MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 21 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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.