# Smallest Fourier coefficient divisible by a prime

If $$f$$ is a cusp form of weight $$k$$ and level $$\Gamma_1(N)$$. For simplicity, suppose that the Fourier coefficients of $$f$$ at $$\infty$$ are in $$\mathbb{Z}$$. Let $$\ell$$ be a prime which does not divide the level $$N$$. Then it is a classical result of Serre that almost all (outside a set of density 0) Fourier coefficients are divisible by $$\ell$$.

Question : Is there any known results on the smallest index $$n$$ such that $$\ell | a_n(f)$$?

Of course I am expecting an answer in terms of the weight and level of $$f$$.

How I approached the problem : Serre's proof is basically an application of the Chebatrov density theorem and some efficient counting.

If $$f$$ were an eigen form, then the smallest Fourier coefficient that I am looking for is the smallest prime $$p$$ such that $$\ell | a_p(f)$$ is the smallest prime in the Chebatrov sequence (analogous to smallest primes in arithmetic progressions), for which there are really strong conditional upper bounds. But the problem here is that these upper bounds depend on the number field $$K(f)$$ generated by the Fourier coefficients of $$f$$.

So, I tried to write the cusp form $$f$$ as a linear combination of eigenforms and use a variant of the argument given in the previous paragraph. But as mentioned above, this again leads to considering the field $$K(f_i)$$ for every eigen form $$f_i$$ of that particular weight and level, on which I have very limited control.

Any help is greatly appreciated.

Thank you

An answer that applies when $$f$$ is a non-CM newform with trivial nebentypus and integral Fourier coefficients can be found in Theorem 1.5 of a paper by Thorner and Zaman (arXiv version). As you suggested, this result uses techniques that descend from work on bound the least prime in an arithmetic progression.

If $$\ell\nmid N$$ is a sufficiently large prime (where "sufficiently" is specified in the paper, and in particular only finitely many $$\ell$$ are omitted, the exceptional set depending on $$f$$) and $$\gcd(k-1,\ell-1)=1$$, then for any residue class modulo $$a\pmod{\ell}$$, there exists a prime $$p\nmid \ell N$$ such that $$a_f(p)\equiv a\pmod{\ell}$$ and

$$p\ll \ell^{4600\ell+1200\omega(N)\ell}\mathrm{rad}(N)^{2100\ell}$$.

The implied constant is absolute and effectively computable, $$\omega(N)$$ is the number of distinct prime divisors of $$N$$, and $$\mathrm{rad}(N) = \prod_{p|N}p$$. If I recall correctly, this improves to a bound roughly of the form $$\ll \ell^4 (\log N\ell)^2$$ under GRH for the $$L$$-functions of Hecke characters.

For nontrivial nebentypus, the arguments in Thorner--Zaman should still work, but obnoxious complications might arise.

• In that paper however they are considering non CM new forms with integer coefficients at infinity, which is quite restrictive for what I need. Is this the current state of our knowledge on the matter? Jun 10, 2021 at 16:05
• @Krishnarjun Yes, as far as I understand. I'd like to think that the corresponding results for cusp forms is not too far off (both with and without GRH), but I don't know for sure. I will update the post to reflect the assumption that $f$ is a newform. Jun 10, 2021 at 16:09
• @Krishnarjun Your concern should be a nonissue when the level $N$ is prime, but that might still be too restrictive for your needs. Jun 10, 2021 at 16:15