Finiteness of the Witt ring 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?
 A: 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 $2$.

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*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.


*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 $\# W(\mathbb{F}_q) = 4$.)
