Let $f=\sum_{n\ge1} a_nq^n$ be an Hecke eigenform cusp form of integral weight $k$ on $\Gamma_0(N)$ with character $\varepsilon\pmod N,$ and let $M_f$ be the subfield of $\mathbb{C}$ generated by the image of $\varepsilon$ and all $a_p$ ($p$ runs on primes) $$M_f:=\mathbb{Q}(\{a_p\}_p,\varepsilon)$$ On what condition we have $\dfrac{a_p}{p^{(k-2)/2}}\in M_f$ ?
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$\begingroup$ If $k$ is even, $p^{(k-2)/2} \in \mathbb Q$, so the answer is "always" in this case. So basically you're interested in the case of odd weight with necessarily non-trivial $\epsilon$. Is it how you intended the question? Or perhaps did you mean it with $(k-2)$ replaced by $(k-1)$? $\endgroup$– JoëlFeb 19, 2017 at 14:12
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$\begingroup$ @Joël, I am interested in the case when $k$ is odd, and $\varepsilon$ non-trivial. Thanks for pointing this out. $\endgroup$– User 101794987Feb 19, 2017 at 14:17
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When $k$ is even the answer is always since $p^{(k-2)/2} \in \mathbb Q$. When $k$ is odd the answer is never since if this was true, then for all $p$ (prime not divdidig $N$), we would have $p^{1/2} \in M_f$, so $M_f$ would contain the field of infinite degree over $\mathbb Q$ generated by the square roots of those prime, contradicting the fact that $M_f$ is a finite extension of $\mathbb Q$.
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1$\begingroup$ Tiny nitpick: this doesn't quite work as stated for all $p \nmid N$, but just for those where $a_p \ne 0$; this set can be somewhat thinner than all $p\nmid N$ (CM forms) but of course it is still infinite, which is all you need for the conclusion. $\endgroup$ Feb 20, 2017 at 9:51
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$\begingroup$ @DavidLoeffler, Thank you very much for pointing this out. $\endgroup$ Feb 21, 2017 at 0:42