Hello,


for any given function $F$ of the Selberg class $\mathcal{S}$, let $A\_{F}$ be the set of coefficients $a\_{n}$ of the Dirichlet series defining $F(s)$ for $\Re(s)>1$, and let $A=\bigcup\_{F\in\mathcal{S}}A\_{F}$. Is it true that $\mathbb{Q}(A)=\mathbb{C}$? Same question for $F$ running through $\mathbb{P\_{\mathcal{S}}}$, where $\mathbb{P\_{\mathcal{S}}}$ is the set of all primitive functions of $\mathcal{S}$.
Thank you in advance.



The "maximal" field associated to the Selberg class
Take any Dirichlet series $L(s)=\sum\_{n \geq 1} \frac{a\_n}{n^s}$ in the [Selberg class](http://en.wikipedia.org/wiki/Selberg_class). If $L(s)$ has no pole at $s=1$, then for any $\theta \in \mathbf{R}$, the additive twist $L\_{\theta}(s)=L(s+i\theta) = \sum\_{n \geq 1} \frac{a\_n n^{-i\theta}}{n^s}$ still belongs to the Selberg class. It follows that your set $A$ contains all complex numbers of absolute value $1$ and that $\mathbf{Q}(A)=\mathbf{C}$.


So you probably want to ask your question in a more precise form, by considering Dirichlet series only up to additive twist.


As pointed out by Emerton, it is believed that all functions in the Selberg class come (in a sense which might not be formulated very precisely at the moment) from arithmetic automorphic forms, see for example the abstract of this [preprint](http://arxiv.org/abs/math/9204217) of J. B. Conrey and A. Ghosh. Since the set of all arithmetic automorphic forms is (provably) countable, the set of coefficients of the Dirichlet series which are (conjecturally) associated to them is countable, so cannot generate the field $\mathbf{C}$, whose transcendance degree over $\mathbf{Q}$ is the cardinality of the continuum.


But, as pointed out by Paul Garrett, the question in terms of elements the Selberg class is likely to be completely intractable (like many open questions on the Selberg class).


