On formally real fields

We can define formally real fields entirely algebraically.

• The field is characteristic zero
• For all sums of squares equal to zero $x^2+y^2=0$ it is the case that $x,y$=0

A proper zero sum of squares is a sum of squares $x^2+y^2=0$ such that $x,y\ne 0$. These are used to define formally real fields. This is reminiscent of the idea of a zero divisor $ab=0$ such that $a$=$b$ used to define integral domains. All formally real fields are of course integral domains as well, so they also have no proper zero divisors.

The simplest counter example is the set of Gaussian rationals $\mathbb{Q}(i)$ which is not a formally real field because $i^2 + 1^2 = 0$ That is to say that {1,i} together form a proper zero sum of their squares. The complex numbers $\mathbb{C}$ of course also do not form a formally real field for the same reason. It is easy to see that if a given field has a subfield that is not formally real, then it is not formally real itself. This means that being formally real is subfield closed.

Theorem. no algebraically closed field is formally real

This follows from that fact that $\mathbb{Q}(i)$ is a non-real subfield of any characteristic zero algebraically closed field. Since the property of being formally real is subfield closed, this demonstrates that the field is not closed.

The purpose of spending so much time on this concept, is that it is fundamental to the definition of an ordered field. A given field can only be ordered only if it is formally real. As a result, we have a concept of order theory that emerges purely from commutative algebra.

The possibility of different orderings: if a given field is an algebraic extension of its prime subfield then it can ordered in a unique manner. The ordering of the formally real field in this case is determined by its embedding within the totally ordered field of real numbers. The resulting ordered field is archimedean by this embedding. The rational numbers $\mathbb{Q}$ as well as real algebaric number fields like $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{3})$, $\mathbb{Q}(\sqrt{2},\sqrt{3})$, and so on are all standard ordered in this way.

Therefore in order for a given formally real field to have multiple orderings it must have at least one transcendental element in it with respect to a subfield. The simplest example is $\mathbb{Q}(x)$ the field of rational functions over the rational numbers. This is a transcendental extension because the variable $x$ is transcendental. Two different orderings are $\mathbb{Q}(e)$ and $\mathbb{Q}(\infty)$. These are a isomorphic as fields, but different as ordered fields as one is archimedean and the other is not. Everyone knows that $e$ is a transcendental number, but not everyone knows that $\infty$ and related numbers are also transcendental in an ordered field. This similarity is seen here, as an infinite transcendental and a finite transcendental number are indistinguishable algebraically and only distinguished through the ordering.

As a result we see three different cases: (1) fields that cannot be ordered like $\mathbb{Q}(i)$, $\mathbb{C}$, etc which are not formally real, (2) fields that can be uniquely ordered which are algebraic extensions of the rational numbers, and (3) other fields that can be ordered in multiple different ways like the field of rational functions over the rational numbers. This addresses the issue of determining how algebraic properties determine orderings.