# 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.

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An interesting field that properly contains your field is the field of Euclidean constructible numbers. –  André Nicolas Feb 1 '13 at 19:55
@AndréNicolas: Even though constructible numbers appear to be not minimal, they make for interesting reading. Thanks for providing that name! –  MvG Feb 1 '13 at 20:12

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

$$\lim_{\leftarrow}_{n \in \mathbb N} \left(\mathbb Z_2\right)^n.$$ 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.

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@JonasMeyer, Yes my mistake I didn't read the question thoroughly and not taking primes was just my way of being lazy. Fixed now. –  JSchlather Feb 1 '13 at 19:57
These $k$-powerful and $k$-maximal fields take some getting used to, particularly since I couldn't yet locate any documents on these in Wikipedia or MathWorld. Are there some other terms I could use to search for this? But your construction of $L$ is nice since $L$ is obviously a field, looks pretty minimal, and therefore gives a good idea of what elements I require to make my $S$ into a field. If only I had a name for this. I'll leave this question active for a while, see if someone can provide that name, but if not will gladly accept your answer. –  MvG Feb 1 '13 at 20:25
@MvG You might ask Ewan Delanoy, but I believe its a name that he invented. So I don't know of any other terminology for it. If you happen to stumble on any references to them I'd be glad to read more about them. We still haven't decided whether or not $3$-maximal or rather $p$-maximal fields are unique up-to isomorphism. I have a related question on the matter here. –  JSchlather Feb 1 '13 at 20:30
Amusing: after writing (a draft of) my answer for 2+ hours I realized that implicitly used the same construction, the injective limit of an infinite sequence of field extensions. –  Incnis Mrsi Dec 14 at 23:28

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, $$x^{-1} = \frac{B - A\sqrt{p_{\rm max}(x)}}{B^2 - p_{\rm max}(x)A^2}.$$

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