Let $\ell \ge 5$ be a prime and $G$ be a closed subgroup of $\operatorname{GL}_2 \mathbb{Z}_\ell$ whose image in $\operatorname{GL}_2 \mathbb{F}_\ell$ is $\operatorname{GL}_2 \mathbb{F}_\ell$. Then $G = \operatorname{GL}_2 \mathbb{Z}_\ell$?

In Serre's "Abelian $l$-adic representations and elliptic curves", the author shows its $\operatorname{SL}_2 \mathbb{Z}_\ell$-version. How can I show the highlighted statement from it?

It suffices to show that $H := G \cap \operatorname{SL}_2 \mathbb{Z}_\ell$ surjects onto $\operatorname{SL}_2 \mathbb{F}_\ell$.


In comments YCor pointed that this is false. I ask this question for the following:

Let $E$ be an elliptic curve over a number field $K$, $\ell \ge 5$ a prime. Assume that the mod $\ell$ representation $G_K \to \operatorname{GL}_2 \mathbb{Z}/\ell$ is surjective. Then is the $\ell$-adic representation $G_K \to \operatorname{GL}_2 \mathbb{Z}_\ell$ surjective?

In the proof of lemma 2.4 of González-Jiménez and Najman - An algorithm for determining torsion growth of elliptic curves, the authors say that this is true by the lemma of Serre that I mentioned above.

How can I show this without first highlighted statement?

  • 6
    $\begingroup$ It fails in $\mathrm{GL}_1$, so the answer is expected to be no. And indeed, consider a subgroup $H$ of $\mathbf{Z}_\ell^*$ of cardinal $\ell-1$ (this exists). Consider the subgroup of $\mathrm{GL}_2(\mathbf{Z}_\ell)$ generated by $\mathrm{SL}_2(\mathbf{Z}_\ell)$ and the diagonal matrices $(t,1)$ with $t\in H$. That is the same as the set of $2\times 2$ matrices with entries in $\mathbf{Z}_\ell$ with determinant in $H$; it surjects onto $\mathrm{GL}_2(\mathbf{F}_\ell)$. $\endgroup$
    – YCor
    Commented Sep 8, 2021 at 14:42
  • $\begingroup$ @YCor Thanks for your comment. I want to show that for an elliptic curve $E$ over a number field, if the mod $\ell$ representation $G_K \to \operatorname{Aut}E[\ell]$ is surjective, then so are $G_K \to \operatorname{Aut}E[\ell^n]$ for all $n$. In a paper arxiv.org/abs/1904.07071, the authors say this true. (lemma 2.4) (And also they use the highlighted statement in the proof of lemma 2.3.) Is this also false? $\endgroup$
    – zom
    Commented Sep 8, 2021 at 16:16
  • $\begingroup$ Possibly (I haven't checked), the quoted lemma allows to show you that the image contains $\mathrm{SL}_2(\mathbf{Z}_\ell)$. If so, you are reduced to proving that composing with the determinant map $G_K\to\mathrm{GL}_2(\mathbf{Z}_\ell)\to\mathbf{Z}_\ell^*$ is surjective, which sounds easier to check (or contradict). $\endgroup$
    – YCor
    Commented Sep 8, 2021 at 16:47
  • $\begingroup$ In your modified question, what is $K$, and what is the relationship of $G_K$ to $E$? $\endgroup$
    – LSpice
    Commented Sep 8, 2021 at 16:51
  • 1
    $\begingroup$ (1) Lemma 2.3 states for an arbitrary number field but weirdly uses in the statement the action of the Galois group of $\mathbb Q$, which doesn't exist in general. I think the proof of Lemma 2.3 is fine if you just replace $\mathbb Q$ everywhere with $K$. (2) Yeah, I think the statement in your question is true assuming also surjectivity of the determinant. Let me think of a good proof... $\endgroup$
    – Will Sawin
    Commented Sep 9, 2021 at 12:41

1 Answer 1


The statement you are trying to prove is false.

In particular, in the comments YCor gave an example of an open proper subgroup $H$ of ${\rm GL}_{2}(\mathbb{Z}_{\ell})$ that surjects onto ${\rm GL}_{2}(\mathbb{F}_{\ell})$. If you take an elliptic curve $E/\mathbb{Q}$ whose $\ell$-adic image is ${\rm GL}_{2}(\mathbb{Z}_{\ell})$ (like $y^{2} + y = x^{3} - x$), let $\rho_{E,\ell^{\infty}}$ be the $\ell$-adic Galois representation attached to $E$, and let $K$ be the fixed field of $\rho_{E,\ell^{\infty}}^{-1}(H)$. Then the map $G_{K} \to {\rm Aut}~ E[\ell]$ is surjective, but $G_{K} \to {\rm Aut}~E[\ell^{2}]$ is not.

  • $\begingroup$ The example I provided is not open since it's the kernel of a homomorphism from $\mathrm{GL}_2(\mathbf{Z}_p)\to\mathbf{Z}_p$. However composing with the canonical surjection $\mathbf{Z}_p\to\mathbf{Z}/p\mathbf{Z}$ we get an open one. $\endgroup$
    – YCor
    Commented Sep 9, 2021 at 7:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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