What is the smallest field which contains all square roots of positive rational numbers? I guess I mean “smallest” in terms of set inclusion, i.e. the minimal one with regard to the “$\subseteq$” relation. The smallest field I know about would be the real algebraic numbers, but I guess that by restricting the degree of minimal polynomials, a smaller field might be possible.
To express this as a formula, I'm looking for the field generated by the set $$ S = \left\{\pm\sqrt x\;\middle\vert\;x\in\mathbb Q\right\} $$
If you know about a field smaller than the algebraic reals which contains $S$, I'd like to know its name and its structure. If you have an argument why the algebraic reals are the smallest field containing $S$, then I'd like to hear the argument or a reference to it.
Update: Answers below indicate that there is such a field containing $S$ and smaller than the real algebraic numbers. So the main issue is finding an established name for this, if there is one.
If you read this question then any $2$-maximal field $\mathbb M$ contains every square root. But such a field is not the real algebraic numbers since it contains no elements of order $3$ such as $\sqrt[3]{2}$. Of course such a field is probably larger than you want. Instead let $a_n$ be an enumeration of the positive integers (or positive primes) and let $L_0=\mathbb Q$, $L_i=L_{i-1}(\sqrt{a_i})$ then
$$L=\bigcup_{i=0}^\infty L_i$$ is your desired field. Notice that $[L_i:L_{i-1}]=1,2$. Every element of $L$ is necessarily has a power of $2$ since each for each finite $n$, we have that $L_n$ is a power of $2$ extension of $L$. We also have that $L$ is not $2$-maximal because $x^2+1$ does not split over it. Notice that $L$ is an abelian extension as well, since it is generated by its degree $2$ subfields which are necessarily abelian and contained in $\mathbb Q^{\mathrm{ab}}$. So its Galois group should be
$$\lim_{\overleftarrow{n \in \mathbb N}} \left(\mathbb Z_2\right)^n.$$ I've never been that good at inverse limits but I imagine that this is $\prod_{n \in \mathbb N} \mathbb Z_2$, if someone could confirm this I would appreciate it.