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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?

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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.

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It's odd extensions of $\Bbb R$ that cannot exist. There are certainly examples of irreducibles in $\Bbb Q[x]$ of odd order, one example would be $x^3 - 2$. The intermediate value theorem does say that all such polynomials have real roots and therefore the corresponding extensions are isomorphic to subfields of $\Bbb R$. Thus, this result does not yield any exotic examples of formally real fields unless you use a base field other than $\Bbb Q$.

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