# What is the ring of integers of the Pythagorean field?

Following Hilbert, we call the complex numbers constructible via compass and straight-edge the field of Euclidean numbers, and the totally real such numbers the field of Pythagorean numbers. (Among other possible definitions, an algebraic number is totally real if its minimal polynomial has all real roots). For a reference, Richard Alperin gives a description of these and related fields from a constructibility viewpoint in his paper "Trisections and Totally Real Origami."

There is a remarkably nice characterization of the Pythagorean numbers -- the Pythagorean field is the smallest field containing the rationals and closed under the operation $x\rightarrow \sqrt{1+x^2}$. Or, from an only slightly different viewpoint, it is the Pythagorean closure of $\mathbb{Q}$, in the sense of

http://planetmath.org/encyclopedia/PythagoreanField.html

Because it's a nice "hands-on" intro to this field, let me include in the question Daniel Litt's comment below that since $\sqrt{2}=\sqrt{1+1^2}$, and $\sqrt{3}=\sqrt{1+\sqrt{2}^2}$, and so on, the Pythagorean field contains $\sqrt{n}$ for all $n\geq 0$, and hence contains the compositum of all real quadratic fields.

My Question:

What is the ring of integers of the Pythagorean field?


Note that the most naive guess of it being the smallest subring of algebraic integers closed under the operation $x\rightarrow \sqrt{1+x^2}$ is incorrect -- this ring does not include $\frac{1+\sqrt{5}}{2}$, which is certainly a totally real Euclidean algebraic integer. I suspect/hope (though this may just be the second most naive guess) that there's some description of the form "smallest subring of the algebraic integers closed under $x\rightarrow \sqrt{1+x^2}$ and division by 2 when certain conditions are met." I've done a little bit of a literature search on rings of integers of totally real multiquadratic extensions of $\mathbb{Q}$, but haven't found anything even remotely inspiring something of this form.

I don't have much to offer in terms of motivation, except that I have come across a variety of rings of integers in my research, and I'm trying to decide if any are exactly the ring of Pythagorean integers. It would be nice to be able to compare them to the Pythagorean integers just by seeing whether or not one of these rings satisfies certain closure operations.

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Nope, definitely not. I debated on that tag...pretty sure I've seen some authors refer to infinite (algebraic) extensions still as number fields, but this is probably less standard. No opposition if someone feels strongly that it should be deleted. – Cam McLeman Jul 27 '10 at 18:59
Quick note: This is not actually a number field, as it is not a finite extension of $\mathbb{Q}$; in particular, it contains $1$ and thus $\sqrt{2}$, and thus $\sqrt{3}$, and so on. That is, it contains $\sqrt{n}$ for all $n\in mathbb{N}$. – Daniel Litt Jul 27 '10 at 19:00
I googled "ring of integers" "pythagorean field" and it gave me page 260 in Lam's book Introduction to quadratic forms over fields, all of chapter 8 is related books.google.com/… – Will Jagy Jul 27 '10 at 19:32
I looked through the chapter -- very interesting, but I don't think it addresses the question at hand. The unique reference to "ring of integers" is to $\mathbb{Z}$ itself. – Cam McLeman Jul 27 '10 at 22:45
This is an interesting question. FWIW, my initial opinion is that hoping for an "iterative" or "closure-theoretic" description of the union of the rings of integers of all Pythagorean number fields is overly optimistic. But that's just an impression, and either way the problem is worth thinking about. – Pete L. Clark Jul 28 '10 at 7:40

The first idea I tried to look at was replacing the iteration by $$x \mapsto x' = \frac{1 + \sqrt{1+4x^2}}{2},$$ which eliminates the problem with $\frac{1+\sqrt{5}}{2}$; on the other hand I cannot see that e.g. the integral element $\frac{1 + \sqrt{3}}{\sqrt{2}}$ lies in the ring generated by this iteration.
By modifying this approach we can eliminate all obvious counterexamples. Define $v_0 = 2$ and $$v_1 = \sqrt{2}, v_2 = \sqrt{2+\sqrt{2}}, ...., v_{n+1} = \sqrt{2+v_n}, ... .$$ Given an algebraic integer $x$ in the Pythagorean ring, let $v(x)$ denote the "maximal" product of elements $v_i$ dividing both $2$ and $x$ (I guess that a proof of the existence of $v(x)$ is not too hard; Krull, Herbrand and Scholz studied the arithmetic in these "infinite number fields" in the 1930s, and their methods should suffice). Then the iteration $$x \mapsto x' = \frac{1+x + \sqrt{1+x^2}}{v(x)}$$ sends algebraic integers to algebraic integers, and this map has at least a chance of generating the ring of all integers in the Pythagorean field.
I'm trying to wrap my head around this definition of $v(x)$. Is $v(x)$ something like "compute the gcd of 2 and $x$ in the compositum of $Q(x)$ with the cyclotomic 2-extension of the rationals?" – Cam McLeman Sep 30 '10 at 17:19