# Level of a commutative ring and its quotient field

Reading Lam's Introduction to Real Algebra, he remarks that:

1. For a Dedekind domain $$A$$ with quotient field $$F$$, then $$s(A)$$ is either $$s(F)$$ or $$s(F) + 1$$. Furthermore, $$s(A)$$ is either $$\infty$$, $$2^{n}$$ or $$2^{n} + 1$$ for integers $$n \geq 0$$. For $$n \geq 2$$ there are no examples in literature of an $$A$$ such that $$s(F) = 2^n$$ and $$s(A) = 2^n +1$$.
2. For a regular local ring, the relation between $$s(A)$$ and $$s(F)$$ is known(equality) for low dimensions, and he conjectures that this holds for higher dimensions.

As Lam's paper was written in 1984 I was hoping to find out whether there have been any development in respect to these two questions. I have tried searching for it, but haven't found anything related to it.

Background: The level of a commutative ring, denoted $$s(A)$$, is the smallest natural number $$n$$ such that $$-1$$ can be expressed as a sum of $$n$$ squares in $$A$$. if $$-1$$ is not a sum of squares we define the level of $$A$$ to be $$\infty$$.

## 2 Answers

Concerning question 1, such examples have been found independently by David Leep (unpublished) and J. K. Arason and R. Baeza:

Arason, J. K.; Baeza, R. On the level of principal ideal domains. Arch. Math. (Basel) 96 (2011), no. 6, 519–524

http://link.springer.com/article/10.1007%2Fs00013-011-0253-2

Let me assume that $$2$$ is invertible in $$A$$.

Regarding question 2, for a regular local ring $$A$$ with fraction field $$F$$ and $$s(F)=2^k$$, the equality $$s(A)=s(F)$$ follows from the Gorthendieck--Serre conjecture for the (split) group scheme $$\mathrm{SO}(2^{k+1})$$ over $$A$$. This case of the conjecture is known for many regular local rings, e.g., when $$A$$ contains a field (see below), but is open in general.

The Grothendieck--Serre conjecture states that if $$A$$ is a regular local ring with fraction field $$F$$ and $$G$$ is a reductive group scheme over $$A$$, then the base-change map of pointed etale cohomology sets $$\mathrm{H}^1(A,G)\to \mathrm{H}^1(F,G)$$ has trivial kernel. In the special case where $$G$$ is $$\mathrm{SO}(2n)$$, i.e., the group of isometries of a $$2n$$-dimensional hyperbolic quadratic form over $$A$$, the set $$\mathrm{H}^1(A,G)$$ is in bijection with isomorphism classes of (regular) $$2n$$-dimensional quadratic forms with trivial discriminant over $$A$$, so in this case, the conjecture says that a quadratic form over $$A$$ that becomes hyperbolic over $$F$$ is already hyperbolic over $$A$$.

Suppose that the Grothendieck-Serre conjecture holds for $$\mathrm{SO}(2^{k+1})$$ and consider the Pfister quadratic form $$q:=2^{k+1}\times \langle 1\rangle =\langle\langle 1\rangle\rangle^{\otimes (k+1)}$$ over $$A$$. Then $$q_F$$ is isotropic, hence hyperbolic, so our assumption implies that $$q\cong 2^{k}\times \langle 1,-1\rangle$$. It is known that Witt's Cancellation Theorem holds over local rings, so this means that $$2^k\times\langle 1\rangle\cong 2^k\times \langle -1\rangle$$ as quadratic forms over $$A$$. The right hand side represents $$-1$$, so $$-1$$ is a sum of $$2^k$$ squares in $$A$$, i.e., $$s(A)\leq 2^k$$. As $$s(A)\geq s(F)$$, we conclude that $$s(A)=2^k$$.

The general Grothendieck-Serre conjecture is known to hold in many cases. Most notably, it holds when $$A$$ contains a field; this is due to Fedorov-Panin and Panin. To my best knowledge, what is presently known for the particular group $$\mathrm{SO}(2n)$$ is that it satisfies the conjecture when:

• $$A$$ is unramified, i.e., $$A/pA$$ is regular with $$p$$ be the characteristic of the residue field of $$A$$ (this includes the case where $$A$$ contains a field), or

• $$A$$ has Krull dimension at most $$4$$.

The first case is due to Cesnavicius (who proved the conjecture for quasi-split groups over unramified local rings), and the second follows from the work of Balmer and Walter on the Gersten-Witt complex, which implies in particular that the base-change map between the Witt groups $$W(A)\to W(K)$$ is injective when $$A$$ is regular local of Krull-dimension $$\leq 4$$.