A field is called *formally real* if -1 cannot be expressed as a sum of squares in . Let be a formally real field and let be irreducible over of odd degree with a root . Show that is formally real.

Suppose to the contrary that there exist formally real fields and irreducible, odd degree polynomials over with roots such that is not formally real. Choose and from among these fields such that the degree of is minimal, let be a root of , and consider (1) .

Since (as we assume) is not formally real, we have for some . Let be the residue in corresponding to under the isomorphism (1); then we have in , so that in , for some .

Note that, since each is a residue mod (using the division algorithm), each has degree strictly less than the degree of . Since is a domain, has even degree less than . So that , has even degree, and since has odd degree, has odd degree, and in fact the degree of is strictly less than the degree of . Moreover, since the degree of is the sum of the degrees of its irreducible factors, some irreducible factor of , say , must also have odd degree (less than that of ).

Let be a root of . In , we have that . So is formally real, irreducible over , a root of , and is not formally real – but has degree (over ) strictly less than the degree of , 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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“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.