## Odd degree extensions of formally real fields are formally real

A field $F$ is called formally real if -1 cannot be expressed as a sum of squares in $F$. Let $F$ be a formally real field and let $p(x)$ be irreducible over $F$ of odd degree with a root $\alpha$. Show that $F(\alpha)$ is formally real.

Suppose to the contrary that there exist formally real fields $F$ and irreducible, odd degree polynomials $p$ over $F$ with roots $\alpha$ such that $F(\alpha)$ is not formally real. Choose $F$ and $p$ from among these fields such that the degree of $p$ is minimal, let $\alpha$ be a root of $p$, and consider (1) $F(\alpha) \cong F[x]/(p(x))$.

Since (as we assume) $F(\alpha)$ is not formally real, we have $-1 = \sum \theta_i$ for some $\theta_i \in F(\alpha)$. Let $p_i$ be the residue in $F[x]$ corresponding to $\theta_i$ under the isomorphism (1); then we have $-1 \equiv \sum p_i(x)^2$ in $F[x]/(p(x))$, so that $-1 + p(x)q(x) = \sum p_i(x)^2$ in $F[x]$, for some $q(x)$.

Note that, since each $p_i$ is a residue mod $p$ (using the division algorithm), each $p_i$ has degree strictly less than the degree of $p$. Since $F$ is a domain, $\sum p_i(x)^2$ has even degree less than $2 \mathsf{deg}\ p$. So that $pq$, has even degree, and since $p$ has odd degree, $q$ has odd degree, and in fact the degree of $q$ is strictly less than the degree of $p$. Moreover, since the degree of $q$ is the sum of the degrees of its irreducible factors, some irreducible factor of $q$, say $t$, must also have odd degree (less than that of $p$).

Let $\beta$ be a root of $t(x)$. In $F(\beta) = F[x]/(t(x))$, we have that $-1 \equiv \sum p_i(x)^2$. So $F$ is formally real, $t$ irreducible over $F$, $\beta$ a root of $t$, and $F(\beta)$ is not formally real – but $F(\beta)$ has degree (over $F$) strictly less than the degree of $F(\alpha)$, a contradiction.

So no such fields and polynomials exist, and in fact every odd extension of a formally real field is formally real.