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Here's a statement from Lam's First Course in Noncommutative Rings. (Paraphrased)

Let $k$ be a field. Then the following conditions are equivalent. $$(\forall a,b,c,d\in k)\;\;(a,b,c,d)\neq 0\implies a^2+b^2+c^2+d^2\neq 0;\tag 1$$ $$-1\text{ is not a sum of two squares.}\tag2$$

I can't prove this fact. If I replace $(2)$ with

$$-1\text{ is not a sum of three squares,}\tag3$$

then the equivalence between $(1)$ and $(3)$ is easy to prove. Indeed, suppose $(1)$ is false. Then we may assume without loss of generality that there are $a,b,c,d\in k$ with $a\neq 0$ such that $$a^2+b^2+c^2+d^2=0.$$ But then $$-1=\left(\frac ba\right)^2+\left(\frac ca\right)^2+\left(\frac da\right)^2,$$

and so $(3)$ is false. Conversely, if $(3)$ is false, then there exist $x,y,z\in k$ such that $$-1=x^2+y^2+z^2$$ and therefore $$0=x^2+y^2+z^2+1^2,$$ which means that $(1)$ is false.

I can also prove $(1)\implies (2).$ Suppose $(2)$ is false. Then $$-1=x^2+y^2,$$ whence $$0=x^2+y^2+1^2+0^2,$$ which implies that $(1)$ is false.

But I can't prove $(2)\implies (1)$ and I'm starting to suspect that it may actually be false. But I can't find a counterexample either. It would require finding a field in which $-1$ is a sum of three squares but not a sum of two squares. But in finite fields, every element is a sum of two squares, as proven here. In the subfields of $\mathbb R$ on the other hand, $-1$ is not a sum of three squares. And in $\mathbb C,$ it is a sum of two squares: $-1=0^2+i^2.$ In general, an example cannot be algebraically closed, which leaves me with the fields of rational functions as the only other fields I know. Also, the example cannot have characteristic $2$ as then $-1=1=1^2+0^2.$

Edit: The theorem Gerry Myerson cites in his answer is proven here. A Wikipedia article on the Stufe is here. Both links have been given by joriki in the comments.

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    This is an example of a post which is well written. I have been discussing this with the user on chat. I did not have this clear frame in my mind--that I read this post makes me organise what I know. +1 for the question... –  Apr 19 '12 at 12:12

2 Answers2

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The "stufe" of a field is the smallest $m$ such that $-1$ can be expressed as a sum of $m$ squares. It is a theorem of Pfister that the stufe (if it exists) is always a power of 2.

Gerry Myerson
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  • Thank you! So if I understand this correctly, Lam's statement is correct. Is is difficult to prove? It would be strange for Lam to make such a statement without proof if it requires advanced field theory... –  Apr 19 '12 at 12:42
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    There's a proof here. There's also a German Wikipedia article on the "Stufe", but it presupposes some of the facts about Pfister forms that the other text proves. – joriki Apr 19 '12 at 12:52
  • @joriki Thank you very much for the links. I can't quite see the proof of Gerry Meyerson's statement in the first one. (I can't read German so I can't use the second one unfortunately...) –  Apr 19 '12 at 13:09
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    @ymar: Yes, sorry, I just realized that that paper contains only a part of this slide show that has the whole proof (part III beginning on p. 81). I was trying to avoid linking to a slide show :-) – joriki Apr 19 '12 at 13:14
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    I'm sorry Gerry Myerson for misspelling you name in the previous comment. –  Apr 19 '12 at 13:16
  • @joriki It looks like it's the same link as before. :) –  Apr 19 '12 at 13:18
  • @ymar: Sorry, I thought I'd pasted the other one :-) I've corrected it. – joriki Apr 19 '12 at 13:19
  • There's an English Wikipedia article on Pfister forms, and it cites Lam (2004), Introduction to quadratic forms over fields. So that might explain why Lam found this obvious enough to use without proof ;-) – joriki Apr 19 '12 at 13:20
  • Random fact: "Stufe" is the German word for "step", though it sometimes is also used similar to "level". – Desiato Apr 19 '12 at 13:42
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    I've created an English Wikipedia article on Albrecht Pfister (translated from the German article). – joriki Apr 19 '12 at 13:46
  • @joriki May I add your link to the question so it's not hidden in a comment? –  Apr 19 '12 at 13:55
  • @ymar: Sure, go ahead. I'm also translating the German Wikipedia article on Stufe; I'll let you know when it's done. – joriki Apr 19 '12 at 14:17
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    @ymar: Here's the new English Wikipedia article on Stufe. – joriki Apr 19 '12 at 15:22
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Hint $\ $ If $\rm\: -1 = a^2 + b^2 + c^2\:$ then either $\rm\:c^2 = -1\:$ or $\rm\:(a^2+b^2)/(c^2+1) = -1,\:$ which may be rewritten as a sum of two squares, since such sums $\ne 0$ form a group, using Brahmagupta for multiplication, and inversion via $\rm\ 0\ne z = x^2\! + y^2\:$ $\Rightarrow$ $\rm\:1/z = (x/z)^2 + (y/z)^2.$

Remark $\ $ The proof generalizes. Replacing the Brahmagupta composition identity by identities discovered by Pfister for quadratic forms in $\rm\:2^n\:$ variables, the proof shows that if $-1$ is a sum of $\rm\:m\:$ squares in a field, then the least value of $\rm\:m\:$ is necessarily a power of $2$, which is called the level (Stufe in German) of the field. Further, every power of $2$ is the level of some field.

Bill Dubuque
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  • Thanks a lot! I was having a bit of trouble lessening the generality of Pfister's argument given in a comment by joriki. Now everything is clear! –  Apr 19 '12 at 14:12