While giving the first of eight lectures on introductory model theory and its applications yesterday, I stated Hilbert's 17th problem (or rather, Artin's Theorem): if $f \in \mathbb{R}[t_1,\ldots,t_n]$ is positive semidefinite -- i.e., non-negative when evaluated at every $x = (x_1,\ldots,x_n) \in \mathbb{R}^n$ -- then it is a sum of squares of rational functions. One naturally asks (i) must $f$ be a sum of squares of polynomials, and (ii) do we know how many rational functions are necessary? The general answers here are no (Motzkin) and no more than $2^n$ (Pfister). Then I mentioned that the case of $n=1$ is a very nice exercise, because one can prove in this case that indeed $f(t)$ is positive semidefinite iff it is a sum of two (and not necessarily one, clearly) squares of polynomials. Finally I muttered that this was a sort of function field analogue of Fermat's Two Squares Theorem (F2ST).

So I thought about how to prove this result, and I was able to come up with a proof that follows the same recipe as the Gaussian integers proof of F2ST. Then I realized that the key step of the proof was that a monic irreducible quadratic polynomial over $\mathbb{R}$ is a sum of two squares, which can be shown by...completing the square.

But then today I went back to the general setup of a "Gaussian integers" proof, and I came up with the following definition and theorem.

**Definition**: An integral domain $R$ is *imaginary* if $-1$ is a square in its fraction field; otherwise it is *nonimaginary*. (In fact I will mostly be considering Dedekind domains, hence integrally closed, and in this case if $-1$ is a square in the fraction field it's already a square in $R$, so no need to worry much about that distinction.) Note that nonimaginary is a much weaker condition than the fraction field being formally real.

(Definition: An element $f$ in a domain $R$ is a sum of two squares up to a unit if there exist $a,b \in R$ and $u \in R^{\times}$ such that $f = u(a^2+b^2)$.)

**Theorem**: Let $R$ be a nonimaginary domain such that $R[i]$ ($= R[t]/(t^2+1)$) is a PID.

a) Let $p$ be a prime element of $R$ (i.e., $pR$ is a prime ideal). Then $p$ is a sum of two squares up to a unit iff the residue field $R/(p)$ is imaginary.

b) Suppose moreover that $R$ is a PID. Then a nonzero element $f$ of $R$ is a sum of two squares up to a unit iff $\operatorname{ord}_p(f)$ is even for each prime element $p$ of $R$ such that $R/(p)$ is nonimaginary.

[**Proof**: Introduce the "Gaussian" ring $R[i]$ and the norm map $N: R[i] \rightarrow R$. Follow your nose, referring back to the proof of F2ST as needed.]

**Corollaries**: 1) F2ST. 2) Artin-Pfister for $n = 1$. 3) A characterization of sums of two squares in a polynomial ring over a nonimaginary finite field (a 1967 theorem of Leahey).

~~4) Let $p \equiv 3,7 \pmod{20}$ be a prime number. Then $p$ ~~ *is* a sum of two squares up to a unit in $\mathbb{Z}[\sqrt{-5}]$ but is not (by F2ST) a sum of two squares in $\mathbb{Z}$.

Finally the questions:

Have you seen anything like this result before?

I haven't, explicitly, but somehow I feel subconsciously that I may have. It's hard to believe that this is something new under the sun.

What do you make of the strange situation in which $R$ is not a PID but $R[i]$ is?

Note that one might think this impossible, but $R = \mathbb{R}[x,y]/(x^2+y^2-1)$ is an example. [Reference: Theorem 12 of http://math.uga.edu/~pete/ellipticded.pdf.] Do you have any idea about how one might go about producing more such examples, e.g. with $R$ the ring of integers of a number field (or a localization thereof)?

**Addendum**: As I commented on below, a good answer to the first question seems to be the paper

MR0578805 (81h:10028) Choi, M. D.; Lam, T. Y.; Reznick, B.; Rosenberg, A. Sums of squares in some integral domains. J. Algebra 65 (1980), no. 1, 234--256.

In this paper, they prove the theorem above with slightly different hypotheses: $R$ is a nonimaginary UFD such that $R[i]$ is also a UFD. Looking back at my proof, the only reason I assumed PID was not to worry about the distinction between $R/pR$ and its fraction field. Just now I went back to check that everything works okay with PID replaced by UFD. So the second question becomes more important: what are some examples to exploit the fact that $R[i]$, but not $R$, needs to be a UFD?

notan example since the ring of integers of ${\mathbf Q}(\sqrt(-5),i)$ is not $R[i]$ but something bigger: the ratio $(i + sqrt(-5))/2$ is an algebraic integer. The ring of integers in fact is ${\mathbf Z}[(i+\sqrt(-5))/2]$. If $K$ is a quadratic field unramified at 2 then the ring of integers of $K(i)$ is $R[i]$ where $R$ is the ring of integers of $K$. So you need $K = {\mathbf Q}(\sqrt{d})$ with $d \equiv 1 \bmod 4$. I don't believe the 4th corollary now except perhaps if you allow 1/2's, but I don't have a specific counterexample, say for $p=3$. $\endgroup$ – KConrad Jul 8 '10 at 5:507more comments