formally real field
A field $F$ is called formally real if $1$ can not be expressed as a sum of squares (of elements of $F$).
Given a field $F$, let ${S}_{F}$ be the set of all sums of squares in $F$. The following are equivalent^{} conditions that $F$ is formally real:

1.
$1\notin {S}_{F}$

2.
${S}_{F}\ne F$ and $\mathrm{char}(F)\ne 2$

3.
$\sum a_{i}{}^{2}=0$ implies each ${a}_{i}=0$, where ${a}_{i}\in F$

4.
$F$ can be ordered (There is a total order^{} $$ which makes $F$ into an ordered field)
Some Examples:

•
$\mathbb{R}$ and $\mathbb{Q}$ are both formally real fields.

•
If $F$ is formally real, so is $F(\alpha )$, where $\alpha $ is a root of an irreducible polynomial of odd degree in $F[x]$. As an example, $\mathbb{Q}(\sqrt[3]{2}\omega )$ is formally real, where $\omega \ne 1$ is a third root of unity.

•
$\u2102$ is not formally real since $1={i}^{2}$.

•
Any field of characteristic nonzero is not formally real; it is not orderable.
A formally real field is said to be real closed if any of its algebraic extension^{} which is also formally real is itself. For any formally real field $k$, a formally real field $K$ is said to be a real closure of $k$ if $K$ is an algebraic extension of $k$ and is real closed.
In the example above, $\mathbb{R}$ is real closed, and $\mathbb{Q}$ is not, whose real closure is $\stackrel{~}{\mathbb{Q}}$. Furthermore, it can be shown that the real closure of a countable^{} formally real field is countable, so that $\stackrel{~}{\mathbb{Q}}\ne \mathbb{R}$.
Title  formally real field 

Canonical name  FormallyRealField 
Date of creation  20130322 14:22:22 
Last modified on  20130322 14:22:22 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  20 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 12D15 
Related topic  PositiveCone 
Related topic  RealRing 
Defines  formally real 
Defines  real closed 
Defines  real closure 