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A Pythagorean field is one in which every sum of two squares is again a square. $\mathbb Q$ is not Pythagorean, which is easy to see. I have read a theorem online which says that every field has a unique (up to isomorphism) Pythagorean closure. I haven't found the proof so I thought I should start with the most familiar field, $\mathbb Q$.

I was thinking if it would be possible to somehow imagine or characterize the Pythagorean closure $\mathbb P$ of $\mathbb Q.$ I know that in the field $\mathbb E$ of all constructible numbers, every positive number is a square because it is possible to construct square roots of already constructed numbers (by drawing a certain right triangle and its altitude). So $\mathbb E$ must be Pythagorean. $(1)$ But is it equal to $\mathbb P?$ If it's not, then what is the $\mathbb P$-dimension of $\mathbb E$?

Surely, for any $q_1,q_2,\ldots,q_n\in \mathbb Q,$ we must have $\sqrt{q_1^2+q_2^2+\cdots + q_n^2}\in \mathbb P$ by a simple induction. $(2)$ Is this all we have to adjoin to $\mathbb Q$ to obtain $\mathbb P?$ It looks like it is but I haven't done anything with infinite extensions and I don't know how to handle this.

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I imagine "unique Pythagorean closure" could mean unique up to isomorphism. This leaves a question of whether two distinct but isomorphic subfields of $\mathbb{R}$ could be Pythagorean closures of $\mathbb{Q}$. But certainly there's one that you get just by closing under the usual field operations and taking square roots. – Michael Hardy Mar 5 '12 at 18:25
@CamMcLeman I'm looking for a basis. I think the two equations I wrote down may be helpful in finding it because they give a way to think about less complicated expressions. – user23211 Mar 5 '12 at 22:05
@CamMcLeman Could you please comment here? I think if I reply here it will create a mess. – user23211 Mar 5 '12 at 22:13
@ymar: Already each non-negative rational is the sum of four squares of rationals, so the $n$ above need not be large! – André Nicolas Mar 5 '12 at 22:31
@ymar: Look at $a/b=ab/b^2$. Express $ab$ as the sum of the squares of the integers $s$, $t$, $u$, $v$. Then $a/b=(s/b)^2+(t/b)^2+(u/b)^2+(v/b)^2$. – André Nicolas Mar 5 '12 at 23:01
up vote 7 down vote accepted

No, the two fields are not the same: It is a result of Hilbert that the Pythagorean field is the maximal totally real subfield of the field of constructible numbers. So any constructible number which is not totally real (i.e., its minimal polynomial has complex roots) gives a non-Pythagorean Euclidean number. An easy example is $\sqrt{1+\sqrt{2}}$ (with non-real conjugate $\sqrt{1-\sqrt{2}}$). Generalizing this example, it's easy to see that $\mathbb{E}$ is infinite-dimensional over $\mathbb{P}$.

I believe your second question is answer in the affirmative by using that $\mathbb{P}$ is the smallest subfield of the Euclidean numbers closed under the operation of $x\rightarrow \sqrt{1+x^2}$, and inducting on the number of such operations you'd have to apply. I'd take a look at Roger Alperin's series of papers on trisections and origami for a good discussion of the fields in question (and others!).

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Thank you very much. Could you please take a look at my edited question? – user23211 Mar 5 '12 at 21:51
@ymar: You're welcome! Though I think it makes more sense to ask a separate follow-up question than to append to the current one (to avoid confusion about which answers are answering which parts). In any case, I agree that it's clear that $(\mathbb{P}:\mathbb{Q})$ is infinite. – Cam McLeman Mar 5 '12 at 21:59
OK, I'll cut-paste this to a new question. – user23211 Mar 5 '12 at 22:07

I treat the concept of Pythagorean closure in these notes. In fact, I speak briefly about a class of fields being "closable" and give the Pythagorean closure as an example of that.

Let $F$ be a field. Here are two key (but easy) observations:
(CC1) Any algebraically closed field extension of $F$ is Pythagorean.
(CC2) If $\{K_i\}_{i \in I}$ is a family of Pythagorean field extensions of $F$ -- all contained in some large field $L$ -- then also $K = \bigcap_{i \in I} K_i$ is a Pythagorean field extension of $F$.

It follows that inside any algebraic closure $\overline{F}$ of $F$ there is a literally unique Pythagorean closure: the intersection of all Pythagorean algebraic extensions of $F$. (When $F = \mathbb{Q}$, Cam McLeman has given a spot-on description of it.)

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They seem to be fantastic notes. I love that you included references. I've never seen lecture notes with references before! I will try to read them, but I may have to return to them later when I learn more about field theory. – user23211 Mar 7 '12 at 16:52
@ymar: thanks. Actually they are not really lecture notes in the sense that I have not (at least not yet) taught any corresponding course. They are more like "study notes" for me and my students...hence the references. (Although I do try to include references in my lecture notes as well...) – Pete L. Clark Mar 7 '12 at 17:19
(Same as my answer, but you explicitly treated uniqueness.) I might have written something like "If $\mathcal{K}$ is a family of Pythagorean field extensions of $F$ -- all contained in some large field $L$ -- then also $k=\bigcap\limits_{K\in\mathcal{K}} K$ is a Pythagorean field extension of $F$." Do the indices $i\in I$ add anything? – Michael Hardy Mar 8 '12 at 13:50
@Michael: Yes, my answer is quite similar to yours...but the uniqueness of the Pythagorean closure was explicitly part of the OP's question, so I felt it was worth it to nail down an answer to that. "Do the indices $i \in I$ add anything?" Not really, no. The way you might have written it seems just as good... – Pete L. Clark Mar 8 '12 at 13:58

When I think of the question of how to prove that every field has a unique Pythagorean closure, the first thing that comes to mind is that I might start with the fact that every field has a unique algebraic closure. Once you've got that, you wouldn't have to construct a field from scratch; you'd just look at the intersection of all subfields of the algebraic closure that include the field you started with and are closed under $(a,b)\mapsto\sqrt{a^2+b^2}$. The algebraic closure is closed under that operation, so you'd get something.

Then you'd have the problem of proving uniqueness. I expect that means unique up to isomorphism. So could there be distinct but isomorphic subfields of the algebraic closure that are Pythagorean closures of the one you started with? I suspect not.

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Thank you very much! – user23211 Mar 5 '12 at 22:28

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