Definition: Let $R$ be a commutative ring. The level of $R$, denoted $s(R)$, is the least positive integer $s$ such that $-1$ can be written as the sum of $s$ many squares in $R$. Set $s(R)=\infty$ if no such $s$ exists.
You might recall that a field $F$ is formally real if and only if $s(F)=\infty$.
Question: Let $F_n$ be the field of fractions of the integral domain $$\mathbb{Q}[x_1,\ldots,x_n]/(x_1^2+\ldots+x_n^2+1).$$ What is $s(F_n)?$ (It is clearly bounded above by $n$.)
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I have been reading some literature about this and the question of possible levels for fields was worked out by the following result of Pfister in 1965.
Theorem:
- If $F$ is a field with $s(F)<\infty$ then $s(F)$ is a power of 2.
- Let $m\geq 0$ and suppose $2^m\leq n<2^{m+1}$. Let $F=\mathbb{Q}(x_1,\ldots,x_n)$ and set $d=x_1^2+\ldots+x_n^2\in F$. Then $s(F(\sqrt{-d}))=2^m$.
So the powers of 2 are exactly the positive integers that can occur as the level of a field.
For general commutative rings there are more possibilities. In particular, there is this result of Dai, Lam, Peng from 1980.
Theorem: Let $R_n=\mathbb{Q}[x_1,\ldots,x_n]/(x_1^2+\ldots+x_n^2+1)$. Then $s(R_n)=n$.
The proof (at least the one I read) of that last one is much easier than Pfister's result, and is interesting in the sense that it uses the famous Borsuk-Ulam Theorem from topology.
So my question asks about the level of the field of fractions of $R_n$, which again is bounded above by $n$. But by Pfister's result it must be a power of 2. For example it should be easy to show $s(F_3)=2$. In light of part two of Pfister's theorem, maybe it's reasonable to expect $s(F_n)$ to be the greatest power of 2 that is still not more than $n$. Maybe there is a way of altering the proof of part two of Pfister's theorem to make it work for $F_n$.
The link I have been reading for all of this is the following: http://www.math.tifr.res.in/~anands/Level.pdf
