Discrete Subrings of Complex Numbers: Topology of Rings

[Paul Colby]

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.

 

[andrewkirk]

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 $n\ge M\Rightarrow |y-s_n|<\epsilon$.

 

[Svein]

The rational numbers are dense i ℝ.

 

[Paul Colby]

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.

 

[Paul Colby]

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.

 

[Office_Shredder]

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.

 

[Paul Colby]

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.

 

[Office_Shredder]

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!

 

[mathwonk]

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.

 

[Paul Colby]

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

 

[mathwonk]

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?

 

[Paul Colby]

Neat. That certainty works.