A formally real field is a field $K$ such that $-1$ is not a sum of squares in $K$. Clearly subfields of $\mathbb{R}$ are formally real. I also know finite fields and algebraically closed fields are never formally real. Then the next example I can think of is a field of rational functions of a formally real field, for example $\mathbb{Q}(t)$. But then mapping $t$ to some transcendental number in $\mathbb{R}$, e.g. $\pi$, I get an isomorphism to a subfield of the reals, in this case $\mathbb{Q}(\pi)$. Is there any example where I cannot get an isomorphism into a subfield of $\mathbb{R}$? Is this possible in division rings/ skew-fields?
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3$\begingroup$ What about a formally real field of cardinality $>2^{\aleph_0}$? $\endgroup$– Noah SchweberCommented Mar 22, 2022 at 21:47
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$\begingroup$ Wouldnt $Q(\zeta_3)$ be formally real ? $\endgroup$– TestcaseCommented Mar 22, 2022 at 21:48
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1$\begingroup$ @Testcase $(\zeta_3)^2+(\zeta_3^2)^2=\zeta_3^2+\zeta_3=-1$. $\endgroup$– mathmaCommented Mar 22, 2022 at 21:56
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2$\begingroup$ Or $\mathbb{Q}((t_a)_{a\subset \mathbb{R}})$ should be a “too large” formally real field. $\endgroup$– AphelliCommented Mar 22, 2022 at 23:11
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2$\begingroup$ Otherwise $K=\bigcup_{n\ge 1} \Bbb{R}((t^{1/n}))$ its algebraic closure is $K[i] \cong \Bbb{C}$ (need the axiom of choice to prove $\cong$) but it is neither isomorphic to $\Bbb{R}$ nor to a subfield of $\Bbb{R}$. The order on $K$ is to compare the first non-zero Laurent series coefficient. $\Bbb{R}(t)$ and $\Bbb{R}((t))$ are ordered as well and again they are not isomorphic to a subfield of $\Bbb{R}$. $\endgroup$– reunsCommented Mar 22, 2022 at 23:20
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