# Principal Ideal Domain

• Jan 25th 2010, 04:40 AM
TiRune
Principal Ideal Domain
I've run into this question that should be easy somehow, but I can't get a hold of:

$R = \{ \frac{a}{3^k} \in \mathbb{Q} : a \in \mathbb{Z}, k \in \mathbb{Z}_{\geq 0} \}$

Is R a PID?

Naturally, $\mathbb{Q}$ is a PID since it's a field. But since R isn't an ideal in $\mathbb{Q}$, I can't use that in an argument in any way. (It's not, because 1/2*r with r in R isn't in it by default, and 1/2 is in $\mathbb{Q}$).
Not every element has a multiplicative inverse in R, since a can be negative, so it's also surely not a field (which would also imply it is a PID)...
So how do I go about it? R is isomorphic with $\mathbb{Z} \times \mathbb{Z}_{\geq 0}$. Would that form an argument somehow?

Also, can anyone confirm for me that R/2R is isomorphic to $\mathbb{Z}/2\mathbb{Z}$ and R/3R is isomorphic to $\mathbb{Z}/3\mathbb{Z}$ for me?
My argument is that you can make a hormomorphism f: $R \to \mathbb{Z}/2\mathbb{Z}$ by $r \mapsto 0$ if a is even, and $r \mapsto 1$ if a is odd. This way only all elements in 2R can be found in $Ker(f)$. So by the isomorphism theorem $R/2R \cong \mathbb{Z}/2\mathbb{Z}$.
And the same argument applies for R/3R, by sending any r with a that is p mod 3 to p.
• Jan 25th 2010, 03:14 PM
clic-clac
Hi

First show that $R$ is an integral domain. What are its invertibles?

$i:\mathbb{Z}\rightarrow R:\ n\mapsto \frac{n}{1}$ is a ring morphism. (why?) (We can see $\mathbb{Z}$ as a subring of $R$)

Let $I$ be an ideal of $R,\ k\mathbb{Z}$ its preimage by $i$ ( $\mathbb{Z}$ is a PID and the preimage of an ideal by a morphism is an ideal).

Clearly, $kR\subseteq I.$ Let $r$ be an element of $I,$ $r=\frac{a}{b}$ with hypotheses on $b;$ show that $r\in I$ iff $i(a)\in I.$ Therefore $a\in k\mathbb{Z}.$ Conclude.
• Jan 25th 2010, 03:33 PM
TiRune
Hmm, think I get the point now. with I an ideal, any $3^n$ in the denominator with any arbitrary numerator is in it, due to the ideal property and $\frac{1}{3^n} \in R$. So we can ignore that, and the upper part is basically just $\mathbb{Z}$ so it's a PID. Not mathematically very well put, but I see that's the idea.

How bout the second part btw, is my "proof" correct?
• Jan 26th 2010, 01:47 AM
clic-clac
I can't confirm that $\mathbb{Z}/3\mathbb{Z}\cong R/3r$: $3$ is invertible in $R,$ so $3R$ is .... whereby $R/3R$ is .... while $\mathbb{Z}/3\mathbb{Z}$ is not.
The error comes from:
Quote:

And the same argument applies for R/3R, by sending any r with a that is p mod 3 to p
indeed, for instance, $\frac{2}{1}=\frac{6}{3}.$ (actually, what you defined is not an application; an important thing, when there are possible equivalence classes, is to check first that the morphism you're defining is an application! It's not always as simple as it seems)

It works for $2$ :) : multiplying by an element of $\{3^n; n\geq 0\}$ does not change the parity. You could also have used the canonical injection $i$.