The First Newform on Γ_{0}(N) such that Q(a_{n}) ≠ Q(…, a_{m}, …) for all n

William A. Stein

July 16, 2002

Abstract We compute the minimum N such that there is a newform f = ∑ a_{n}q^{n} ∈ S_{2}(Γ_{0}(N)) with the odd property that Q(a_{n}) ≠ Q(…, a_{m}, …) for all n.^{1}

1 The Newform

Using modular symbols, we computed the first few terms of each newform f = ∑ a_{n}q^{n} on S_{2}(Γ_{0}(N)) for N ≤ 512. For each newform of level < 512, we found that there exists an integer n such that the field K_{f} = Q(…, a_{m}, …) generated by all Fourier coefficients of f equals the field Q(a_{n}) generated by a single coefficient. This lead me to suspect that we always have equality.

We do not always have equality. The characteristic polynomial of T_{3} on S_{2}^{new}(Γ_{0}(512)) is (x^{2} − 6)^{2}(x^{2} − 2)^{4}(x^{2} + 4x + 2)(x^{2} − 4x + 2), and if we let f = ∑ a_{n}q^{n} be one of the newforms in the four-dimensional kernel V of T_{3}^{2} − 6, then

f = q − \sqrt{6}q^{3} + 2\sqrt{3}q^{5} + 2\sqrt{2}q^{7} + 3q^{9} − \sqrt{6}q^{11} − 2\sqrt{3}q^{13} − 6\sqrt{2}q^{15} + ⋯

provides an example in which K_{f} ≠ Q(a_{n}) for some n.

We will use the following theorem (see [1, Prop. 3.64]):

Theorem 1.1 (Shimura). Let N, r, s, k > 0 be integers such that s|N and let M be the least common multiple of N, r^{2}, and rs. Let χ (resp. Ψ) be a primitive character mod r (resp. s). If f = ∑ a_{n}q^{n} ∈ S_{k}(Γ_{0}(N), Ψ) then

∑ χ(n)a_{n}q^{n} ∈ S_{k}(Γ_{0}(M), Ψχ^{2}).

1Henri Cohen told me that he discoved this example many years ago, and Elkies “explained” it to him. This example is not in the literature, and I found it indepedently of Cohen and Elkies.

Lemma 1.2. Suppose n ≥ 6 and let ε be a primitive Dirichlet character of conductor dividing 8. Then the map f ↦ f ⊗ ε preserves S_{k}(Γ_{0}(2^{n})).

Proof. Set r = 8 in Theorem 1.1, and note that ε^{2} = 1.

It is clear from the q-expansion of f that Q(\sqrt{2}, \sqrt{3}) ⊂ K_{f}. The newform f and its companions lie inside of the kernel V of T_{3}^{2} − 6. A modular symbols computation shows that this kernel has dimension 4, which proves that K_{f} = Q(\sqrt{2}, \sqrt{3}).

Proposition 1.3. There is no integer n such that K_{f} ≠ Q(a_{n}).

Proof. Let χ be a character of conductor dividing 8. By Lemma 1.2, f ⊗ χ lies in S_{2}(Γ_{0}(512)). We check computationally that f ⊗ χ is one of the four Galois-conjugates of f, and that each conjugate is of the form f ⊗ χ for some χ. Let n > 2 be a positive integer and let σ be an automorphism. Then we have just showed that σ(a_{n}) = χ(n)a_{n} for some character χ of order at most 2, so χ(n) ∈ \{±1\}. Thus the action of Gal(\overline{Q}/Q) on a_{n} factors through {±1}, so Q(a_{n}) has degree at most 2.

Acknowledgment: It is a pleasure to thank Kevin Buzzard and Ken Ribet for helpful comments.

References

[1] G. Shimura, Introduction to the Arithmetic Theory of Automorphic Functions, Princeton University Press, (1994).