Stufe (algebra)

If ${\displaystyle s(F)\neq \infty }$ then ${\displaystyle s(F)=2^{k}}$ for some natural number ${\displaystyle k}$.[1][2]

Proof: Let ${\displaystyle k\in \mathbb {N} }$ be chosen such that ${\displaystyle 2^{k}\leq s(F)<2^{k+1}}$. Let ${\displaystyle n=2^{k}}$. Then there are ${\displaystyle s=s(F)}$ elements ${\displaystyle e_{1},\ldots ,e_{s}\in F\setminus \{0\}}$ such that

${\displaystyle 0=\underbrace {1+e_{1}^{2}+\cdots +e_{n-1}^{2}} _{=:\,a}+\underbrace {e_{n}^{2}+\cdots +e_{s}^{2}} _{=:\,b}\;.}$

Both ${\displaystyle a}$ and ${\displaystyle b}$ are sums of ${\displaystyle n}$ squares, and ${\displaystyle a\neq 0}$, since otherwise ${\displaystyle s(F)<2^{k}}$, contrary to the assumption on ${\displaystyle k}$.

According to the theory of Pfister forms, the product ${\displaystyle ab}$ is itself a sum of ${\displaystyle n}$ squares, that is, ${\displaystyle ab=c_{1}^{2}+\cdots +c_{n}^{2}}$ for some ${\displaystyle c_{i}\in F}$. But since ${\displaystyle a+b=0}$, we also have ${\displaystyle -a^{2}=ab}$, and hence

${\displaystyle -1={\frac {ab}{a^{2}}}=\left({\frac {c_{1}}{a}}\right)^{2}+\cdots +\left({\frac {c_{n}}{a}}\right)^{2},}$

and thus ${\displaystyle s(F)=n=2^{k}}$.

Any field ${\displaystyle F}$ with positive characteristic has ${\displaystyle s(F)\leq 2}$.[3]

Proof: Let ${\displaystyle p=\operatorname {char} (F)}$. It suffices to prove the claim for ${\displaystyle \mathbb {F} _{p}}$.

If ${\displaystyle p=2}$ then ${\displaystyle -1=1=1^{2}}$, so ${\displaystyle s(F)=1}$.

If ${\displaystyle p>2}$ consider the set ${\displaystyle S=\{x^{2}:x\in \mathbb {F} _{p}\}}$ of squares. ${\displaystyle S\setminus \{0\}}$ is a subgroup of index ${\displaystyle 2}$ in the cyclic group ${\displaystyle \mathbb {F} _{p}^{\times }}$ with ${\displaystyle p-1}$ elements. Thus ${\displaystyle S}$ contains exactly ${\displaystyle {\tfrac {p+1}{2}}}$ elements, and so does ${\displaystyle -1-S}$. Since ${\displaystyle \mathbb {F} _{p}}$ only has ${\displaystyle p}$ elements in total, ${\displaystyle S}$ and ${\displaystyle -1-S}$ cannot be disjoint, that is, there are ${\displaystyle x,y\in \mathbb {F} _{p}}$ with ${\displaystyle S\ni x^{2}=-1-y^{2}\in -1-S}$ and thus ${\displaystyle -1=x^{2}+y^{2}}$.

The Stufe s(F) is related to the Pythagoras number p(F) by p(F) ≤ s(F) + 1.[4] If F is not formally real then s(F) ≤ p(F) ≤ s(F) + 1.[5][6] The additive order of the form (1), and hence the exponent of the Witt group of F is equal to 2s(F).[7][8]

• The Stufe of a quadratically closed field is 1.[8]
• The Stufe of an algebraic number field is ∞, 1, 2 or 4 (Siegel's theorem).[9] Examples are Q, Q(√−1), Q(√−2) and Q(√−7).[7]
• The Stufe of a finite field GF(q) is 1 if q ≡ 1 mod 4 and 2 if q ≡ 3 mod 4.[3][8][10]
• The Stufe of a local field of odd residue characteristic is equal to that of its residue field. The Stufe of the 2-adic field Q₂ is 4.[9]

• Lam, Tsit-Yuen (2005). Introduction to Quadratic Forms over Fields. Graduate Studies in Mathematics. 67. American Mathematical Society. ISBN 0-8218-1095-2. Zbl 1068.11023.
• Milnor, J.; Husemoller, D. (1973). Symmetric Bilinear Forms. Ergebnisse der Mathematik und ihrer Grenzgebiete. 73. Springer-Verlag. ISBN 3-540-06009-X. Zbl 0292.10016.
• Rajwade, A. R. (1993). Squares. London Mathematical Society Lecture Note Series. 171. Cambridge University Press. ISBN 0-521-42668-5. Zbl 0785.11022.

• Knebusch, Manfred; Scharlau, Winfried (1980). Algebraic theory of quadratic forms. Generic methods and Pfister forms. DMV Seminar. 1. Notes taken by Heisook Lee. Boston - Basel - Stuttgart: Birkhäuser Verlag. ISBN 3-7643-1206-8. Zbl 0439.10011.