# Archimedean ordered fields are real

In this entry, we shall show that every Archimedean^{} ordered field is isomorphic^{} to
a subfield^{} of the field of real numbers. To accomplish this, we shall construct an
isomorphism^{} using Dedekind cuts.

As a preliminary, we agree to some conventions. Let $\mathbb{F}$ denote an ordered
field with ordering relation $>$. We will identify the integers with multiples of
the multiplicative identity^{} of fields. We further assume that $\mathbb{F}$ satisfies
the Archimedean property: for every element $x$ of $\mathbb{F}$, there exists an
integer $n$ such that $$.

Since $\mathbb{F}$ is an ordered field, it must have characteristic zero. Hence, the subfield generated by the multiplicative identity is isomorphic to the field of rational numbers. Following the convention proposed above, we will identify this subfield with $\mathbb{Q}$. We use this subfield to construct a map $\rho $ from $\mathbb{F}$ to $\mathbb{R}$:

###### Definition 1.

For every $x\mathrm{\in}\mathrm{F}$, we define

$$\rho (x)=(\{y\in \mathbb{Q}\mid y>x\},\{y\in \mathbb{Q}\mid y\le x\})$$ |

###### Theorem 1.

For every $x\mathrm{\in}\mathrm{F}$, we find that $\rho \mathit{}\mathrm{(}x\mathrm{)}$ is a Dedekind cut.

###### Proof.

Because, for all $y\in \mathbb{Q}$, we have either $y>x$ or $y\le x$, the two sets of $\rho (x)$ form a partition of $\mathbb{Q}$. Furthermore, every element of the latter set is less than every element of the former set. By the Archimedean property, there exists an integer $n$ such that $$; hence the former set is not empty. Likewise, there exists and integer $m$ such that $$, or $x>-m$, so the latter set is also not empty. ∎

Having seen that $\rho $ is a bona fide map into the real numbers, we now show that
it is not just any old map, but a monomorphism^{} of fields.

###### Theorem 2.

The map $\rho \mathrm{:}\mathrm{F}\mathrm{\to}\mathrm{R}$ is a monomorphism.

###### Proof.

Let $p$ and $q$ be elements of $\mathbb{F}$; set $(A,B)=\rho (p)$, set $(C,D)=\rho (q)$, and set $(E,F)=\rho (p+q)$. Since $a>p$ and $b>q$ implies $a+b>p+q$ for all rational numbers $a$ and $b$, it follows that $a\in A$ and $b\in C$ implies that $a+b\in E$. Likewise, since $a\le p$ and $b\le q$ implies $a+b\le p+q$ for all rational numbers $a$ and $b$, it follows that $a\in B$ and $b\in D$ implies $a+b\in F$. Hence, $\rho (p)+\rho (q)=\rho (p+q)$.

Since a rational number is positive if and only if it is greater than $0$, it follows that $\rho (0)=0$. Together with the fact proven in the last paragraph, this implies that $\rho (-x)=-\rho (x)$ for all $x\in \mathbb{F}$.

Suppose that $p$ and $q$ are positive elements of $\mathbb{F}$. As before, set $(A,B)=\rho (p)$, set $(C,D)=\rho (q)$, and set $(E,F)=\rho (p\cdot q)$. Since $a>p$ and $b>q$ implies $a\cdot b>p\cdot q$ for all rational numbers $a$ and $b$, it follows that $a\in A$ and $b\in C$ implies that $a\cdot b\in E$. Likewise, since $a\le p$ and $b\le q$ implies $a\cdot b\le p\cdot q$ for all rational numbers $a$ and $b$, it follows that $a\in B$ and $b\in D$ implies $a\cdot b\in F$. Hence, $\rho (p)\cdot \rho (q)=\rho (p\cdot q)$.

By using the fact demonstrated previously that $\rho (-x)=-\rho (x)$, we may extend
what was shown above to the statement that $\rho (p\cdot q)=\rho (p)\cdot \rho (q)$
for all $p,q\in \mathbb{F}$. Thus, $\rho $ is a morphism^{} of fields. Since $\mathbb{F}$
is Archmiedean, if $p\ne q$, there must exist a rational number $r$ between $p$ and $q$,
hence $\rho (p)\ne \rho (q)$, so $\rho $ is a monomorphism.
∎

Since $\rho $ is a morphism of fields, its image is a subring of $\mathbb{R}$. Since $\rho $
is a monomorphism, its restriction^{} to this image is an isomorphism, hence $\mathbb{F}$ is
isomorphic to a subfield of $\mathbb{R}$.

Title | Archimedean ordered fields are real |
---|---|

Canonical name | ArchimedeanOrderedFieldsAreReal |

Date of creation | 2013-03-22 17:26:22 |

Last modified on | 2013-03-22 17:26:22 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 12 |

Author | rspuzio (6075) |

Entry type | Theorem |

Classification | msc 12D99 |

Related topic | GelfandTornheimTheorem |