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Is there some slick proof of the fact that for a field $F$, the Witt ring $W(F)$ is finite if and only if $-1$ is a sum of squares and $F^\times/F^{\times 2}$ is finite?

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up vote 4 down vote accepted

I will assume that we are in the context of the "classical" algebraic theory of quadratic forms, i.e., that the characteristic of $F$ is not $2$.

1) Suppose $W(F)$ is finite.
a) Let $I$ be the fundamental ideal of $W(F)$. Then $W(F)/I \cong \mathbb{Z}/2\mathbb{Z}$ and $I/I^2 \cong F^{\times}/F^{\times 2}$. Thus $F^{\times}/F^{\times 2}$ is finite iff $W(F)/I^2$ is finite, which it certainly is if $W(F)$ is finite.
b) If $F$ were formally real -- i.e., if $-1$ is not a sum of squares in $F$ -- then for all $n$ the form $\langle 1,\ldots,1 \rangle$ is anisotropic, and $W(F)$ would be infinite. So $F$ is not formally real.

2) Suppose $F$ is not formally real and $F^{\times}/F^{\times 2}$ is finite. By Pfister's Local-Global Principle -- see e.g. Theorem 28 of these notes -- $W(F)$ is a $2$-torsion abelian group. But $W(F)$ is additively generated by the finite set $F^{\times}/F^{\times 2}$, so $(W(F),+)$ is a finitely generated torsion abelian group, hence finite.

Added: In a comment below, the OP suggests the following proof, which avoids PLGP. Since $F$ is not formally real, there exists $N \in \mathbb{Z}^+$ such that $q_N = [1,\ldots,1]$ ($N$ times) is isotropic. Hence so is $a \cdot q_N$ for any $a \in F^{\times}$, and since a form containing an isotropic subform is isotropic, this implies that for any $[a_1,\ldots,a_n]$ if we have at least $N$ instances of the same square class $a_i$, then the form is isotropic. But if $K = \# F^{\times}/F^{\times 2} < \aleph_0$, this shows that there are at most $N^K$ anisotropic quadratic forms over $F$. (Note that this bound is sharp for a finite field $\mathbb{F}_q$ with $q \equiv 1 \pmod 4$: we have $N = K = 2$ and $\# W(\mathbb{F}_q) = 4$.)

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Thanks. (2) is pretty neat, although in Lam's book there's a hint that (2) can be done by using the pigeon hole principle, so there should be a slightly more elementary argument than that. Need to think about it. – dst Oct 25 '11 at 1:59
@dst: well, you said "slick", not "elementary" so...I gave the first argument that I could come up with. I also believe there should be a more elementary way to go. (By the way, I learned PLGP from Lam's book, of which the linked to notes are little more than a partial gloss.) – Pete L. Clark Oct 25 '11 at 3:14
Yes, your notes are great. I found them through google about a week ago. I just started reading Lam's book and your notes clarify many things that Lam is not being too clear about. – dst Oct 25 '11 at 3:25
I came up with the following proof for (2) which is elementary and shorter. If $-1$ is a sum of squares, then there's some largest $n$ s.t. $n\langle 1\rangle$ is anisotropic. If $S$ are the representatives of $F^\times/F^{\times 2}$, then all anisotropic forms are of the form $n_1\langle a_1\rangle +\ldots +n_k\langle a_k\rangle$, $|n_i|\leq n$ and $a_i\in S$. If $S$ is finite, then there are finitely many such, so $W(F)$ is finite. – dst Oct 25 '11 at 5:09

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