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 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?

• I don't know of an example with $p>2$. Maybe the group theorists can tell us examples of $H^1(H, V)\neq 0$ for subgroups $H$ of $\operatorname{SL}_2(\mathbb{F}_p)$ with $V$ the 2-dimensional vector space over $\mathbb{F}_p$. Say with $p=3$ or $p=5$. – Chris Wuthrich Nov 11 '14 at 13:20

Fix elements $\zeta$ and $\alpha$ with $\zeta$ a primitive third root of unity and $\alpha^3 = -4$. These generate a field $K = \Bbb Q(\zeta,\alpha)$ which is the splitting field of $x^3 + 4$, with Galois group $G$ the symmetric group on three letters.

Consider the elliptic curve $y^2 = x^3 + 1$. Unless I have miscalculated, the $3$-torsion points on this curve are the points $(x,y)$ with $x^4 + 4x = 0$. In particular, the points $(0,1)$ and $(\alpha,2\zeta+1)$ are independent 3-torsion points on this curve, so $K = \Bbb Q(E[3])$. Taking these as a basis, the resulting image of the Galois group into $GL_2(\Bbb F_3)$ must be $$\begin{bmatrix}1 & * \\ 0 & *\end{bmatrix}$$ because $(0,1)$ is fixed and the map must be injective.

Let $H < G$ be the subgroup of order three. Since the coefficient group $E[3]$ is $3$-torsion, a transfer argument implies that the restriction $H^1(G;E[3]) \to H^1(H;E[3])$ is injective with image the invariants under $G/H \cong \Bbb Z/2$.

If $$A = \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$$ represents the generator $\tau$ of $H$, then the group $H^1(H;E[3])$ is $ker(1 + A + A^2) / Im(1 - A)$, which is generated by the column vector $\left[\begin{smallmatrix}0 \\ 1\end{smallmatrix}\right]$. (This describes an element of $H^1$ by where an associated $1$-cocycle sends a chosen generator of $H$.)

As a $1$-cocycle, this is represented by the map $f:H \to E[3]$ with $$f(\tau^k) = (1 + A + \cdots + A^{k-1})\left[\begin{smallmatrix}0 \\ 1\end{smallmatrix}\right].$$ The action of the element $\sigma = \left[\begin{smallmatrix}1 & 0 \\ 0 & -1\end{smallmatrix}\right]$ on this cocycle is given by $$({}^\sigma f)(\tau) = \sigma \cdot f(\sigma^{-1} \tau \sigma) = \sigma \cdot f(\tau^2) = \left[\begin{smallmatrix}1 \\ 1\end{smallmatrix}\right]$$ which shows that the two $1$-cocycles ${}^\sigma f$ and $f$ represent the same element of $H^1$. Therefore, this element of $H^1(H;E[3])$ is invariant under $G/H$ and lifts to a nontrivial element of $H^1(G;E[3])$.

• You are right. More generally, If $G$ is the group of all matrices of the form $(\begin{smallmatrix} 1 & * \\ 0 & * \end{smallmatrix})$, then $H^1(G,E[p])= \mathbb{F}_p$ if $p =3$ and it is zero if $p>3$. Your example, and many other curves with a 3-torsion point rational over $\mathbb{Q}$, has indeed this group. – Chris Wuthrich Nov 11 '14 at 22:20
• @ChrisWuthrich Even worse, if I'm calculating correctly (based on your argument) there only appears to be a candidate subgroup of $GL_2(\Bbb F_p)$ which can support a nonzero cohomology group if $p \not \equiv 1 \mod 3$; the subgroup needs to be the set of matrices of the form $(\begin{smallmatrix}c^2 & * \\ 0 & c\end{smallmatrix})$. I don't know whether $p=5$ supports a curve with this type of torsion. – Tyler Lawson Nov 11 '14 at 23:33
• @TylerLawson thanks for this very interesting example! – Ahmed Matar Nov 12 '14 at 8:56
• Yes that is possible for $p=5$. It turns out to be equivalent to be a quadratic twist by 5 of a curve with a rational 5-torsion point. – Chris Wuthrich Nov 12 '14 at 14:37

If $E[p]$ is generated by $P,Q$ with $P$ invariant (i.e. rational) and the action of $G$ is generated by $Q \mapsto Q+P$, so $G$ is isomorphic to $<P>$, then $H^1(G,E[p])$ contains $H^1(G,<P>) = Hom(G,<P>)$ which is non-trivial. Of course there is the question of realizing this over $\mathbb{Q}$ which won't be possible for large $p$ by Mazur but probably can be for $p=3$.

• Over $\mathbb{Q}$, the determinant $G\to \mathbb{F}_p^{\times}$ must be surjective. So your $G$ won't appear as a group for an elliptic curve over $\mathbb{Q}$. – Chris Wuthrich Nov 11 '14 at 12:10
• That's true. It only works over the rationals for $p=2$. – Felipe Voloch Nov 11 '14 at 12:22

$\newcommand{\FF}{\mathbb{F}}\DeclareMathOperator{\SL}{SL}$ OK, let me try.

Write $M=E[p]$. If the order of $G$ is coprime to $p$, then $H^1(G,M)=0$. Assume that $p$ divides the order of $G$. Now by Prop 15 in Serre's "Propriétés galoisiennes... " Invent Math 15, $G$ either contains $\SL_2(\FF_p)$ or it is contained in a Borel subgroup. According to the question, we may assume that $G$ is contained in a Borel, say upper triangular matrices.

Let $H=G\cap \SL_2(\FF_p)$. Restriction shows that $H^1(G,M)$ is the $G/H$-fixed part of $H^1(H,M)$. Now $H$ is a subgroup of $(\begin{smallmatrix} a & b \\ 0 & 1/a\end{smallmatrix})$ with invertible $a$ and arbitrary $b$. By assumption $H$ contains the subgroup $K$ generated by $h = (\begin{smallmatrix} 1 & 1\\ 0 & 1\end{smallmatrix})$. Again by restriction-inflation, $H^1(H,M)$ is contained in $H^1(K,M)$. The latter can be computed as usual and we find that a cocyle is determined by its image on $h$ which has to belong to $\FF_p(1, 0)$.

Now we consider again the action of $G/H$ on $H^1(K,M)$. If I am not mistaken, then the class $\bar g$ of matrices of determinant $d$ acts by $(\bar g * \xi)(h) = d \cdot \xi(h)$. If so, then there are no elements in $H^1(H,M)$ fixed by $G/H$ as soon as there is an element of determinant $\neq 1$ in $G$. By the Weil pairing, this must be the case when $p>2$. Hence $H^1(G,M)=0$.

Edit: However, I am mistaken as the example of Tyler Lawson shows. The action of $G/H$ may be trivial.