I came across this question in a calculus book.

Is it possible to prove that an ordered field must be infinite? Also - does this mean that there is only one such field?


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    $\begingroup$ $0 \lneq 1 \lneq 1+1 \lneq 1+1+1 \lneq ...$ $\endgroup$ – jspecter Jun 10 '12 at 17:18
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    $\begingroup$ I have a feeling that I answered that like twice before. Too bad it's easier to write another answer instead of finding a duplicate (the search function stinks!) $\endgroup$ – Asaf Karagila Jun 10 '12 at 17:19
  • $\begingroup$ Spivak's book? ${}$ $\endgroup$ – Michael Hardy Jun 10 '12 at 17:58

Recall that in an ordered field we have:

  1. $0<1$;
  2. $a<b\implies a+c<b+c$.

Suppose that $F$ is an ordered field of characteristic $p$, then we have in $F$ that $$\underbrace{1+\ldots+1}_{p\text{ times}} = 0$$

Therefore: $$0<1<1+1<\ldots<\underbrace{1+\ldots+1}_{p\text{ times}} = 0$$

Contradiction! Therefore the characteristic of $F$ is $0$ and therefore it is infinite, since it contains a copy of $\mathbb Q$.

Few fun facts on the characteristic of a field:

Definition: The characteristic of a field $F$ is the least number $n$ such that $\underbrace{1+\ldots+1}_{n\text{ times}}=0$ if it exists, and $0$ otherwise.


  1. If a field has a positive characteristic $n$ then $n$ is a prime number.
  2. If $F$ is a finite field then its characteristic is non-zero (Hint: the function $x\mapsto x+1$ is injective, start with $0$ and iterate it $|F|$ many times and you necessarily got $0$ again.)
  3. If $F$ is finite and $p$ is its characteristic then $p$ divides $|F|$.
  • $\begingroup$ What does "field of characteristics $p$" mean? thanks $\endgroup$ – yotamoo Jun 10 '12 at 17:22
  • $\begingroup$ @yotamoo: It means that $p$ is the least natural number such that adding $1$ to itself $p$ many times gives $1+1+\ldots+1=0$. One good exercise is to prove that $p$ is either $0$ or a prime number. Fields of characteristics $0$ can be ordered ($\mathbb Q$ is, but $\mathbb C$ is not); and every finite field has to have a positive characteristics (because $x\mapsto x+1$ is an injective function in a field). $\endgroup$ – Asaf Karagila Jun 10 '12 at 17:23
  • $\begingroup$ I usually hear it said "a field of characteristic $p$" with no plural. Also, $\mathbb C$ has characteristic 0 but can't be ordered, so that comment is kind of misleading. $\endgroup$ – Ben Millwood Jun 10 '12 at 17:40
  • $\begingroup$ @benmachine: Possible, I usually get confused about that. Feel free to edit, I am on the iPhone now... Also I mentioned that the complex numbers have char. zero but cannot be ordered. I don't know how that is misleading... $\endgroup$ – Asaf Karagila Jun 10 '12 at 17:50
  • $\begingroup$ @AsafKaragila: that doesn't sound like what you are saying, is all: you say "fields of characteristic 0 can be ordered ($\mathbb Q$ is, but $\mathbb C$ is not)" – I guess I thought it misleading because it's kind of ambiguous if you meant is/is not ordered, or is/is not characteristic 0. I guess I mentally inserted "all" before "fields", where you meant "only". $\endgroup$ – Ben Millwood Jun 10 '12 at 17:53

An ordered field must be infinite. Notice that each field has a subset of numbers that behave like the natural numbers, with $0<1<1+1<1+1+1\dots$

However, not every ordered field is isomorphic to all other ordered fields. Notice that both the rational numbers and real numbers are ordered fields.


Hint $\ $ Linearly ordered groups are torsion-free: $\rm\: 0\ne n\in \mathbb N,$ $\rm\:g>0 \:\Rightarrow\: n\cdot g = g +\cdots + g > 0,\:$ since positives are closed under addition. Conversely, a torsion-free commutative group can be linearly ordered (Levi 1942).

  • $\begingroup$ Even more: a lattice ordered (nor necessarily commutative) group is always torsion free. $\endgroup$ – Leo Jul 29 '12 at 22:15

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