# Does weight $2$ cuspidal Bianchi modular form have infinitely many zero Fourier coefficient at prime ideals?

Let $$K$$ be an imaginary quadratic field. Let $$f \in S_2(\mathfrak{n})$$ be a weight $$2$$ cuspidal cof level $$\Gamma_0(\mathfrak{n})$$ over $$K$$ (for definitions one can see http://www.lmfdb.org/knowledge/show/mf.bianchi.bianchimodularforms ), consider its $$L$$-function

$$L(f,s)=\sum_{\mathcal{P} \not =0 \in Spec(O_K)} c(\mathcal{P}) (N\mathcal{P})^{-s}$$

here $$c(\mathcal{P})$$ are the Fourier coefficients at the cusp $$\infty$$ and we assume they are all rational numbers.

The question is, do we know that $$c(\mathcal{P})=0$$ for infinitely many prime ideals $$\mathcal{P}$$? How about $$c(\mathcal{P})=0 \mod N(\mathcal{P})$$?

Motivation: if we are considering similar problems over $$\mathbb Q$$, then we know $$a_p=0$$ for infinitely many $$p$$ because the corresponding elliptic curve over $$\mathbb Q$$ must have supersingular reduction at infinitely many primes.

Edit: as the answer shows, we shall expect $$c(\mathcal{P})=0 \mod p$$ happens for infinitely many times, rather than $$\mod N(\mathcal{P})$$.

• Do these (conjecturally) correspond to elliptic curves over imaginary quadratic field? I don't think there is a single non-CM elliptic curve over a field with no real places where we know there are infinitely many supersingular primes. If you pick your favorite no-CM elliptic curve over an imaginary quadratic field, and check it is modular, then I think we have no clue how to check it has infinitely many vanishing Fourier coefficients. – Will Sawin Sep 11 '19 at 19:30
• @Will Sawin I hope there is an automorphic approach to this problem, because modularity is so difficult.. – sawdada Sep 11 '19 at 23:32
• I'm trying to argue that there is no automorphic approach. Checking modularity for a given elliptic curve is easy nowadays (I think). But there is no progress on this problem in either the geometric or automorphic settings. – Will Sawin Sep 12 '19 at 1:40
• There is indeed no automorphic approach, AFIAK, which is why the corresponding statement is completely unknown for elliptic modular forms of weight $> 2$. – David Loeffler Sep 12 '19 at 9:43

The answer to your first problem is that we do not know, since the result is not even true. Take a modular elliptic curve $$E$$ over $$K$$ (say a base change from Q) such that the 4-torsion is defined over $$K$$, certainly possible when $$K=\mathbf{Q}(i)$$. Then the action of Frobenius on the $$4$$-torsion is trivial and so one has the congruence $$c(\mathcal{P}) \equiv 2 \bmod 4$$, and so can never be zero. In particular, any supersingular prime cannot have prime norm (this is related to the difficulty of generalizing Elkies theorem to this context.) The second problem (congruence rather than equality) is open for the reasons described in the comments.
• The congruence version also has a negative answer for these curves, because sawdada put the congruence modulo $N(\mathcal P)$ and not modulo the characteristic of $p$. Because $|c (\mathcal P )| \leq 2 \sqrt{ N (\mathcal P)}$, the congruence can never happen for nonzero $c(\mathcal P)$ unless $N(\mathcal P)\leq 4$. – Will Sawin Sep 20 '19 at 0:00
• I give the OP the benefit of the doubt and assume they meant mod $p$ since otherwise as you correctly say they are equivalent for large p (in the context of Elliptic curves). – Orac Sep 20 '19 at 0:07
• That's probably a good idea! Then let me make the comment I was going to make before I noticed that, which is that for these curves, the mod $p$ congruence can only happen for inert primes when $c(\mathcal P)$ is exactly $\pm 2 p$, in other words at the very ends of the possible range. We still conjecture infinitely many here, but they should be so rare it seems hard to see how to prove it. – Will Sawin Sep 20 '19 at 0:45
• Thank you. You're right, one shall mean mod $p$, otherwise the question is false for your examples. I notice in a recent paper arxiv.org/abs/1909.07473 where they prove every 2-dimensional abelian scheme over $O_K$ ($K$ can be any number field) admits infinitely many places of geometrically non simple reductions. So it seems interesting that one can't say much for elliptic curves over imaginary quadratic fields. – sawdada Sep 20 '19 at 2:06