. Then what results ?
$endgroup$
– reuns
Jan 21 at 20:15
$begingroup$
The result I have in mind of Lang is from the paper "Algebraic groups over finite fields" and the statement I am interested in says (something to the effect) that a connected commutative algebraic group over a finite field is cohomologically trivial. I am paraphrasing what I need from a paper of Mazur, where he quotes this theorem.
$endgroup$
– debanjana
Jan 21 at 20:34
$begingroup$
$mathbb{Q}_infty$ is the unique cyclotomic $mathbb{Z}_p$ extension, ie the $mathbb{Z}_p$ extension of $mathbb{Q}$ contained in $mathbb{Q}(mu_{p^infty})$.
$endgroup$
– debanjana
Jan 21 at 20:05
$begingroup$
$mathbb{Q}_infty$ is the unique cyclotomic $mathbb{Z}_p$ extension, ie the $mathbb{Z}_p$ extension of $mathbb{Q}$ contained in $mathbb{Q}(mu_{p^infty})$.
$endgroup$
– debanjana
Jan 21 at 20:05
1
1
$begingroup$
So you meant $mathbb{Q}_infty=mathbb{Q}(mu_{p^infty})^{langle sigma rangle}$ where $sigma(mu_{p^k}) = mu_{p^k}^{zeta_{p-1}bmod p^k}$ and $zeta_{p-1}$ is the primitive root in $p$-adic integers. Then what results ?
$endgroup$
– reuns
Jan 21 at 20:15
$begingroup$
So you meant $mathbb{Q}_infty=mathbb{Q}(mu_{p^infty})^{langle sigma rangle}$ where $sigma(mu_{p^k}) = mu_{p^k}^{zeta_{p-1}bmod p^k}$ and $zeta_{p-1}$ is the primitive root in $p$-adic integers. Then what results ?
$endgroup$
– reuns
Jan 21 at 20:15
$begingroup$
The result I have in mind of Lang is from the paper "Algebraic groups over finite fields" and the statement I am interested in says (something to the effect) that a connected commutative algebraic group over a finite field is cohomologically trivial. I am paraphrasing what I need from a paper of Mazur, where he quotes this theorem.
$endgroup$
– debanjana
Jan 21 at 20:34
$begingroup$
The result I have in mind of Lang is from the paper "Algebraic groups over finite fields" and the statement I am interested in says (something to the effect) that a connected commutative algebraic group over a finite field is cohomologically trivial. I am paraphrasing what I need from a paper of Mazur, where he quotes this theorem.
$endgroup$
– debanjana
Jan 21 at 20:34
add a comment |
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$begingroup$
$mathbb{Q}_infty$ is the unique cyclotomic $mathbb{Z}_p$ extension, ie the $mathbb{Z}_p$ extension of $mathbb{Q}$ contained in $mathbb{Q}(mu_{p^infty})$.
$endgroup$
– debanjana
Jan 21 at 20:05
1
$begingroup$
So you meant $mathbb{Q}_infty=mathbb{Q}(mu_{p^infty})^{langle sigma rangle}$ where $sigma(mu_{p^k}) = mu_{p^k}^{zeta_{p-1}bmod p^k}$ and $zeta_{p-1}$ is the primitive root in $p$-adic integers. Then what results ?
$endgroup$
– reuns
Jan 21 at 20:15
$begingroup$
The result I have in mind of Lang is from the paper "Algebraic groups over finite fields" and the statement I am interested in says (something to the effect) that a connected commutative algebraic group over a finite field is cohomologically trivial. I am paraphrasing what I need from a paper of Mazur, where he quotes this theorem.
$endgroup$
– debanjana
Jan 21 at 20:34