Square classes of a real closed field

Question

The group of square classes $k^*/k^{*2}$ $\simeq$ $H^1(k,\mu_2)$ for $k$ the real numbers, is well known to be $\mu_2$ with the non trivial element represented by $(-1)$.

Question: Assume $k$ is some real closed field. Is $k^*/k^{*2} \simeq \mu_2$ ?

Answer 1

"The" algebraic closure $\bar k$ of any real closed field $k$ is $k(\sqrt -1)$. This follows from two properties of $k$ : any positive element of $k$ is a square, and every irreducible polynomial of $k[X]$ of odd degree has a root in $k$ (see e.g. Lang's "Algebra", chap. XI, §2, thm. 1). The proof proceeds exactly as in the case of $\mathbf C/\mathbf R$, using Galois theory Galois over $k$.

Back to your question. Let $G=Gal(\bar k/k)$ and take the $G$-cohomology of the exact sequence $1\to \mu_2\to{\bar k}^*\to {{\bar k}^*}^2={\bar k}^* \to 1$, where the rightmost map is the "squaring" map. This yields the exact sequence $1\to \mu_2\to k^*\to k^*\to H^1(G,\mu_2)\to H^1(G,{\bar k}^*)=0 $, i.e. $k^*/{k^*}^2 \cong H^1(G,\mu_2)$, and the latter group is $\cong \mu_2$ because $G$ is cyclic of order $2$.