We can define an ordered field axiomatically as a totally ordered field that has two conditions (1) additive monotonicity which is that if $a < b$ then for all $x$ it is the case that $a+x < b+x$ and (2) multiplicative positivity preservation which means that for all $0 < a$ and $0 < b$ then $0 < ab$. We will prove several theorems that describe other properties of ordered fields.
Theorem. the sum of positive elements is positive
Let $a$ and $b$ be two positive elements. By positivity we have that $0 < a$ and $0 < b$. By the fact that $0 < a$ we can add $b$ to both sides to get $b < a+b$. Then connecting that with the previous inequality $0 < b$ we get $0 < b < a+b$ which by transitivity means that $0 < a+b$.
Theorem. the additive inverse of a negative number is positive
Let $a$ be a negative number. By negativity we have that $a < 0$. By additive monotonicity we know that $-a$ is an additive monotone element which can be added to both sides to get $a+(-a) < 0+(-a)$ which equals $0 < (-a)$ which means that the additive inverse $(-a)$ is greater then zero and therefore positive.
Theorem. addition must be torsion-free
In order for a group to be torsion-free it must not have any elements that are of finite order. Suppose that we have a field that has a non-identity additive torsion element, then there are two cases (1) the element is positive in which case the fact that the element can be added to itself to get zero defies additive positivity preservation or (2) the element is negative in which case its additive inverse is a positive torsion element which can be added to itself to get zero which also defies additive positivity preservation. So by contradiction we know that the additive group of the ordered field must be torsion-free.
Previously we touched on the two essentially types of properties of algebraic operations (1) an element being non-periodic and (2) an element being torsion-free. Being torsion-free is the weaker condition. These two properties are related to rather or not an algebraic structure can be ordered. It is actually intuitive that a non torsion-free group cannot be ordered, because it has some element that exhibits unordered cyclical behavior. So we can get a special case of this theorem, which is that any ordered group must be torsion-free.
Corollary. the characteristic of the field is zero
Theorem. the square of any non-zero number is positive.
By the total ordering of the field we can split this up into two cases (1) the element is positive and (2) the element is negative. In the first case that the element is positive we know that its square is positive by positivity preservation which is axiomatic. In the second case that the element is negative, by a previous theorem we know that its additive inverse is a positive number $p$. By the fact that the additive inverse is an involution this means that this number can be expressed as $-p$. Then consider the square as $(-p)^2$. This is equal to $(-1)^2*p^2$. We can cancel out the $(-1)^2$ to get $p^2$. By positivity preservation this again a positive number. This means that the product of any non-zero number with itself is positive.
Theorem. there are no proper zero sums of squares in an ordered field.
By the fact that the square of a non-zero number is positive and that the sum of positive numbers is again a positive number, we know that the sum of the squares of non-zero numbers is positive and therefore not equal to zero. This means that there are no proper zero sums of squares.
Conclusion. in order for a field to be orderable it must be formally real
Tuesday, March 17, 2020
Monday, March 16, 2020
On formally real fields
We can define formally real fields entirely algebraically.
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.
- 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
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.
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