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S Jul 29, 2021 at 19:35 history bounty ended Rdrr
S Jul 29, 2021 at 19:35 history notice removed Rdrr
Jul 29, 2021 at 19:35 vote accept Rdrr
Jul 29, 2021 at 13:37 vote accept Rdrr
Jul 29, 2021 at 19:35
Jul 28, 2021 at 1:06 answer added Bjorn Poonen timeline score: 5
Jul 24, 2021 at 14:16 answer added Yuri Zarhin timeline score: 4
Jul 23, 2021 at 21:23 comment added Will Sawin The way I see how to do it is to observe that the endomorphism algebra is a quaternion algebra, and (by the $\ell$-adic Tate module) split at each prime $\ell$ not $p$. If $\mathbb Q(\alpha)$ is split at $p$ then it is split everywhere, hence a matrix algebra, thus contains nilpotents, which is absurd. Is this the argument of Lang you mention? I don't see a better way.
Jul 23, 2021 at 19:50 comment added Rdrr That's exactly where I get stuck; I would like to know what extra information could finish off the proof.
Jul 23, 2021 at 19:15 comment added Will Sawin I don't think it's possible to finish the argument from where you are because you need to rule out polynomials like $x^2 - \sqrt{q} x + q$ for $p$ congruent to $1$ mod $3$ and $q$ a square, but roots $\alpha$ of these polynomials satisfy the condition $\alpha^N \in \mathbb Z$. So you need more information about elliptic curves.
Jul 23, 2021 at 18:52 history edited Rdrr CC BY-SA 4.0
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S Jul 23, 2021 at 18:36 history bounty started Rdrr
S Jul 23, 2021 at 18:36 history notice added Rdrr Authoritative reference needed
Jul 21, 2021 at 19:26 history edited YCor CC BY-SA 4.0
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Jul 21, 2021 at 18:59 history edited Rdrr CC BY-SA 4.0
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Jul 21, 2021 at 18:01 history asked Rdrr CC BY-SA 4.0