Existence of a maximal field It is well known that $(\Bbb Q^+,\cdot)$ is a free $(\Bbb Z,+)$-module, one of its bases being the set $\Bbb P$ of the prime numbers.
Also, the set $\{r^a:r\in\Bbb Q^+,a\in\Bbb Q\}$ along with the usual product is a $\Bbb Q$-vector space. $\Bbb  P$ is also a basis for this space.
If we allow real exponents and scalars, $\Bbb P$ is no longer a basis. In fact, the dimension of the space would be $1$:
$$pq^{-\log_q p}=1$$
Questions:
Is there a maximal field $F\supset \Bbb Q$ such that $\Bbb P$ is a basis for the $F$-vector space $\{r^a:r\in\Bbb Q^+,a\in F\}$? What would it be?
Is there any field $F$ such that the above mentioned vector space has finite dimension greater than $1$?
EDIT: A claim of no global knowledge about this matter will be appreciated.
 A: It seems that you have to allow finite $\mathbb{Q}$-multiplicative span on the set $\{r^a\}$, to have a vector space. Equivalently, the space that we are interested in is for $\log \mathbb{P}=\{\log 2, \log 3, \ldots, \log p, \ldots \}$, 
$$
\mathrm{span}_{\mathbb{Q}}(\log\mathbb{P})=\mathrm{span}_{\mathbb{Q}} \{ \log 2, \log 3, \ldots, \log p, \ldots \}.
$$
The fact that $\log \mathbb{P}$ provides a basis for the above space, is that $\log \mathbb{P}$ is linearly independent over $\mathbb{Q}$ and the proof is by the fundamental theorem of arithmetic. 
For the first question, I am not sure if it would be maximal, but a result by Baker(1968) states that
$$
\{1\}\cup\log \mathbb{P} \ \textrm{ is linearly independent over }\overline{\mathbb{Q}}.
$$
So, the field $F$ may contain up to the algebraic closure of $\mathbb{Q}$, namely $\overline{\mathbb{Q}}$, to make sure $\log \mathbb{P}$ is linearly independent over $F$. 
For the second question, let $p_1, \ldots, p_n$ be the first $n$ primes. We may consider $F=\mathbb{Q}(\log\mathbb{P}-\{\log p_1, \ldots, \log p_n\})$. This extension $F$ guarantees that 
$$
\dim \mathrm{span}_{F}(\log\mathbb{P}) \leq n.$$
I am not aware of any method that can prove the equality above. It may be an open problem. 
