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There appear to be two senses of the qualifier "Archimedean" for fields. One is for ordered fields, and one is for "valued fields" (fields with an absolute value function defined). In the first case, the field is said to be "Archimedean" iff there exist no elements $x$ for which $nx < 1$ for every natural number $n$ and which are yet not equal to zero. Intuitively, we can think of such an element, if one existed, as being "infinitely small". If such elements exist, then the field is non-Archimedean.

The other sense deals with so-called valued fields. Instead of an order, we have an "absolute value function" which returns a real number and satisfies properties you'd expect such a thing to satisfy. Once again we have another formulation of the "Archimedean" property: the field is Archimedean iff there are no elements $x$ such that $|nx| < 1$ for every natural number $n$ and which are yet not equal to zero, and otherwise is non-Archimedean. Yet what is the intuition behind this, when the absolute value, which one may think of as a sort of "size measurement" of an element in some sense, must be a good ol' real number, and so surely can't be "infinitely small" and yet nonzero? In a non-Archimedean valued field, the absolute value function satisfies the inequality $|x + y| \le \max\{|x|, |y|\}$, which makes its behavior very strange... in particular, given a sequence of $|x|$, $|x + x|$, $|x + x + x|$, ..., the absolute value must at the very least not grow (the above inequality tells us $|x + x| \le |x|$, and induction does the rest), and may even shrink! Very weird indeed!

The_Sympathizer
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3 Answers3

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I guess the only way I can help you to get the intuition behind the non-archimedean idea is to say that it is the way we can order or measure numbers or things that cannot be aligned as we do with the real numbers. The best way to see this is to use an example.

Example: Consider $\mathbb{R}[x]$, the ring of polynomials with coefficients in $\mathbb{R}$. Consider the field of rational functions $\mathbb{R}(x)=\{p/q:p,q\in\mathbb{R}[x], q\neq0\}$.

  1. $\mathbb{R}(x)$ as non-Archimedean valued field: For any $p\in\mathbb{R}[x]$, $p\neq0$ let $deg(p)$ be the degree of the polynomial $p$ and put $|p|:=2^{deg(p)}$ and $|0|:=0$. Then on $\mathbb{R}(x)$ the map $|\cdot|$ defined by $|p/q|:=|p|/|q|$ is a non-Archimedean valuation. Notice that $|x|=|x+x|=|nx|=2$ for all $n\in\mathbb{N}$ and that $|a|=1$ for all $a\in\mathbb{R}, a\neq0$. Also, $|3+5x^7|=2^7$, $|x^{-1}|=2^{-1}$, etc,

  2. $\mathbb{R}(x)$ as non-Archimedean ordered field: Define an order on $\mathbb{R}[x]$ as follows: for constant polynomials consider the order of $\mathbb{R}$. For $p=a_0+a_1x+\cdots+a_nx^n\in\mathbb{R}[x]$, with $a_n\neq0$, put $p>0$ whenever $a_n>0$. Now, in the field of rational functions $\mathbb{R}(x)$ consider the order defined as follows: $p/q>0$ whenever $pq>0$ in $\mathbb{R}[x]$. Here, for all $f,g\in\mathbb{R}(x)$, $f<g$ whenever $0<g-f$.

This field is an extension of $\mathbb{R}$ as ordered field where $a<x^2$ for all $a\in\mathbb{R}$. Therefore $\mathbb{R}(x)$ is a non-Archimedean ordered field.

Epic facts: (a) The topology on $\mathbb{R}(x)$ induced by the metric $D(f,g)=|f-g|$ coincides with the order topology induced by the order in $\mathbb{R}(x)$. Let's denote this topology by $\tau$.

(b) The topology on $\mathbb{R}$ as a topological subspace of $(\mathbb{R}(x),\tau)$ is the discrete topology.

This is just another example of the following:

Remark: Let $(A,<)$ be an ordered set equipped with the corresponding order topology $\tau_A$ and let $B$ be a subset of $A$. The order in $A$ induces an order in $B$ which induces a topology $\tau_<$ on B. In general, the subspace topology $\tau_A\cap B$ may be different than the topology $\tau_<$. For other examples of this situation, see page 90 of the book: Topology, James R Munkres, Prentice Hall, 2000.

Proof of epic facts: The proof of the statement (a) is consequence of the following inclusions which can be verified straightforward for all $n\in\mathbb{N}$:

$$\{f\in\mathbb{R}(x):|f|<2^n\}\subset(-x^n,x^n)\subset \{f\in\mathbb{R}(x):|f|<2^{n+1}\}$$ $$\{f\in\mathbb{R}(x):|f|<2^{-n-1}\}\subset(-x^{-n},x^{-n})\subset \{f\in\mathbb{R}(x):|f|<2^{-n}\}.$$

The proof of the statement (b) follows from (a) and the fact that the valuation $|\cdot|$ is the trivial valuation when it is restricted to $\mathbb{R}$. An alternative proof is that for all $a\in\mathbb{R}$ and for all $n\in\mathbb{N}$, $$\{a\}=(a-x^{-n},a+x^{-n})\cap\mathbb{R}.$$ Notice that $\{x^{-n}:n\in\mathbb{N}\}$ is a set of infinitesimals.

Chilote
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  • So then basically non-Archimedean in the case of valued fields means in a way the same thing, that is that some elements are infinite with respect to others, but in an "unordered" fashion like how the plane ($\mathbb{R}^2$) is unordered, or perhaps more generally, like a dust of points, although because the absolute value is with codomain $\mathbb{R}$, it cannot assign formally infinite lengths to anything but geometrically in some way it's the same? – The_Sympathizer Dec 18 '17 at 11:35
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    Every time you mention the word "unordered" just refers to an order different to the one on $\mathbb{R}$. For example, $\mathbb{R}^2$ can be ordered with the lexicographic order where $(1,0)>(0,m)$ for all $m\in\mathbb{Z}$. In the case of the non-Archimedean valuation on $\mathbb{R}(x)$, there are so many copies of $\mathbb{R}$ inside the field that the moment we try to assign a "measure" (real number) to each member of $\mathbb{R}(x)$, we need to make whole "lines" to have the same valuation (e.g. $|ax^m|=2^m$ for all $a\in\mathbb{R}$). – Chilote Dec 18 '17 at 16:12
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Full disclosure I landed on this page because I was trying to answer the same question (well asked, by the way!), but let me share what I've come to understand which will hopefully be helpful.

Both concepts are trying to capture the notion that a field is "too big" in a certain sense. A non-Archimedean field in the first sense is one with elements too "large" (or equivalently since, every element has an inverse, too "small") that we can't "get to it" (equivalently "get from it to $1$") by multiplying by some integer.

In the non-Archimedean valuation case, it's too "large" for us to be able to give everything a "size" (in the reals) without overlapping. The fact that $|x|, |x + x|, |x+ x + x|,...$ doesn't "grow" captures the fact that all these elements lie on the same "line" as @Chilote very helpfully described in a comment above. There isn't enough "room" in the reals for them all to fit.

Torsten Schoeneberg
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lukemassa
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  • Wow! I really like this idea. – The_Sympathizer Mar 27 '21 at 23:09
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    Sorry, I do not think this is a good way to look at it. I do not think of $\mathbb Q_p$ as "bigger" than $\mathbb R$ or $\mathbb C$. More strikingly, you can view $\mathbb Q$ with the usual archimedean, but also with the non-archimedean $p$-adic absolute values. So how would $\mathbb Q$ with some $p$-adic value be "bigger" than $\mathbb Q$ with the usual absolute value? – Torsten Schoeneberg Apr 01 '21 at 02:28
  • As for the idea in the last paragraph, well the usual archimedean absolute value on $\mathbb C$ also has a lot of "overlap": on every circle in the complex plane, all numbers have the same absolute value. If we're talking of "lines" in a more, well, linear sense, though in @Chilote's example one can look at $\mathbb R$-subspaces and they are bounded, that is a very specific example. What is a "line" in $\mathbb Q_p$ (or $\mathbb Q$ itself) which would stay bounded? On the contrary, every non-trivial $\mathbb Q$-subspace of $\mathbb Q_p$ has unbounded values! – Torsten Schoeneberg Apr 01 '21 at 02:33
  • What one actually has is that for each element $x$ in a non-archimedean valued field, the $\mathbb Z$-submodule it generates is bounded. But then again that boils down to $\mathbb Z$ itself being bounded. Which boils down to $\mathbb N$ being bounded. Which is more or less the original idea and definition of "non-archimedean". Hence my answer. – Torsten Schoeneberg Apr 01 '21 at 02:48
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I think the one thing that both meanings have in common is, well, the negation of the archimedean axiom. Best expressed like this:

$$\textbf{The natural numbers $\mathbb N$ are bounded.}$$

This is the vague English sentence which one can easily turn into a precise mathematical statement in either of the mathematical usages of the word -- albeit distinctly different ones.

Namely, in a non-archimedean ordered field, for every natural number $n$ there is still a bigger natural number $m$. There is, however, also an element (and then infinitely many) which are bigger than all natural numbers. (Vaguely said, the "bound" happens "after allowing $\mathbb N$ to go to the first infinity", but then there is "more coming beyond that".)

Whereas with respect to a non-archimedean value, the natural numbers already have a finite bound, they do not "go off to infinity" to begin with. E.g. with respect to the $p$-adic value for some prime $p$, all natural numbers not divisible by $p$ have the same size, $1$; whereas those which are divisible by $p$, are smaller than that. (Vaguely said, no $n$ can ever get bigger than $1$ -- but some can get smaller. The bound is at $1$.)

Torsten Schoeneberg
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