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The question is :

Let $R$ be the ring defined by $R = \{a + \frac {b(1 + i{\sqrt {19}})} {2} | a,b\in \mathbb {Z} \}$.Then $R$ is a principal ideal domain but not a Euclidean domain.

I have tried my best.But I fail.Please help me.

Thank you in advance.

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    Please see whether I have edited it correctly. – 2017-01-10
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    This, as far as I know, is a rather difficult thing to prove. Try in Dummit&Foote's "Algebra", where you can find a prove of this with Hasse norms and stuff. – 2017-01-10
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    @DonAntonio do you ensure me that I will find it in Dummit & Foote?My exam is knocking at the door.Please suggest me whether it is better to read it before or after exam. – 2017-01-10
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    @A.Chattopadhyay It is the very last example, before the exercises, in section 8.2 (page 282 in the 3rd edition of the book) – 2017-01-10
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    Ok.Thanks a lot. – 2017-01-10

1 Answers 1

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I denote $R$ your ring. For instance, I will show that $R$ is not euclidian. First, I want to prove that $U(R)$ = {$1,-1$}.

Let $x\in U(R)\Rightarrow \exists y\in U(R) $ such that $x\cdot y =1\Rightarrow |x|^2\cdot|y|^2=1$.

But $|x|^2\in \mathbb{Z}\Rightarrow x\in $ {$1,-1$}.

Let suppose that $R$ is euclidian. Then, there exists $\phi$ an euclidian norm. Let $x\in R$ \ {$1,0,-1$} with $\phi(x)$ minim.

Let $I=xR$ and $y\in R$ \ {$1,0,-1$}. There exists $q,r \in R$ such that $y=x\cdot q+r$ where $r=0$ or $\phi (r)< \phi(x)$. But $\phi(x)$ is minim. So $r=0\Rightarrow y\in I\Rightarrow R/I\subseteq$ {$\hat{1}, \hat{0}, \hat{-1}$}.

I consider $\alpha=\frac{1+i\sqrt{19}}{2}$. I notice that $\alpha^2-\alpha+ 5= 0\Rightarrow \hat{\alpha}^2-\hat{\alpha}+\hat{5}=\hat{0}$ in $R/I$.

But $R/I \subseteq ${$\hat{1}, \hat{0},\hat{-1}$}.

For example, if I take $\hat{\alpha}=\hat{0}$, I obtain $\hat{5}=\hat{0}\Leftrightarrow 5\in I\Leftrightarrow 5=x\cdot z= (a+\frac{b\cdot(1+i\sqrt{19})}{2})\cdot (c+\frac{d\cdot(1+i\sqrt{19})}{2})$, where $a,b,c,d\in\mathbb{Z}$.

I apply the norm:

$100= ((2a+b)^2+19b^2)\cdot((2c+d)^2+19d^2) $, where $a,b,c,d\in \mathbb{Z}$.

It's easy to show that the right side is divisible by $16$, but $100$ isn't. So, I have a contradiction.

Now, I will prove that $R$ is principal.

Lemma: Let $R$ a domain such that there exists $f:R\rightarrow \mathbb{N}$ with properties:

1) $f(a)=0\iff a=0$ and

2) $\forall x,y\in R$, with $y$ doesn't divides $x$ there exists $a,b\in R$ such that $0

Then $R$ is a principal ring.

In your problem $f$ is the norm function. You have to find $a, b$ which satisfy 2). Then you have to apply the lemma.

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    Sorry @alexb I don't understand why $\hat{{\alpha}^2} - \hat{\alpha} + \hat{5} = \hat{0}$ leads a contradiction.Please help me.All other things are ok. – 2017-01-10
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    @A.Chattopadhyay I completed the proof. In my opinion, it's too hard for an exam. – 2017-01-11