Field containing all square roots of rational numbers 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.

Unless I missed something in your question, your field $S$ is obtained by adjoining to $\mathbf Q$ all the square roots of all the positive rational numbers, or equivalently, of all the natural integers, or equivalently, of all the positive rational primes $p$. As a compositum of quadratic fields, $S$ is thus the maximal real subfield contained in $T$ :=  the maximal abelian extension of $\mathbf Q$ of exponent 2, which in turn is the maximal extension of exponent 2 contained in $\mathbf Q^{ab} $ := the maximal abelian extension of $\mathbf Q$ (described by the Kronecker-Weber theorem) . Obviously  $T$ is the compositum of $\mathbf Q (i)$ and $S$, two linearly disjoint extensions of $\mathbf Q$, and $T = S(i)$, where $ i^2 = - 1$. Note that, by Kummer theory, $Gal(T/ \mathbf Q) \cong Hom (\mathbf Q^* / (\mathbf Q^*)^2,(\pm1))$, and the Kummer radical of $Gal(S/ \mathbf Q)$ is an $\mathbf F_2$ - vector space which admits a basis consisting of the classes of  all the positive rational primes. Which  probably "explains" why $S$ has no official name.
