13
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ 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 $p=5$. $\endgroup$ – Chris Wuthrich Nov 11 '14 at 13:20
9
$\begingroup$

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 $H^1(G;E[3])$.

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ – Chris Wuthrich Nov 11 '14 at 22:20
  • $\begingroup$ @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. $\endgroup$ – Tyler Lawson Nov 11 '14 at 23:33
  • $\begingroup$ @TylerLawson thanks for this very interesting example! $\endgroup$ – Ahmed Matar Nov 12 '14 at 8:56
  • $\begingroup$ 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. $\endgroup$ – Chris Wuthrich Nov 12 '14 at 14:37
2
$\begingroup$

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 $p=3$.

$\endgroup$
2
  • 3
    $\begingroup$ 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 $\mathbb{Q}$. $\endgroup$ – Chris Wuthrich Nov 11 '14 at 12:10
  • $\begingroup$ That's true. It only works over the rationals for $p=2$. $\endgroup$ – Felipe Voloch Nov 11 '14 at 12:22
2
$\begingroup$

$\newcommand{\FF}{\mathbb{F}}\DeclareMathOperator{\SL}{SL}$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.