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Let $E$ be an elliptic curve defined over a number field $L$, having CM by by the ring of integers $\mathcal{O}_K$ for $K$ quadratic imaginary. If $K \subseteq L$, then (as constructed in Silverman's Advanced Topics in the Arithmetic of Elliptic Curves) we have an associated Grössencharacter $\psi_{E/L}$. If $K \nsubseteq L$, then we take $L' = LK$ (being a quadratic extension of $L$) and consider the Grössencharacter $\psi_{E/L'}$.

I think that Silverman uses the following without mentioning anything about it, so I wonder if there is a short proof of these facts and I am missing something making it totally obvious.

  1. Assume $K \nsubseteq L$ and let $\mathfrak{P}$ be a prime in $\mathcal{O}_L$ of good reduction for $E$, splitting as $\mathfrak{P'}\mathfrak{P}''$ in $\mathcal{O}_{L'}$. Then we have \begin{align*} \psi_{E/L'}(\mathfrak{P}')\psi_{E/L'}(\mathfrak{P}'') = \sharp(\mathcal{O}_L/\mathfrak{P}). \end{align*}
  2. Assume $K \nsubseteq L$ and let $\mathfrak{P}$ be a prime in $\mathcal{O}_L$ ramifying in $\mathcal{O}_{L'}$ as $\mathfrak{P'}^2$. Then we have \begin{align*} \psi_{E/L'}(\mathfrak{P}') = 0. \end{align*}

Thanks in advance for any comment.

Jupp
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1 Answers1

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For part 2, note that $\psi(\mathfrak{P}) = 0$ when $\psi$ is ramified at $\mathfrak{P}$ is a convention. See Silverman's Advanced Topics, page 173, after Conjecture 10.2 (see also page 177).

For part 1, this follows from the fact that $N_{\mathbb{Q}}^K(\psi_{E/L}(\mathfrak{P})) = N_{\mathbb{Q}}^L(\mathfrak{P})$. See Silverman's Advanced Topics, page 175, Corollary 10.4.1, part (a).

Álvaro Lozano-Robledo
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  • First of all, thanks for your answer. To your suggestion on part 2: If $\mathfrak{P}$ ramifies, then (by 2.31(a) on page 185) $E$ has bad reduction at $\mathfrak{P}$. But I can not apply 9.2.(b) (page 168) to this as there is no Grössencharacter attached to $E/L$, only to $E/L'$. It is not obvious to me why $E$ has bad reduction at $\mathfrak{P'}$. To your suggestion on part 1: I am not quite in the situation of the Corollary you mention. I would have to replace $L$ by the compositum $LK$ (to obtain an inclusion as needed) and then I don't see how to apply this to get the desired result. – Jupp Aug 20 '19 at 18:58
  • Dear @Jupp, my apologies, I read your question too quickly and missed the subtleties of the fields of definition. Where is (2) used in Silverman's book? – Álvaro Lozano-Robledo Aug 20 '19 at 21:19
  • Dear @Álvaro, no problem. I think (2) has to be used for proving that $L(E/L,s) = L(\psi_{E/L'},s)$ (Theorem 10.5. on page 175). Using some stuff discussed before one computes $L(E/L,s)$ - it equals a product of terms corresponding to primes of $L$ which split and primes of $L$ which are inert. On the other hand, computing $L(\psi_{E/L'},s)$ one arrives at the same product except that it has additional terms for primes which ramify - these are of the form $(1-\psi_{E/L'}(\mathfrak{P}')q_{\mathfrak{P}}^{-s})^{-1}$. So I guess it should happen that these equal $1$. – Jupp Aug 20 '19 at 21:31
  • Dear @Álvaro, I am sorry, my last comment contains a typo: the additional factors are of the form $(1-\psi_{E'/L}(\mathfrak{P}')q_{\mathfrak{P}'}^{-s})^{-1}$, where $\mathfrak{P}'$ is the prime above a ramified prime. – Jupp Aug 21 '19 at 06:28
  • I've been thinking a bit more about this, and I can't find a complete proof that an elliptic curve $E/L$ with CM wouldn't become of good reduction in the extension $KL/L$. This doesn't happen for instance when $L=\mathbb{Q}$. In general, for $p>3$, you need to extend roughly by adjoining $\sqrt[12]{-\Delta}$, which means that if good reduction happens in a quadratic extension of $L$ then $K$ was already contained in $L$. But I don't think I have all the complete details for a complete proof at my fingertips. Hopefully someone sees this question and gives you an answer, or try MathOverflow. – Álvaro Lozano-Robledo Aug 21 '19 at 17:20
  • No problem, and thanks for your time, I appreciate your try! – Jupp Aug 21 '19 at 18:25
  • Dear @Álvaro, I would like to come back to your comment: If I understand right, you think that when $L = \mathbb{Q}$, the curve can not attain good reduction in the considered quadratic extension (?) - Do you have further reference for this? – Jupp Aug 28 '19 at 21:02
  • Look at Theorem 2.2 in this paper https://alozano.clas.uconn.edu/wp-content/uploads/sites/490/2014/01/lozano-robledo_uniform_supersingular_REV1_v5.pdf, there you can see that if p>3 is a bad prime, in order to achieve good reduction in a quadratic extension you would have a discriminant whose valuation at p is 6 mod 12. However, I believe that for all E/Q with CM, and p>3, the valuation is 3 or 9, never 6 mod 12 (and the good reduction occurs in an extension of degree at least 4). The situation is different for p=2, where good reduction occurs in a quad. ext'n but not the one you want – Álvaro Lozano-Robledo Aug 30 '19 at 00:39
  • Ah, that’s interesting... I will take a more detailed look at this at some point. Thank you! – Jupp Sep 03 '19 at 18:01
  • This comment box is running out of space... feel free to email me. – Álvaro Lozano-Robledo Sep 04 '19 at 13:47