Is there an alternate definition for $\{ z \in \mathbb{C} \colon \vert z \vert \leq 1 \} $.

Question

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.

Answer 1

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.

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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".

Answer 2

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.

Answer 3

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 $\mathbb{C}$.