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Ahmed Matar
Ahmed Matar
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Suppose that $E$ an elliptic curve defined over $\mathbb{Q}$ and $p$ an odd prime. Let $G=\text{Gal}(\mathbb{Q}(E[p])/\mathbb{Q})$. I am wondering whether the cohomology group $H^1(G, E[p])$ can be nontrivial. If $G=GL_2(\mathbb{F}_p)$ (which is the case for all but finitely many primes $p$ if $E$ does not have complex multiplication) then $H^1(G, E[p])$ is trivial. This can be shown by considering the homotheyhomothety subgroup $Z \le G$ which has order $p-1>1$. One easily sees that $H^i(Z, E[p])=0$ for all $i \geq 0$ and so the result follows from the Hochschild-Serre spectral sequence.

Now suppose that $G$ is a proper subgroup of $GL_2(\mathbb{F}_p)$. Can $H^1(G, E[p])$ be nontrivial?

Suppose that $E$ an elliptic curve defined over $\mathbb{Q}$ and $p$ an odd prime. Let $G=\text{Gal}(\mathbb{Q}(E[p])/\mathbb{Q})$. I am wondering whether the cohomology group $H^1(G, E[p])$ can be nontrivial. If $G=GL_2(\mathbb{F}_p)$ (which is the case for all but finitely many primes $p$ if $E$ does not have complex multiplication) then $H^1(G, E[p])$ is trivial. This can be shown by considering the homothey subgroup $Z \le G$ which has order $p-1>1$. One easily sees that $H^i(Z, E[p])=0$ for all $i \geq 0$ and so the result follows from the Hochschild-Serre spectral sequence.

Now suppose that $G$ is a proper subgroup of $GL_2(\mathbb{F}_p)$. Can $H^1(G, E[p])$ be nontrivial?

Suppose that $E$ an elliptic curve defined over $\mathbb{Q}$ and $p$ an odd prime. Let $G=\text{Gal}(\mathbb{Q}(E[p])/\mathbb{Q})$. I am wondering whether the cohomology group $H^1(G, E[p])$ can be nontrivial. If $G=GL_2(\mathbb{F}_p)$ (which is the case for all but finitely many primes $p$ if $E$ does not have complex multiplication) then $H^1(G, E[p])$ is trivial. This can be shown by considering the homothety subgroup $Z \le G$ which has order $p-1>1$. One easily sees that $H^i(Z, E[p])=0$ for all $i \geq 0$ and so the result follows from the Hochschild-Serre spectral sequence.

Now suppose that $G$ is a proper subgroup of $GL_2(\mathbb{F}_p)$. Can $H^1(G, E[p])$ be nontrivial?

Fixed the actual question -- I think what you really want to know is not only whether someone *knows* whether H^1(G, E[p]) can be nontrivial.
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Stefan Kohl ♦
Stefan Kohl ♦
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Suppose that $E$ an elliptic curve defined over $\mathbb{Q}$ and $p$ an odd prime. Let $G=\text{Gal}(\mathbb{Q}(E[p])/\mathbb{Q})$. I am wondering whether the cohomology group $H^1(G, E[p])$ can be nontrivial. If $G=GL_2(\mathbb{F}_p)$ (which is the case for all but finitely many primes $p$ if $E$ does not have complex multiplication) then $H^1(G, E[p])$ is trivial. This can be shown by considering the homothey subgroup $Z \le G$ which has order $p-1>1$. One easily sees that $H^i(Z, E[p])=0$ for all $i \geq 0$ and so the result follows from the Hochschild-Serre spectral sequence.

Now suppose that $G$ is a proper subgroup of $GL_2(\mathbb{F}_p)$. Does anybody know whetherCan $H^1(G, E[p])$ can be nontrivial?

Suppose that $E$ an elliptic curve defined over $\mathbb{Q}$ and $p$ an odd prime. Let $G=\text{Gal}(\mathbb{Q}(E[p])/\mathbb{Q})$. I am wondering whether the cohomology group $H^1(G, E[p])$ can be nontrivial. If $G=GL_2(\mathbb{F}_p)$ (which is the case for all but finitely many primes $p$ if $E$ does not have complex multiplication) then $H^1(G, E[p])$ is trivial. This can be shown by considering the homothey subgroup $Z \le G$ which has order $p-1>1$. One easily sees that $H^i(Z, E[p])=0$ for all $i \geq 0$ and so the result follows from the Hochschild-Serre spectral sequence.

Now suppose that $G$ is a proper subgroup of $GL_2(\mathbb{F}_p)$. Does anybody know whether $H^1(G, E[p])$ can be nontrivial?

Suppose that $E$ an elliptic curve defined over $\mathbb{Q}$ and $p$ an odd prime. Let $G=\text{Gal}(\mathbb{Q}(E[p])/\mathbb{Q})$. I am wondering whether the cohomology group $H^1(G, E[p])$ can be nontrivial. If $G=GL_2(\mathbb{F}_p)$ (which is the case for all but finitely many primes $p$ if $E$ does not have complex multiplication) then $H^1(G, E[p])$ is trivial. This can be shown by considering the homothey subgroup $Z \le G$ which has order $p-1>1$. One easily sees that $H^i(Z, E[p])=0$ for all $i \geq 0$ and so the result follows from the Hochschild-Serre spectral sequence.

Now suppose that $G$ is a proper subgroup of $GL_2(\mathbb{F}_p)$. Can $H^1(G, E[p])$ be nontrivial?

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Ahmed Matar
Ahmed Matar
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For an elliptic curve $E/\mathbb{Q}$ can the cohomology group $H^1(\text{Gal}(\mathbb{Q}(E[p])/\mathbb{Q}), E[p])$ be nontrivial?

Suppose that $E$ an elliptic curve defined over $\mathbb{Q}$ and $p$ an odd prime. Let $G=\text{Gal}(\mathbb{Q}(E[p])/\mathbb{Q})$. I am wondering whether the cohomology group $H^1(G, E[p])$ can be nontrivial. If $G=GL_2(\mathbb{F}_p)$ (which is the case for all but finitely many primes $p$ if $E$ does not have complex multiplication) then $H^1(G, E[p])$ is trivial. This can be shown by considering the homothey subgroup $Z \le G$ which has order $p-1>1$. One easily sees that $H^i(Z, E[p])=0$ for all $i \geq 0$ and so the result follows from the Hochschild-Serre spectral sequence.

Now suppose that $G$ is a proper subgroup of $GL_2(\mathbb{F}_p)$. Does anybody know whether $H^1(G, E[p])$ can be nontrivial?