Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $f = \sum_n a_n q^n$ be a cuspidal newform of weight $k$ on $\Gamma_0(N)$ for some $N$. Let $K_f$ be the number field generated by the $a_q$ as $q$ runs over all primes.

My question: if we consider the field generated by all but one of these $a_q$, can this field be smaller than $K_f$?

(edited based on Sawin's answer to only look at Fourier coefficients at primes)

share|improve this question
    
For what it's worth this has been asked and answered here before: mathoverflow.net/questions/114145/… –  D. Savitt May 24 '13 at 23:10
    
@Savitt -- Thanks! –  MF1 May 25 '13 at 3:49

2 Answers 2

up vote 9 down vote accepted

No, strong multiplicity one says that all-but-finitely-many of the $a_p$'s determine all the others.

Edit in response to comment/query, and further in response to subsequent comments: ... and, once all the other coefficients are determined by strong multiplicity one (for newforms), invoke Shimura's results (arguably going back to Fricke-Klein in principle) that automorphisms of the complex numbers over $\mathbb Q$ (not merely Galois automorphisms on an algebraic closure of $\mathbb Q$) can be applied to the Fourier coefficients of a holomorphic modular form, producing another. In the case at hand, applying such an automorphism over the field generated by all the given Fourier coefficients, and subtracting, would either give a modular form with only finitely-many non-zero $a_p$'s (impossible), or $0$ (giving the desired result).

[Thanks to MF1 for corrections.]

share|improve this answer
1  
Yes...the missing $a_p$ is uniquely determined by all of the other $a_q$'s for a newform -- but does this imply that this missing $a_p$ lies in the same field as the other $a_q$'s?? –  MF1 May 23 '13 at 17:18
    
@MF1, sorry to be so cryptic. –  paul garrett May 23 '13 at 23:02
    
@paul: Nice...this argument shows that the Galois closure of the field generated by the prime Fourier coefficients is unchanged if I drop one Fourier coefficient. But I don't think there is any reason to expect that the Fourier coefficients of a modular form generate an abelian extension. I think the fields they generate can be fairly complicated. I think it's still open whether the field (and not its Galois closure) can get smaller if one drops a single coefficient. –  MF1 May 24 '13 at 2:27
    
@MF1: you may be right about the field generated by individuals... I was inadvertently thinking about the aggregate... –  paul garrett May 24 '13 at 2:59
    
@paul: actually your argument above is fine (and essentially matches the argument of D. Savitt linked to in the comment above). If every automorphism fixes the missing $a_p$, then $a_p$ is really in the base field (and not its Galois closure). –  MF1 May 25 '13 at 3:51

No. Using all but one of the coefficients, it is easy to compute the Hecke operator eigenvalues using just addition, subtraction, multiplication, and division. Then one can compute the missing coefficient using a Hecke operator.

share|improve this answer
    
I think I see -- I changed the question above to only look at Fourier coefficients at the primes. –  MF1 May 23 '13 at 16:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.