Is there a method of constructing a subset of a reasonably arbitrary ring so that when the construction is applied the $\mathbb{C}$ the result is $B = \{ z \in \mathbb{C} \colon |z| \leq 1 \} $?

My interest is in constructing something like an absolute value for an arbitrary ring. Notice that $\{ zB \colon z\in \mathbb{C} \}$ is, as an ordered set, isomorphic to $[0, \infty)$. The isomorphism is $zB \mapsto |z|$.

Suppose we have a ring $R$ and a $G \subseteq R$ satisfying:

  1. We have $0, 1, -1 \in G$.
  2. For all $x,y \in G$ we have $xy \in G$.
  3. For all $r \in R$ there exists an $s \in R$ with $rG = Gs$.

Then we can define a map $$\Vert \Vert \colon R \rightarrow \{ \sum_{i = i}^{n}a_{i}G \colon a_{i} \in R \} $$ for all $r \in R$ by the assignment $$r \mapsto rG.$$ The image of this map is partially ordered by inclusion. The smallest element is $\Vert 0 \Vert = \{ 0 \} $. For all $r,s \in R$ we have both $\Vert rs \Vert = \Vert r \Vert \Vert s \Vert$ and$\Vert r +s \Vert \subseteq \Vert r \Vert + \Vert s \Vert$. This summation property is why I used a collection of finite sums for the range of the function. The idea is to use such a function a type of absolute value for a arbitrary ring. It would be nice to have a definition for $G$ so that when the construction was applied to the complex numbers the result yielded a structure isomorphic to the usual absolute value for complex numbers. Perhaps such a construction is impossible. If so it would be nice to know that as well.

  • $\begingroup$ You can put an actual absolute value on rings that are algebras over a field (i.e. vector spaces.) You might also be interested in reading about non-Archimedean fields like $\mathbb{Q}_p$, if you haven't yet. This is related to an analogy between the non-units of a valuation ring and the open unit disk, with the units corresponding to the unit circle. None of this is exactly what you're asking for, but all of it could be germane. $\endgroup$ Oct 11, 2012 at 20:49
  • $\begingroup$ Relating more directly to your question, the unit disk is naturally a special case of the unit ball in any normed field or algebra; as for valuation rings, it's perhaps ad hoc, but the unit disk is a valuation subring of $\mathbb{C},$ and it and its reflection across the unit circle are even the maximal ones not equal to $\mathbb{C}$ itself. $\endgroup$ Oct 11, 2012 at 20:54

3 Answers 3


There can't be any such construction that starts only from the ring structure.

To see this, note that there exist non-continuous "wild" field automorphisms of $\mathbb C$, and these do not preserve the closed unit disk. However, any purely algebraic definition of $B$ would necessarily commute with isomorphisms.

On the other hand, if you have a topology on your ring, you could say something like "the smallest closed subset of the ring that contains all roots of unity and is closed under arithmetic means".

  • $\begingroup$ Hm, taking that arithmetic means idea, you could add all elements $x \in R$ to $G$ for which $x+x = a - b$ where $a,b \in G$. $\endgroup$
    – fgp
    Oct 11, 2012 at 21:44
  • $\begingroup$ @fgp: Something like that was what I meant. (Of course arbitrary arithmetic means don't always exist in a ring unless the number of elements to take the mean of happens to be invertible). $\endgroup$ Oct 11, 2012 at 21:47
  • $\begingroup$ Just wanted to emphasize that you don't need a topology for that part. Question is, of course, if it's sufficient for $G$ to be a dense subset of the unit disc in $\mathbb{C}$... $\endgroup$
    – fgp
    Oct 11, 2012 at 21:51
  • $\begingroup$ @fgp: Oh yes, the topology is for taking the closure of the result. $\endgroup$ Oct 11, 2012 at 21:58
  • $\begingroup$ The smallest closed subset of $\mathbb{Q}_p$ that contains all roots of unity and is closed under arithmetic means is $\mathbb{Q}_p$ (any rational number can be written as an arithmetic mean of $0$, $-1$ and $1$). So this definition might not be very interesting in general. Maybe "the closure of the set of points $x$ for which the series $\sum x^n$ converges" works in more cases (I think it yields the unit ball whenever the topology is given by a norm). $\endgroup$
    – Joel Cohen
    Oct 11, 2012 at 23:14

I don't think there's a purely algebraic way to characterize the unit ball in $\mathbb{C}$ as is pointed out by Henning Makholm.

However, I think your approach to try and define some notion of "abstract unit ball" is interesting (but I think your conditions are not sufficient, for example you might want to exclude $G = R$). A more standard approach is simply to call an absolute value on a ring $R$ any function $|| : R \to \mathbb{R}$ satisfying the following properties :

  1. $|x| = 0$ if and only if $x = 0$
  2. $|xy| = |x| |y|$
  3. $|x + y| \le |x| + |y|$

Conditions 1 and 3 insure $d(x,y) = |x-y|$ is a distance on $R$ for which addition is continuous, and condition 2 insures multiplication and inversion (when defined) is continuous. So in a nutshell, it puts a nice topology on such a ring. Two absolute values on $R$ are said to be equivalent if they define the same topology. Well known example of rings with absolute value include $\mathbb{C}$ (or any subring such as $\mathbb{R}$ or $\mathbb{Q}$) with the usual absolute value. Notice that $x \mapsto \sqrt{|x|}$ is also an absolute value, but it is equivalent to $x \mapsto |x|$.

From properties 1 and 2, we see that the existence of an absolute value implies the ring is an integral domain, and the absolute value can be naturally extended to its field of fraction. So it's hopeless to expect an absolute value on an arbitrary ring (to include some additional rings, you might relax condition 2 to $|xy| \le |x| |y|$, and the product would still be continuous) and you might as well assume that $R$ is a field.

Also a given field can have many inequivalent absolute values (i.e. yielding different topologies). It is the case for $\mathbb{Q}$. Fix a prime $p$ and for any $x \in \mathbb{Q}^*$, denote $v_p(x)$ the exponent of $p$ in the decomposition of $x$ as a product of (possibly negative) powers of primes ($v_p(x)$ is called the $p$-adic valuation of $x$). We define the $p$-adic absolute value by $|x|_p = p^{-v_p(x)}$ (and $|0|_p = 0$) and check that this is an absolute value on $\mathbb{Q}$ (actually, we have a stronger version of property 3 : $|x+y|_p \le \max(|x|_p, |y|_p)$ called the ultrametric inequality). The topology induced on $\mathbb{Q}$ by this $p$-adic absolute value is very different from the one you get with the usual absolute value. The unit ball in $\mathbb{Q}$ is the subring of rational numbers $x$ that can be written $x = \frac{a}{b}$ with $a, b \in \mathbb{Z}$ and $b$ coprime to $p$. So you get a different ball for every prime $p$, and also different from the usual unit ball $\{x \in \mathbb{Q}, |x| \le 1\}$. But all of them are unit balls for some absolute value. So even from this point of view, the unit ball in a field is not unique.

Now you might wonder if there are other absolute values on $\mathbb{C}$. The answer is yes : in some sense, we can extend the $p$-adic absolute value to $\mathbb{C}$. The process is quite intricate : first take the completion $\mathbb{Q}$ with regards to $||_p$, call that $\mathbb{Q}_p$, then extend $||_p$ to the algebraic closure $\overline{\mathbb{Q}}_p$ (this step is not obvious, but we can show there a unique way of doing so) and denote $\mathbb{C}_p$ the completion of $\overline{\mathbb{Q}}_p$ with regards to $||_p$. We can show $\mathbb{C}_p$ is algebraically isomorphic to $\mathbb{C}$, so you can "transport" $||_p$ to $\mathbb{C}$ (but it's better to denote $\mathbb{C}_p$ to keep track of the topology we put on it). The unit ball in $\mathbb{C}_p$ is a subring of $\mathbb{C}_p$ that does not contain the number $\frac{1}{p}$, so once again, it's very different from the unit ball in $\mathbb{C}$.

  • $\begingroup$ Henning Malcolm answered the question I asked but your answer provides interesting information as well. The idea for this question was that if the "norm" yields the usual topology for a commonly used topological ring then the "norm" might yield a nice topology for the more arbitrary rings. Thank you. $\endgroup$
    – Jay
    Oct 12, 2012 at 0:36
  • $\begingroup$ Joel, @Jay: Who is this Malcolm you guys keep speaking of? $\endgroup$ Oct 12, 2012 at 9:31
  • $\begingroup$ Henning Makholm Sorry about that. $\endgroup$
    – Jay
    Oct 12, 2012 at 13:24

You could add

  • If $y \in R$ and there is a $n \in \mathbb{N}$ with $y^n \in G$ then $y \in G$.

to your list of requirements. In $\mathbb{C}$, $G$ would then at least be a dense subset of the unit circle (the unit circle plus $0$, actually). I don't see how you can make $G$ be the unit disc in $\mathbb{C}$ with a definition that works on arbitrary rings - it seems that you need some sort of topology for that.


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