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Consider the field of fractions $K$ of the quotient algebra $\mathbb{R}[x,y,z,t]/(x^2+y^2+z^2+t^2+1)$, where $\mathbb{R}$ is the field of real numbers and $x,y,z,t$ are variables. Clearly $-1$ is a sum of 4 squares in $K$. How can one prove that $-1$ is not a sum of 2 squares in $K$?

Serre mentions without proof this (probably known or easy) fact in a letter to Eva Bayer of May 1, 2010, and I am stuck: I cannot prove it.

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I hope it's not impertinent to ask: how do you know about a letter Serre wrote four days ago? –  Graham Leuschke May 5 '10 at 19:48
@Graham: I asked a question to Eva Bayer about Galois cohomology of $G=PSU_n$ and maximal tori of $G$ over field extensions of $\mathbb{R}$, Eva forwarded my question to Serre, Serre answered it in a letter to Eva, and Eva forwarded me his letter. –  Mikhail Borovoi May 6 '10 at 6:17

1 Answer 1

up vote 20 down vote accepted

This is a special case of a theorem of A. Pfister. It is well known to quadratic forms specialists. See e.g. Theorem XI.2.6 in T.Y. Lam's Introduction to Quadratic Forms over Fields.

I believe the original paper is

Pfister, Albrecht, Zur Darstellung von $-1$ als Summe von Quadraten in einem Körper. (German) J. London Math. Soc. 40 1965 159--165.

In this same paper Pfister defines the "stufe" (which Lam has successfully campaigned to be called the "level") of a non-formally real field, namely the least positive integer $n$ such that $-1$ is a sum of $n$ squares. Among his other achievements, he proves that the level is always a power of $2$ (so that Kevin Buzzard's recollection is correct). It is also worth remarking that his work is an insightful and rapid response to previous work of J.W.S. Cassels.

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My favourite "well known to quadratic forms specialists" fact---I hope I've remembered it right---if K is a field and -1 is the sum of three squares in K then it's the sum of two squares in K. –  Kevin Buzzard May 5 '10 at 19:36
The special case Kevin mentioned has a short proof which makes it easy to remember: if $-1=a^2+b^2+c^2$ then $-1-c^2=a^2+b^2$ implying $-1=(a^2+b^2)(1+c^2)/(1+c^2)^2$. Now use the fact that the product of the sum of two squares is also a sum of two squares. –  Keivan Karai May 6 '10 at 9:54

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