# Is this a property of an integral domain that is not a field?

I am working on a specific problem and I've almost got it solved. To solve it, however, I need to prove one last claim (if it is even true):

Consider an integral domain $R$ that is not a field. Let $a$ be an element of $R$ that is not a unit or $0$. I want to prove that there cannot exist nonzero elements $r_n$, $n \in \mathbb{N}$, of $R$ such that

$$r_1 a=r_2a^2=r_3a^3=\cdots= r_na^n=\cdots$$

First of all, is this even true? My intuition says it is. And am I missing something obvious? This is the last step in a long proof, so I hope it is true!

EDIT: It turns out my proof goes through without this being true (which is good since it's false), but definitely thanks for the answers. It has improved my understanding.

-
Well, if you have something like that then you can manufacture a strictly increasing sequence of ideals, so if you know you have a noetherian ring you would have a contradiction. –  Zhen Lin Dec 6 '12 at 18:49
I have a feeling this might fail in a ring which contains an element which cannot be factored into irreducible elements. –  Tobias Kildetoft Dec 6 '12 at 18:53
So this does hold in, a UFD for instance. –  JSchlather Dec 6 '12 at 18:58
The "not a field" requirement seems artificially added to get rid of the obvious finite-field examples like $2 \in \mathbb{Z}_3$. But you could easily extend to, say, $2 \in \mathbb{Z}_3[x]$, and then it's no longer a field but still a counterexample. –  MartianInvader Dec 6 '12 at 19:41
@MartianInvader But 2 is a unit in that ring. –  Three Dec 6 '12 at 19:59

This is false. As Zhen Lin points out, $R$ can't be Noetherian, so let $R$ be my favourite non-Noetherian domain, $\{f(x)\in \Bbb{Q}[x]\mid f(0)\in \Bbb{Z}\}$, and consider $a=2$. Then $$x=2\frac{x}{2}=4\frac{x}{4}=8\frac{x}{8}=\ldots$$ but $2$ is not a unit.

-
Aha! I was trying to extract this from non-standard integers example, but I hadn't realized the ring had such a simple description! –  Hurkyl Dec 6 '12 at 18:58
@Hurkyl, are you saying that the ring provided by Chris Eagle is isomorphic to nonstandard integers? –  PrimeRibeyeDeal Jun 6 '13 at 21:24
@PrimeRibeyeDeal: No, but it is isomorphic to a subring of the nonstandard integers (send $x \to 2^H$). ("subring" is given its standard meaning) –  Hurkyl Jun 7 '13 at 8:00
@Hurkyl, what is $2^H$ in this context? Is there a "nice" description of the nonstandard integers as a ring? –  PrimeRibeyeDeal Jun 9 '13 at 17:23
@PrimeRibeyeDeal: My comments were referring to my answer in this thread. Re your other question, the only "nice" description of the nonstandard integers I know of is that it is a nonstandard model of the integers. –  Hurkyl Jun 9 '13 at 21:30

Let $R = {}^\star \mathbb{Z}$ be the hyper-integers; that is, the ring of integers in a non-standard model of real analysis.

Let $H$ be an infinite hyper-integer.

Then, we can choose $a = 2$ and $r_i = 2^{H - i}$ to get a counter-example.

-
Hum. If I construct the nonstandard reals via an ultraproduct, does this denote the subring of the nonstandard reals which are a.e. equal to an integer? –  Qiaochu Yuan Dec 6 '12 at 19:08
@Qiaochu Yuan: Yes. And in that case, we might as well make an explicit choice of $H$ and set $r_i = ( 2^{-i}, 2^{1-i}, 2^{2-i}, \cdots )$. –  Hurkyl Dec 6 '12 at 19:09

No. Let $R = \mathbb{Z}[\frac{1}{x}, 2x, 2x^2, 2x^3, ...]$, let $r_i = 2x^i$, and let $a = \frac{1}{x}$.

-
$a$ was specified not to be a unit. –  Greg Muller Dec 6 '12 at 18:49
@Greg: fixed. ${}$ –  Qiaochu Yuan Dec 6 '12 at 18:55
Let $R$ be the ring of functions $[0,1]\to \mathbb R$ that are analytic in $(0,1]$ and continuous on $[0,1]$. Let $a$ be the function $x\mapsto x$. It is not a unit because $a(0)=0$.
For each $n\in \mathbb N$, let $$r_n(x)=\begin{cases}x^{-n}e^{-\frac1x}&\text{if }x\ne 0\\0&\text{if }x=0\end{cases}$$ Note that $r_n$ is analytic in $(0,1]$ and is continuos also at $0$. We have $$(a^nr_n)(x)=\begin{cases}e^{-\frac1x}&\text{if }x\ne 0\\0&\text{if }x=0\end{cases}$$