Field containing all square roots of rational numbers

Question

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.

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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.

Answer 1

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.

Answer 2

As was already said, solution is the limit of infinite sequence of field extensions of rational numbers with square roots of all primes. But I’d point out there is more explicit description.

The smallest ring (and ${\mathbb Q}$-algebra) which contains all square roots of positive rational numbers is, obviously, an infinite(countable)-dimensional vector space over ${\mathbb Q}$ generated by square roots of all square-free natural numbers. Hint: any square root of a positive rational is a rational multiple of one of said generators. Not only are said generators linearly independent, but this ring is a field (the same as constructed in JSchlather’s answer).

Consider an element $x$ of $\langle 1,\sqrt 2, \sqrt 3, \sqrt 5, \sqrt 6,\ldots\rangle$, represented by a formal non-zero linear combination of generators. Let’s prove it is invertible (actual linear independence of generators will be entailed). I formulate it as a proof by induction on maximal prime factor $p_{\rm max}(x)$ of all natural numbers under square root (define $p_{\rm max}(x)=1$ if “$\sqrt 1$” is the only one); essentially it is the same as a sequence of extensions.
 Base of induction: $p_{\rm max}(x) = 1$, i.e. $x$ is a non-zero rational number. Invertible, since ${\mathbb Q}$ is a field.
 Step of induction: $p_{\rm max}(x)$ is a prime. Let’s rewrite our element in the form $x = B + A\sqrt{p_{\rm max}(x)}$ where $B, A$ are elements of the same ring and do not contain roots of $p_{\rm max}(x)$ or its multiples (exercise). Then, $$ x (B - A\sqrt{p_{\rm max}(x)}) = B^2 - p_{\rm max}(x)A^2.$$ First of all, $B, A$ do not contain roots of $p_{\rm max}(x)$, higher primes, and their multiples (that implies their $p_{\rm max}$ is less). From divisibility considerations (or “$p$-adic norm” who likes it more) right-hand side is non-zero, hence (due to induction) invertible. Then, $$ x^{-1} = \frac{B - A\sqrt{p_{\rm max}(x)}}{B^2 - p_{\rm max}(x)A^2}.$$

Answer 3

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.