April 3rd, 2018, 09:41 AM  #21  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 284 Thanks: 86 Math Focus: Algebraic Number Theory, Arithmetic Geometry  What exactly does this mean? The left hand side of this is an element of $\mathbb{Z}[\sqrt{d}]$ but the right hand side is not. Quote:
Since $d$ is positive, $\mathbb{Z}[\sqrt{d}]$ is a subset of $\mathbb{R}$, so it is an ordered set.  
April 4th, 2018, 09:48 AM  #22  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 
What does $\displaystyle \equiv$ mean? If you don't like this notation, forget it. Irrelevant. What does $\displaystyle a+b\sqrt{d} > e+ f\sqrt{d}$ mean? What does $\displaystyle a+b\sqrt{d}$ is positive mean? Quote:
Quote:
Or are you saying the d in $\displaystyle Z(\sqrt{d})$ are ordered? So what?, any subset of R is ordered. Last edited by zylo; April 4th, 2018 at 10:01 AM.  
April 4th, 2018, 11:09 AM  #23  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 284 Thanks: 86 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
Here's a trick: if I give you any positive integer $d$ and any integers $a$ and $b$, you can check if $a + b \sqrt{d}$ is positive by typing it into a calculator. For example, $1 + 2 \sqrt{3} \approx 4.46$ is positive, while $4  3\sqrt{5} \approx  2.71$ is negative. $a + b \sqrt{d} > e + f \sqrt{d}$ if and only if $(a  e) + (b  f) \sqrt{d}$ is positive. Quote:
$\mathbb{Z}[\sqrt{d}]$ is a subset of R since all its elements are real numbers. Therefore it is ordered according to what you've just said (which is also what I've been saying the past god knows how many posts lol). Last edited by cjem; April 4th, 2018 at 11:39 AM.  
April 4th, 2018, 02:21 PM  #24 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 
To be specific: $\displaystyle Z[\sqrt{2}]= a+b\sqrt{2}$ defines an ordered pair (a,b) just as a+ib defines an ordered pair (a,b). (a,b)+(c,d) = (a+b,c+d), etc 
April 4th, 2018, 03:20 PM  #25 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 284 Thanks: 86 Math Focus: Algebraic Number Theory, Arithmetic Geometry  If you want to treat elements of $\mathbb{Z}[\sqrt{d}]$ as ordered pairs in this way, multiplication would be $(a,b) . (a', b') = (aa' + dbb', ab' + a'b)$, and the ordering would be $(a,b) < (a',b')$ if and only if (the real number) $a + b \sqrt{d}$ is less than (the real number) $a' + b' \sqrt{d}$.

April 5th, 2018, 07:56 AM  #26  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 
The literal evaluation of a+b$\displaystyle \sqrt{d}$ does not give a member of Z[$\displaystyle \sqrt{d}$] so does not qualify for ordering, which is the same reason that z, z complex, doesn't. Please, just look it up. Apparently you are unaware that an answer to OP has long since been found: Quote:
To show there is no integer between 0 and 1, assume there is. Then by well ordering principleof positive integers the set of integers less than 1 has a least member $\displaystyle 0<m<1$. Then $\displaystyle 0<m^{2}<m$ is a contradiction. (Theorem 3) Z[$\displaystyle \sqrt{2}$] is given as an example of an integral domain, NOT ORDERED, which has nonunit divisors of 1. Standard example by the way. pg 2 BM Birkhoff MacLane, A Survey of Modern Algebra 3rd Ed  
April 5th, 2018, 10:44 AM  #27  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 284 Thanks: 86 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
Now, unless you think this exercise is a trick question, perhaps it's time to rethink your position.  
April 9th, 2018, 10:13 AM  #28  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115  Quote:
"Define "positive" element in the domain $\displaystyle Z[\sqrt{2}]$, and show that the addition, multiplication and trichotomy laws hold." Also from same reference: "An integral domain D is said to be ordered if there are certain elements of D, called the positive elements, which satisfy the addition, multiplication and trichotomy laws for integers." my underlining. "For a given integer a, one and only one of the following alternatives holds: either a is positive, or a=0, or a is positive." Fair enough. Frankly I had given up on the thread. Glad I came back. That's an excellent (educational) question and I would like to know the answer. I note positive is in quotes. When I see an answer I will rethink my postion. Off hand, the only thing that occurs to me is a+b$\displaystyle \sqrt{2}$ is positive iff a is positive. But that defines a PROPERTY of the element, P(z)=a, rather than the element itself, as positive, which is not strictly the definition of positive, it is "positive," so it is a trick question (which may have been the intent) and my position is unchanged. I note that if this were true, the same would hold for complex numbers, which do not have positive members. But I welcome another answer.  
April 9th, 2018, 12:11 PM  #29  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 284 Thanks: 86 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
Recall: $d$ is positive and squarefree (i.e. the only square number dividing $d$ is $1$). Now, the easy and enlightening answer is: say $a + b \sqrt{d} \in \mathbb{Z}[\sqrt{d}]$ is "positive" if $a + b \sqrt{d}$ is a positive real number. Since $\mathbb{Z}[\sqrt{d}]$ is a subdomain of the ordered domain of real numbers, this makes everything work (cf exercise 9, section 1.3). But if you don't like that for some reason, we can do things directly and without referencing $\mathbb{R}$ (but it will take a lot more work). I'll start things off for you. ____________________________________ Say $a + b \sqrt{d} \in \mathbb{Z}[\sqrt{d}]$ is "positive" if either: (i) $a > 0$ and $b > 0$ (ii) $a > 0$, $b \leq 0$ and $a^2 > b^2 d$ (iii) $a \leq 0, b > 0$ and $a^2 < b^2 d$ (Aside: these are precisely the conditions for $a + b\sqrt{d}$ to be positive as a real number  I didn't just pluck them out of thin air!) Let's check that this definition makes $\mathbb{Z}[\sqrt{d}]$ into an ordered domain. 1) Check that the sum of two "positive" numbers is "positive". Let $a + b \sqrt{d}$, $a' + b' \sqrt{d}$ be "positive". If both of them satisfy (i) above, then so does their sum, so it is positive. Suppose $a + b \sqrt{d}$ satisfies (i) (so $a, b > 0$) and $a' + b' \sqrt{d}$ satisfies (ii) (so $a' > 0, b' \leq 0$ and $(a')^2 > (b')^2 d$). Then their sum is $(a + a') + (b + b') \sqrt{d}$. Since $a > 0$ and $a' > 0$, certainly $a + a' > 0$. From here there are two possibilities: either $b > b'$ or $b \leq b'$. In the first case, $b + b' > 0$ and the sum $(a + a') + (b + b') \sqrt{d}$ satisfies (i), so is positive. In the second case, notice that $\begin{align*} (a+a')^2  (b+b')^2 d &= a^2 + 2aa' + (a')^2  b^2 d  2bb'd  (b')^2 d \\ &> a^2 + 2aa'  b^2 d  2bb'd \qquad \qquad \qquad \qquad &\text{(as $\: (a')^2 > (b')^2 d \:$ by assumption)} \\ &= a^2 + 2 a a' + 2 b b' d  b^2 d &\text{(as $ \:a, a', b > 0$, $b' < 0$)} \\ &> 2bd (b'  b) &\text{(as $\: a^2 + 2 a a' > 0$)} \\ &\geq 0 &\text{(as we're in the case $ \: b \leq b')$} \end{align*}$ This shows $(a + a')^2 > (b+b')^2$, so the sum $(a + a') + (b + b') \sqrt{d}$ satisfies (ii) and so is positive. Continue this way: for every other combination of (i), (ii), (iii) $a + b\sqrt{d}$ and $a' + b'\sqrt{d}$ might satisfy, check that their sum also satisfies one of them. I won't bother as it is quite time consuming, but I encourage you to have a go! 2) Check that the product of two "positive" numbers is positive. Same deal here: check that no matter what combination of (i), (ii), (iii) $a + b\sqrt{d}$ and $a' + b'\sqrt{d}$ could satisfy, their product satisfies one of them. 3) Check that every element $a + b\sqrt{d}$ of $\mathbb{Z}[\sqrt{d}]$ satisfies precisely one of the following: $a + b \sqrt{d} = 0$ (i.e. $a = b = 0$), $a + b \sqrt{d}$ is "positive" or $(a + b \sqrt{d}) = a  b \sqrt{d}$ is "positive". As $0 = 0 + 0 \sqrt{d}$ does not satisfy the definition of "positive", it suffices to show the following two statements: (a) if $a + b \sqrt{d}$ is nonzero and not positive, then $a b\sqrt{d}$ is positive (b) if $a + b \sqrt{d}$ is positive, then $a b\sqrt{d}$ is not positive Let's start with (a): Suppose $a + b\sqrt{d} \neq 0$ satisfies none of (i), (ii), (iii). i.e. the three following statements hold: (i)' $a \leq 0$ or $b \leq 0$ (ii)' $a \leq 0$, $b > 0$ or $a^2 \leq b^2 d$ (iii)' $a > 0$, $b \leq 0$ or $a^2 \geq b^2 d$ But this means the following hold (i)'' $a \geq 0$ or $b \geq 0$ (ii)'' $a \geq 0$, $b < 0$ or $a^2 \leq b^2 d$ (iii)'' $a < 0$, $b \geq 0$ or $a^2 \geq b^2 d$ Now $a + b\sqrt{d} \neq 0$ combined with (i)'' implies $a > 0$ or $b > 0$. If $a > 0$ and $b > 0$, then $a  b\sqrt{d}$ satisfies (i) and we're done. Else $a \leq 0$ or $b \leq 0$. If $a \leq 0$, then $b > 0$ (see the paragraph above starting with "Now"), and now by (ii)'', $a^2 \leq b^2 d$. But this inequality must be strict as $d$ is assumed squarefree. So $a  b\sqrt{d}$ satisfies (ii) in this case. The last possibility is that $b \leq 0$. In this case, we must have $a > 0$, and then (iii)''' implies $a^2 \geq b^2 d$. Again, $d$ squarefree means this inequality is strict, so $a  b\sqrt{d}$ satisfies (iii). So in every case, $a b \sqrt{d}$ satisifes one of (i), (ii), (iii), so is positive. This shows (a). Now for (b): A similar kind of argument works. Feel free to do this if you're still not convinced. ____________________________________  
April 10th, 2018, 07:37 AM  #30 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 
cjem points out that a+b$\displaystyle \sqrt{d}$ $\displaystyle \epsilon$ Z$\displaystyle [\sqrt{d}]$ can be ordered by calling those members positive that evaluate to a real positive number For example, 53$\displaystyle \sqrt{2}$ "=," ie, evaluates to, 1.414, which is positive. Addition, multiplication, and trichotomy follow trivially. cjem also noted Z$\displaystyle [\sqrt{d}]$ was ordered because it was a subset of the real numbers. I never felt the breeze. My apologies, and thanks to cjem for his persistence. The next question, well ordering, is relevant to OP. If 1 is an integer, the OP has been answered utilizing the wellordering property of the integers. But what if 1$\displaystyle \equiv$(1,0) is instead the unit of Z$\displaystyle [\sqrt{d}]$? Are the only units of Z$\displaystyle [\sqrt{d}]$ (divisors of 1) $\displaystyle \pm$1? No, as cjem showed. Last edited by zylo; April 10th, 2018 at 07:54 AM. 