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.

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  • Andrew  On January 22, 2014 at 5:11 pm

    “Since F is a domain, \sum{p_i(x)^2} has even degree” is not immediately obvious. It might be worth elaborating:

    If the sum had odd degree, then sum of the leading coefficients of the maximal degree p_i ‘s would be zero.

    Say the p_i’s of maximal degree have leading coefficients a_1, …, a_k.
    Then (a_1)^2 + … + (a_k)^2=0, so multiplying by the inverse of a_1 writes -1 as a sum of squares, contradicting that F is formally real.

    This contradiction shows that the sum must have even degree.

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