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This is a follow-up question to this one. I asked it there first but moved it here following the advice from Cam McLeman.

I tried to prove that $(\mathbb P:\mathbb Q)=\aleph_0$ and I think I succeeded: it's clear that the square of every natural number is in $\mathbb P$. Indeed, $$\sqrt n=(\underbrace{1^2+1^2+\ldots+1^2}_n)^{1/2}\in\mathbb P.$$ In this question I learned how to prove that the squares of primes are linearly independent over $\mathbb Q.$ Thus $\{\sqrt p\,|\,p \text{ is prime}\}\subset \mathbb P$ is an infinite linearly independent set.

Can we construct a $\mathbb Q$-basis of $\mathbb P?$ We have for $p_1,p_2,\ldots,p_m,q_1,q_2,\ldots,q_n\in \mathbb Q$

$$

\frac 1 {\sqrt{p_1^2+p_2^2+\ldots+p_m^2}}=\frac{1}{p_1^2+p_2^2+\ldots+p_m^2}\cdot\sqrt{p_1^2+p_2^2+\ldots+p_m^2} $$ and $$ \sqrt{p_1^2+p_2^2+\ldots+p_m^2}\cdot \sqrt{q_1^2+q_2^2+\ldots+q_n^2}=\sqrt{\sum_{1\leq i\leq m,1\leq j\leq n}(p_iq_j)^2}.

$$

So I think every number in $\mathbb P$ should be possible to write as a linear combination of numbers of the form $\sqrt{p_1^2+p_2^2+\ldots+p_m^2}.$ But the set of such numbers cannot possibly be linearly indepentent over $\mathbb Q.$ (Although I think it's a step in the right direction, isn't it?)

EDIT I'll try to clarify why I think the above equations are relevant. I think I know that every number in $\mathbb P$ is a result of iterating field operations on the set of numbers of the form $\sqrt{p_1^2+\ldots +p_n^2}.$ The above equations tell me that I don't need muliplication and multiplicative inversion in this procedure. So every number in $\mathbb P$ is a result of iterating addition and additive inversion on the set of numbers of the form $\sqrt{p_1^2+\ldots +p_n^2}.$ These two happen to be vector space operations. So this set generates the whole $\mathbb P$ as a vector space.

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  • I think there's some confusion between a basis and a minimal adjoining set. For example, your last equations show how elements of the Pythagorean field relate multiplicatively, but this doesn't say anything about linear independence. For example, $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{6}$ are $\mathbb{Q}$-linearly independent, but $\sqrt{6}\in\mathbb{Q}(\sqrt{2},\sqrt{3})$. – Cam McLeman Mar 05 '12 at 22:16
  • @CamMcLeman I agree that this says nothing about linear independence but I think it says something about generating, which is the second thing a basis has to do. I'm simply saying that the numbers of the form $\sqrt{p_1^2+\ldots+p_n^2}$ generate $\mathbb P$ as a vector space over $\mathbb Q$ because inverting and multiplying such numbers doesn't lead us outside the set of such numbers multiplied by scalars from $\mathbb Q.$ –  Mar 05 '12 at 22:26
  • Inverting and multiplying are not things you do when generating something as a vector space. – Gerry Myerson Mar 05 '12 at 22:33
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    @ymar: If a certain field $F$ is obtainable by adding finitely many algebraic numbers to $\mathbb{Q}$, then $F$ has finite dimension as a vector space over $\mathbb{Q}$. The square roots of the primes, indeed any positive integers, are in $\mathbb{P}$, so since you already know the square roots of the primes are linearly independent over $\mathbb{Q}$, you are finished. More delicate would be to prove the undoubtedly true fact that the field of (say real) constructible numbers has infinite dimension over $\mathbb{P}$. – André Nicolas Mar 05 '12 at 22:45
  • @GerryMyerson I have edited the question. –  Mar 05 '12 at 22:45
  • @AndréNicolas Yes, that was part of my previous question. Cam McLeman has given a hint and I'll try to do this later. But what about a $\mathbb Q$-basis of $\mathbb P$? –  Mar 05 '12 at 22:48
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    @ymar: Any such basis is, by my comment above, also by your work, countably infinite. I would no more know how to construct a nice basis for $\mathbb{P}$ as a vector space over $\mathbb{Q}$ than how to construct a nice basis for the algebraic numbers over $\mathbb{Q}$. One can cheat and enumerate $\mathbb{P}$, then climb up the enumeration, keeping a number if it is not a linear combination of previous ones, discarding it otherwise. – André Nicolas Mar 05 '12 at 22:58
  • @AndréNicolas But that would be very difficult to do constructively, wouldn't it? We would have to know which numbers we are discarding. And the very enumeration seems to be problematic to find constructively. I suspected that it might be hard to construct a basis of this space, but is there an argument for that? Some infinitely-dimensional spaces do have concrete bases. (Although the only examples I know are those which are defined as the spaces spanned by a given linearly independent set.) –  Mar 05 '12 at 23:08
  • @ymar: Constructively is likely not a problem. (I am sure of the details only for enumerating the algebraic numbers.) Explicitly, whatever that may mean, is a different matter, though a computer program that does it can be explicitly given. – André Nicolas Mar 05 '12 at 23:18
  • @AndréNicolas Could you please refer me to an article or a book with those details? I'd like to learn more. –  Mar 05 '12 at 23:21
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    @ymar: I did it once long ago, to win an argument about whether Cantor's proof of the existence of transcendentals is constructive (it is). Do not know a standard source for the details. – André Nicolas Mar 05 '12 at 23:28

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