Let $f = \sum_{n=1}^{\infty} a_n(f) q^n$ be a normalized eigenform in $S_2(\Gamma_0(N))^{\mathrm{new}}$, i.e., $f$ is a newform of level $N \geq 2$. Assume that $f$ does not have complex multiplication. Let $K_f$ be the number field generated by these Fourier coefficients. For every prime $p$, fix a prime $\mathcal{P}$ above $p$ in $\mathcal{O}_{K_f}$, the integral closure of $\mathbb{Z}$ in $K_f$ .

Is it possible that $$a_p(f) \equiv 1 \mod \mathcal{P}$$ for all primes $p \gg 0$?

For example, if one knows the existence of infinitely many non-ordinary primes for $f$, then the above congruence will not hold. This is the case when $K_f = \mathbb{Q}$, by the result of Elkies.


1 Answer 1


These primes are called anomalous (they occur in some work of Mazur). When $f$ comes from an elliptic curve $E$, if you fix $n>2$ for which there are infinitely many primes that split completely in $K(E[n])$, then for such a prime to be anomalous, it would mean $pn | \#E(\mathbb{F}_p)$, which is not possible by the Weil bound. Maybe a similar argument holds in general, a congruence modulo $n$, together with the congruence modulo $p$ are incompatible with $|a_p| \le 2\sqrt{p}$, but I don't see how to put it all together now.

  • $\begingroup$ Dear Voloch, Thanks for your comments. $\endgroup$
    – N. Kumar
    Jan 30, 2013 at 6:04

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