# sum of squares in ring of integers

Lagrange proved that every (positive) rational integer is a sum of 4 squares.

Are there general results like this for ring of integers of a number field? Is this class field theory?

Explicity, suppose a number field is formally real. Denote its ring of integers as Z. Is it true for every algebraic integer x in Z, that either x or -x is a sum of squares?

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This is the quadratic case of Waring's problem for number fields. See C. L. Siegel's paper: jstor.org/pss/2371900 – S. Carnahan Feb 7 '10 at 4:42
Thank you KConrad for clarifying the distinction between the integral and the fractional. Now I realize I meant to ask about the integral situation. – user2529 Feb 11 '10 at 8:57

To address the particularities of this question for number fields, the basic theorem is attributed to Hilbert, Landau and Siegel. First of all, any nonzero sum of squares in a number field has to be totally positive (that is, it is positive in all real embeddings). Hilbert (1902) conjectured that in any number field, a totally positive element is a sum of 4 squares in the number field. This was proved by Landau (1919) for quadratic fields and by Siegel (1921) for all number fields.

This sounds superficially like a direct extension of Lagrange's theorem, but there is a catch: it is about field elements, not algebraic integers as sums of squares of algebraic integers. A totally positive algebraic integer in a number field $K$ need not be a sum of 4 squares of algebraic integers in $K$. The Hilbert-Landau-Siegel theorem only says it is a sum of 4 squares of algebraic numbers in $K$.

For instance, in $\mathbf{Q}(i)$ all elements are totally positive in a vacuous sense (no real embeddings), so every element is a sum of four squares. As an example, $$i = \left(\frac{1+i}{2}\right)^2 + \left(\frac{1+i}{2}\right)^2.$$ This shows $i$ is a sum of two squares in $\mathbf{Q}(i)$. It is impossible to write $i$ as a finite sum of squares in ${\mathbf Z}[i]$ since $$(a+bi)^2 = a^2 - b^2 + 2abi$$ has even imaginary part when $a$ and $b$ are in $\mathbf{Z}$. Thus any finite sum of squares in $\mathbf{Z}[i]$ has even imaginary part, so such a sum can't equal $i$. Therefore it is false that every totally positive algebraic integer in a number field is a sum of 4 squares (or even any number of squares) of algebraic integers.

Here are some further examples:

1. In $\mathbf{Q}(\sqrt{2})$, $5 + 3\sqrt{2}$ is totally positive since $5+3\sqrt{2}$ and $5-3\sqrt{2}$ are both positive. So it must be a sum of at most four squares in this field by Hilbert's theorem, and with a little fiddling around you find $$5 + 3\sqrt{2} = (1+\sqrt{2})^2 + \left(1 + \frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2.$$ It is impossible to write $5 + 3\sqrt{2}$ as a sum of squares in the ring of integers $\mathbf{Z}[\sqrt{2}]$ because of the parity obstruction we saw for $i$ as a sum of squares in $\mathbf{Z}[i]$: the coefficient of $\sqrt{2}$ in $5 + 3\sqrt{2}$ is odd.

2. In $\mathbf{Q}(\sqrt{2})$, $\sqrt{2}$ is not totally positive (it becomes negative when we replace $\sqrt{2}$ with $-\sqrt{2}$), so it can't be a sum of squares in this field. But in the larger field $\mathbf{Q}(\sqrt{2},i)$, everything is totally positive in a vacuous sense so everything is a sum of at most four squares in this field by the Hilbert-Landau-Siegel theorem. And looking at $\sqrt{2}$ in $\mathbf{Q}(\sqrt{2},i)$, we find $$\sqrt{2} = \left(1 + \frac{1}{\sqrt{2}}\right)^2 + i^2 + \left(\frac{i}{\sqrt{2}}\right)^2.$$

Hilbert made his conjecture on totally positive numbers being sums of four squares as a theorem, in his Foundations of Geometry. It is Theorem 42. He says the proof is quite hard, and no proof is included. A copy of the book (in English) is available at the time I write this as http://math.berkeley.edu/~wodzicki/160/Hilbert.pdf. See page 83 of the file (= page 78 of the book).

Siegel's work on this theorem/conjecture was done just before the Hasse-Minkowski theorem was established in all number fields (by Hasse), and the former can be regarded as a special instance of the latter.

Indeed, for nonzero $\alpha$ in a number field $K$, consider the quadratic form $$Q(x_1,x_2,x_3,x_4,x_5) = x_1^2+x_2^2+x_3^2+x_4^2-\alpha{x}_5^2.$$ To say $\alpha$ is a sum of four squares in $K$ is equivalent to saying $Q$ has a nontrivial zero over $K$. (In one direction, if $\alpha$ is a sum of four squares over $K$ then $Q$ has a nontrivial zero over $K$ where $x_5 = 1$. In the other direction, if $Q$ has a nontrivial zero over $K$ where $x_5 \not= 0$ then we can scale and make $x_5 = 1$, thus exhibiting $\alpha$ as a sum of four squares in $K$. If $Q$ has a nontrivial zero over $K$ where $x_5 = 0$ then the sum of four squares quadratic form represents 0 nontrivially over $K$ and thus it is universal over $K$, so it represents $\alpha$ over $K$.) By Hasse-Minkowski, $Q$ represents 0 nontrivially over $K$ if and only if it represents 0 nontrivially over every completion of $K$.

Since any nondegenerate quadratic form in five or more variables over a local field or the complex numbers represents 0 nontrivially, $Q$ represents 0 nontrivially over $K$ if and only it represents 0 nontrivially in every completion of $K$ that is isomorphic to ${\mathbf R}$. The real completions of $K$ arise precisely from embeddings $K \rightarrow {\mathbf R}$. For $t \in {\mathbf R}^\times$, the equation $x_1^2+x_2^2+x_3^2+x_4^2-t{x}_5^2 =0$ has a nontrivial real solution if and only if $t > 0$, so $Q$ has a nontrivial representation of 0 in every real completion of $K$ if and only if $\alpha$ is positive in every embedding of $K$ into ${\mathbf R}$, which is what it means for $\alpha$ to be totally positive. (Strictly speaking, to be totally positive in a field means being positive in every ordering on the field. The orderings on a number field all arise from embeddings of the number field into $\mathbf R$, so being totally positive in a number field is the same as being positive in every real completion.)

Siegel's paper is "Darstellung total positiver Zahlen durch Quadrate, Math. Zeit. 11 (1921), 246--275, and can be found online at http://gdz.sub.uni-goettingen.de/en/dms/loader/img/?PPN=PPN266833020_0011&DMDID=DMDLOG_0022.

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@KC: If you decide to say even more (which would be great -- there seems to be plenty more to say), consider adding it as a separate answer. On the one hand it is nice if any given answer does not discuss too many different examples, and on the other hand you can get rewarded (in the MO sense of rep) again for your additional good work. – Pete L. Clark Feb 7 '10 at 22:28
I wanted to keep my answer in one piece rather than spread it across multiple submissions (you refer to my later adding the further examples). I was going to include "LATER EDIT" in my response to indicate the place where I was adding something later, but decided not to since 3 weeks from now what's the difference that some part was added? – KConrad Feb 7 '10 at 23:45

K. Conrad's answer shows that one must, in general, make a distinction between rational representations of sums of squares and integral representations of sums of squares and that the latter is significantly more subtle.

Still, a lot of work has been done. (I myself am familiar with only a little of it.) The following classic paper gives comprehensive results for imaginary quadratic fields:

Niven, Ivan Integers of quadratic fields as sums of squares. Trans. Amer. Math. Soc. 48, (1940). 405--417.

http://math.uga.edu/~pete/Niven40.pdf

Niven shows that the obstruction pointed out by Conrad is essentially the only one to representing integers in an imaginary quadratic field as sums of squares. More precisely:

Let $m$ be a squarefree positive integer, and put $K = \mathbb{Q}(\sqrt{-m})$, $\mathbb{Z}_K$ the ring of integers of $K$.

Case 1: $m \equiv 1 \pmod 4$. In this case $\mathbb{Z}_K = \mathbb{Z}[\sqrt{-m}]$ and there is an obstruction as above. Namely, an element $a + b \sqrt{-m}$ is a sum of squares in $\mathbb{Z}_K$ iff it is a sum of $3$ squares in $\mathbb{Z}_K$ iff $b$ is even.

Case 2: $m \equiv 3 \pmod 4$. In this case $\mathbb{Z}_K = \mathbb{Z}[\frac{1+\sqrt{-m}}{2}]$ and the obstruction of the previous case disappears: every element of $\mathbb{Z}_K$ is a sum of 3 squares in $\mathbb{Z}_K$.

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@few_reps: I have repaired the link, so you can see for yourself. – Pete L. Clark Sep 24 '15 at 21:57
Thank you very much. – few_reps Sep 24 '15 at 22:04

For explicit results, see

• Cohn, Harvey: Decomposition into four integral squares in the fields of $2^{1/2}$ and $3^{1/2}$, Amer. J. Math. 82, 301-322 (1960) (A different proof was given by J. Deutsch, An alternate proof of Cohn's four squares theorem; J. Number Theory 104, No. 2, 263-278 (2004))

• Colliot-Thélène, Jean-Louis; Xu, Fei: Brauer-Manin obstruction for integral points of homogeneous spaces and representation by integral quadratic forms Compos. Math. 145, No. 2, 309-363 (2009) (This one is a lot deeper and connects the representability with the Brauer-Manin obstruction; I have only seen the review.)

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