Call a field formally real if there exists an order on it making it an ordered field. Let $L$ be a field with characteristic $0$. As discussed in this post, there exists a subfield $K$ which is maximal formally real, which means that there is no intermediate field of $L/K$ which is formally real. The extension $L/K$ must be algebraic: if $a\in L\setminus K$ is transcendental over $K$, then $K(a)$ (being isomorphic to univariate function field over $K$) is formally real.
Conjecture: If $L$ is quadratically closed with characteristic $0$, meaning that every element in $L$ is a square, then every maximal formally real subfield $K$ must have $[L:K]=2$, so $L=K(\sqrt{-1})$.
Let $K$ be a maximal formally real subfield of $L$ and pick an order on $K$. If $x\in K_{\ge 0}$, we can pick $y\in L$ such that $y^2=x$ since $L$ is quadratically closed. I'm quite sure that $K(y)$ is also formally real, by defining $$ a+by\ge 0\Longleftrightarrow (a,b\ge 0)\vee(a\ge 0,b\le 0,a^2\ge (-b)^2y)\vee (a\le 0,b\ge 0,b^2y\ge (-a)^2). $$ It would be tedious to verify, but I'm confident that this defines a positive cone over $K(y)$. Since $K$ is maximal formally real we must have $y\in K$, so every nonnegative element in $K$ has a square root in $K$ (such field is called Euclidean field). This implies that the only quadratic extension of $K$ is $K(\sqrt{-1})$.
But I do not know if we must have $K(\sqrt{-1})=L$. Why can't $K$ have a higher-degree extension within $L$? Thank you for your help.
