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Sep 13, 2021 at 1:39 comment added LgF Forgot to mention there is a separate argument for $l=2$
Sep 12, 2021 at 15:40 comment added Will Sawin I think this is basically right, but extra care must be taken for $\ell=2$ because then the comutator subgroup of $GL_2(\mathbb Z_\ell)$ is smaller than $SL_2(\mathbb Z_\ell)$ (as you can see just by looking mod $2$).
Sep 12, 2021 at 15:30 history edited LeechLattice
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S Sep 12, 2021 at 15:08 review First questions
Sep 12, 2021 at 15:31
S Sep 12, 2021 at 15:08 history asked LgF CC BY-SA 4.0