TITLE: Example of formally real field $F$ with irreducible polynomial of odd degree in $F[x]$
QUESTION [0 upvotes]: Let $F$ be a formally real field; i.e. $-1$ cannot be expressed as the sum of squares. Let $K/F$ be a field extension of odd order. Hence, there exists $\alpha\in K$ which satisfies an irreducible equation of that same odd order in $F$. Prove that $K=F(\alpha)$ is also formally real.

I know that although $\mathbb{Q}$ is formally real, any field extension of odd order would be $\mathbb{Q}$ itself, as no polynomial of odd degree is irreducible in $\mathbb{Q}$. Could someone please give me an example of a formally real field which does have an irreducible polynomial of odd degree?

REPLY [2 votes]: Your statement is not true of $\mathbb Q$; there are infinitely many irreducible polynomials which are irreducible over $\mathbb Q$ of odd degree. For example $x^3-2$ is irreducible over $\mathbb Q$.
The statement you reference is true for $\mathbb R$.
Regardless, supposing that $K$ is not totally real, let $\beta\in K$ such that $\beta^2 = -1$. Then $\beta$ satisfies $x^2 +1 \in F[x]$, and $F(\beta) \subset K$.
But $[F(\beta):F] =2$, since $F$ is formally real  (hence, $x^2+1$ is irred over $F$).
Also, $[K:F] = [K:F(\beta)][F(\beta):F]$ shows that $K$ is an even degree extension, contradicting the assumption of odd degree.