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April 10th, 2018, 04:00 PM   #41
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Quote:
Originally Posted by cjem View Post
By well-ordered domain, I mean an ordered domain whose positive elements form a well-ordered set.
Perfectly reasonable.
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April 11th, 2018, 06:21 AM   #42
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$\displaystyle Z[\sqrt{d}]$ is not well-ordered: (0,1) doesn't have a least element.

$\displaystyle -\varepsilon < a-b\sqrt{d} < \varepsilon\\
-\varepsilon /b < a/b - \sqrt{d} < \varepsilon/b < \varepsilon$

and I can choose a/b greater than and as close to $\displaystyle \sqrt{d}$ as I like (but never equal).
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April 11th, 2018, 07:24 AM   #43
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Let's compromise.
$\displaystyle Z[\sqrt{d}]\equiv (a,b)=z$ is an Abelian Group.
(a,b)+(c,e)=(a+c,b+e)
assoc and comm ok
e=(0,0)
z+e=z
z+(-z)=e
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April 11th, 2018, 08:54 AM   #44
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Quote:
Originally Posted by zylo View Post
Let's compromise.
$\displaystyle Z[\sqrt{d}]\equiv (a,b)=z$
Would you mind defining precisely what you mean by $\equiv$ here?

Anyway, of course the group $(\mathbb{Z}[\sqrt{d}],+)$ is isomorphic to the group $(\mathbb{Z} \times \mathbb{Z},+)$ of ordered pairs of integers. But the groups are not literally equal: the elements of the first group are all real numbers, while the elements of the second are all ordered pairs of integers.
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April 12th, 2018, 07:02 AM   #45
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Is Z[sqrtD] isomorphic to the rational numbers/

An interesting question (isomorphism):

Is there a 1:1 mapping from $\displaystyle Z[\sqrt{D}]\equiv a+b\sqrt{D}$ to the rational numbers expressed as an ordered pair $\displaystyle (a,b)$ st
$\displaystyle (ef)'=e'f'$
where unprimed are the rationals. Try:
$\displaystyle e=(a,b), e'=a+b\sqrt{D}$. Then
$\displaystyle ef=(a,b)(c,d)=(ac,bd)$, and $\displaystyle (ef)'=ac+bd\sqrt{D}$
$\displaystyle e'f'= (ac+bdD)+(ad+bc)\sqrt{D}\neq (ef)'$

I didn't think it would work. I was just setting up a train of thought to find a mapping that works. And don't forget addition:
(a,b)+(c,d) = (ad+bc,bd).

Looks daunting.

(I do not like formal set-theoretic language when unnecessary)
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April 12th, 2018, 07:29 AM   #46
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Quote:
Originally Posted by zylo View Post
An interesting question (isomorphism):

Is there a 1:1 mapping from $\displaystyle Z[\sqrt{D}]\equiv a+b\sqrt{D}$ to the rational numbers expressed as an ordered pair $\displaystyle (a,b)$ st
$\displaystyle (ef)'=e'f'$
where unprimed are the rationals. Try:
$\displaystyle e=(a,b), e'=a+b\sqrt{D}$. Then
$\displaystyle ef=(a,b)(c,d)=(ac,bd)$, and $\displaystyle (ef)'=ac+bd\sqrt{D}$
$\displaystyle e'f'= (ac+bdD)+(ad+bc)\sqrt{D}\neq (ef)'$

I didn't think it would work. I was just setting up a train of thought to find a mapping that works. And don't forget addition:
(a,b)+(c,d) = (ad+bc,bd).

Looks daunting.

(I do not like formal set-theoretic language when unnecessary)
The rationals are a field, $\mathbb{Z}[\sqrt{D}]$ isn't. So it can't exist.
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