Discrete Subrings of Complex Numbers: Topology of Rings

In summary, the conversation discusses discrete subrings containing 1 of the complex numbers and their properties. The question is when does a subring, such as ##\mathbb{R}##, become dense in the reals. It is suggested that the answer is when it contains an element, ##|x|<1##. The ring generated by 1 and ##\sqrt{2}## is considered and it is shown that every element of the ring is an accumulation point. The next step is to show that any real number is also an accumulation point of some sequence in the ring. It is then discussed that the rational numbers are dense in ##\mathbb{R}## and that all discrete subrings of
  • #1
Paul Colby
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Hi,

I've[1] recently become interested in discrete subrings containing 1 of the complex numbers. Being complex numbers these rings have all sorts of properties but my question may be formed in terms of the reals. The question is; when does a subring, say of the reals, ##\mathbb{R}##, becomes dense in the reals? I think the answer is when it contains an element, ##|x|<1##. Let's look at the ring generated by 1 and ##\sqrt{2}##, ##R=\langle 1,\sqrt{2}\rangle##. Any member of this ring, ##r\in R##, may be written,

##r = n+m\sqrt{2}##​

where ##n,m\in \mathbb{Z}##. Okay, consider ##x = \sqrt{2}-1##. We have that ##|x| < 1## so the sequence, ##x_n = x^n## where ##n\in\mathbb{N}## accumulates at ##x=0##. Since the ring is closed under addition, every element of the ring is an accumulation point. The next step is showing that any real ##y## is also an accumulation point of some sequence in the ring. My thought was given ##y\notin R## there is always an ##r\in R## such that ##|y-r| < 1##. Given this ##r## we may always find a small ##x\in R## about 0 we can add so that, ##|y - r - x| < |y - r|##. So there is no smallest ##x## we can always find more ring elements near ##y##.

[1] My training is in physics.
 
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  • #2
Yes that works. In fact we don't even need the axiom of choice, since the ring is countable. Assume an index mapping of the positive integers to the ring. We then use the fact that the ring accumulates at 0.
Given ##y\notin R##, define a sequence ##s_n## with ##s_0=0## and ##s_{n+1}## calculated as follows.
We assume wlog that ##y>0##, since if the ring accumulates at ##y## it must also accumulate at ##-y## by just taking the negative of any sequence that accumulates at ##y##.
Assume ##s_n\in R \wedge s_n\ge 0##.
Let ##x## be the least-indexed, positive ring element that is less than ##y-s_n##. Then define:
$$s_{n+1} = s_n + \left\lfloor \frac{y-s_n}{x} \right\rfloor x$$
which we see must be in ##R## being an integer combination of ring elements.
Consideration of cases based on whether ##x## is greater or less than ##\tfrac12 (y-s_n)## (it cannot equal it, else ##y## would be in ##R##) shows us that ##s_{n+1}\ge s_n + \tfrac12(y - s_n)## so that the distance remaining from ##y## at least halves at every step of the sequence.
Hence for any ##\epsilon>0## we can find ##M## such that ##n\ge M\Rightarrow |y-s_n|<\epsilon##.
 
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  • #3
The rational numbers are dense i ℝ.
 
  • #4
andrewkirk said:
Yes that works. In fact we don't even need the axiom of choice, since the ring is countable.
Thanks. I've known for years that the rationals are dense in ##\mathbb{R}## but never expected ##n+m\sqrt{2}## to be so. I know this must be very well known but it was kind of startling. Made me doubt the proof I was constructing. So, if ##R\subset \mathbb{R}## is a ring and ##\mathbb{Z}\subset R## is a proper subset, then ##R## is dense in ##\mathbb{R}##. This is actually quite useful in attacking the discrete subrings of ##\mathbb{C}##. Really limits what one may do.
 
  • #5
Actually, after some reflection a much stronger statement is true. All discrete sub rings, ##R\subset\mathbb{R}##, of the reals are contained in the integers, ##R\subset\mathbb{Z}##.

proof: Let ##R\subset\mathbb{R}## which we assume discrete. If ##R-\mathbb{Z}## is empty, we’re done. So assume we’re not done and pick ##a## from this set. We may construct sequences in ##R## of the form $$r_n(a,m)=a(a-m)^n$$ where, ##n\in\mathbb{N}## and ##m\in\mathbb{Z}##. Clearly, ##a-m## need not be in ##R##, however, expanding ##(a-m)^n## yields only a single integer term, ##m^n##, which may lie outside of ##R##. The overall factor of ##a## assures, ##r_n(a,m)\in R##.

Okay, we may choose ##m## such that ##0 < \epsilon= a-m <1##. The sequence $$|r_n(a-m)| = |a||\epsilon|^n$$ accumulates at zero for large ##n## which contradicts ##R## being discrete. Therefore, all discrete sub rings of the reals are contained in the integers.
 
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  • #6
I don't understand the proof. Since the ring contains the integers, ##a-m\in R##. Also there is no reason for ##R## to be an ideal of the reals, so the a factor is not sufficient (nor necessary, since ##R## contains 1, and hence all integers)

And if you did not intend to restrict to rings with 1, I think the set of even numbers is a counterexample.
 
  • #7
Office_Shredder said:
I think the set of even numbers is a counterexample.
The even numbers[1] are discrete and therefore contained in the integers which is the conclusion obtained in #5.

I think you’re confused by the posts prior to #5 which did assume ##1\in R## and hence contain the integers. The proof in #5 does not make this assumption. As, I said in #5, it’s a much stronger statement.

[1] I assume by number you mean integers.
 
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  • #8
Sorry I misread your post as claiming the ring must equal the integers.

I reread your proof and it makes more sense to me now - ##a(a-m)^n## is a sum of terms of the form ##a^k n## for some integer ##n## and ##k>0##, so all must lie in ##R##.

Cool proof!
 
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  • #9
nice observation. I initially thought this was wrong, but you are absolutely correct. I was confused as to the difference between a submodule and a subring of the reals. Does this work?: If S is discrete and has a non zero element, then it has a least positive element a. Then S consists precisely of the reals of form am where m is an integer. But then a^2 = ak for some integer k, whence a=k.
 
  • #10
mathwonk said:
Does this work?: If S is discrete and has a non zero element, then it has a least positive element a. Then S consists precisely of the reals of form am where m is an integer. But then a^2 = ak for some integer k, whence a=k.

I think this may hinge on ##S## being a Euclidian ring[1]. For ##S## to be Euclidian one needs ##d(a)=|a|## to be an integer which is quite an assumption.

[1] "Topics in Algebra" Herstein 1964 - pages 104-105
 
  • #11
I think it works without any assumption. In fact it is essentially your argument in line -4 of post #5, i.e. if there is any element of S that is not of form ma, then some element x of S lies between say ma and (m+1)a. But ma < x < (m+1)a implies that 0 < x-ma < a, a contradiction to a being the smallest positive element of S. is this ok?
 
  • #12
Neat. That certainty works.
 

1. What are discrete subrings of complex numbers?

Discrete subrings of complex numbers are subsets of the complex numbers that form a ring under the operations of addition and multiplication. They are called "discrete" because they do not contain any accumulation points, meaning that they are separated from each other by a finite distance.

2. How are discrete subrings different from other subrings of complex numbers?

Unlike other subrings of complex numbers, discrete subrings have a unique topology, or way of measuring distance between elements. This topology is defined by the discrete metric, where the distance between any two distinct elements is always 1. This makes discrete subrings behave differently than other subrings in terms of convergence, continuity, and connectedness.

3. What is the significance of studying the topology of discrete subrings?

The topology of discrete subrings has important implications in number theory, algebraic geometry, and other areas of mathematics. It helps us understand the structure and behavior of these subrings, and can also be used to prove theorems and solve problems related to discrete subrings.

4. Can every subring of complex numbers be considered a discrete subring?

No, not every subring of complex numbers can be considered a discrete subring. In order for a subring to be discrete, it must satisfy the properties of being separated from each other by a finite distance and having a unique topology. Some subrings, such as the set of all algebraic integers, do not meet these criteria and therefore cannot be classified as discrete subrings.

5. Are there any applications of discrete subrings in real-world problems?

Yes, there are many real-world applications of discrete subrings, particularly in cryptography and coding theory. Discrete subrings are used to construct error-correcting codes and to encrypt data in secure communication systems. They are also used in the study of prime numbers and other number-theoretic problems.

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