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Associates and their Norms

Hello PhysicsForums:

I have two things I'd like to learn more about:

1. I was noticing that the norms of +-1+-i (which are associates) are the same. The same goes for +-(1-2i). Is there a general reason explaining why any two integers which are associates have the same norm?

2. Assume that 'a' is a set of quadratic integers in Q[Sqrt(-3)] so that $$\bar{a}$$ and 'a' are associates. Can someone help describe what this set would be like when this occurs?

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Recognitions:      Gold Member  1. It isn't true in general that associates have the same norm. For example, consider Q($\sqrt{2}$). Then u = 1 + $\sqrt{2}$ is a unit. Let a = 2 + $\sqrt{2}$. Then ua is an associate of a and ua = 4 + $3\sqrt{2}$. However,  N(a) = (2 + $\sqrt{2}$)(2 - $\sqrt{2}$) = 2,  but  N(ua) = (4 + $3\sqrt{2}$)(4 - $3\sqrt{2}$) = -2  This fails because norms in Q($\sqrt{2}$) can have negative values. However, if d < 0 and squarefree, then norms of elements in Q($\sqrt{d}$) are $\geq$ 0. Then the norm of any unit = 1 and the product formula for norms implies that associates have the same norm.  2. I don't understand what you mean by the phrase so that $\bar{a}$  and 'a' are associates. Could you please elaborate? Thanks.

Quote by Petek 					 				 				1. It isn't true in general that associates have the same norm. For example, consider Q($\sqrt{2}$). Then u = 1 + $\sqrt{2}$ is a unit. Let a = 2 + $\sqrt{2}$. Then ua is an associate of a and ua = 4 + $3\sqrt{2}$. However,  N(a) = (2 + $\sqrt{2}$)(2 - $\sqrt{2}$) = 2,  but  N(ua) = (4 + $3\sqrt{2}$)(4 - $3\sqrt{2}$) = -2  This fails because norms in Q($\sqrt{2}$) can have negative values. However, if d < 0 and squarefree, then norms of elements in Q($\sqrt{d}$) are $\geq$ 0. Then the norm of any unit = 1 and the product formula for norms implies that associates have the same norm.  2. I don't understand what you mean by the phrase so that $\bar{a}$  and 'a' are associates. Could you please elaborate? Thanks.
1. Is saying that any two integers which are associates the same thing as saying two associates? I was referring to the Norm of these aforementioned integers being equal, not the norm of the associates being equal. (I hope that makes sense)

2. I meant to say that $\bar{a}$ and $a$ are associates. If I would re-write it:

Assume that $a$ is a set of quadratic integers in Q[Sqrt(-3)] so that $\bar{a}$ and $a$ are associates. Can someone help describe what this set would be like when this occurs?

Recognitions:
Gold Member

Associates and their Norms

1. You originally observed that (1 + i) and -(1 + i) were associates and had the same norm (that's not exactly what you wrote, but I think is what you meant). You also observed that (1 -2i) and -(1 -2i) were associates and had the same norm. You wondered whether there was an explanation why any two integers that were associates had the same norm. I gave an example that showed this wasn't always true. However, I pointed out that your statement was true if the norm of any integer was non-negative. Here's how to prove that:

First, it's easy to see that, in general, the norm of a unit must be either +1 or -1. Since we're assuming that all norms are non-negative, then any unit must have norm = 1. Now suppose that a and b are associates. Then there exists a unit u such that a = ub. Therefore,

N(a) = N(ub) = N(u) N(b) [Product rule for norms] = N(b)

since N(u) = 1. It remains to show that if d < 0, then the norm of any integer in Q($\sqrt{d}$) is non-negative. That's a simple calculation.

I hope this makes my previous post more clear.

2. $a$ is a set, not a number, so you need to define what you mean by $\bar{a}$ and what you mean by two sets being associates. I could try to guess, but prefer that you give your definition. I'm not trying to be obstinate, but your usage isn't standard, AFAIK.

Quote by Petek 					 				 				1. You originally observed that (1 + i) and -(1 + i) were associates and had the same norm (that's not exactly what you wrote, but I think is what you meant). You also observed that (1 -2i) and -(1 -2i) were associates and had the same norm. You wondered whether there was an explanation why any two integers that were associates had the same norm. I gave an example that showed this wasn't always true. However, I pointed out that your statement was true if the norm of any integer was non-negative. Here's how to prove that:  First, it's easy to see that, in general, the norm of a unit must be either +1 or -1. Since we're assuming that all norms are non-negative, then any unit must have norm = 1. Now suppose that a and b are associates. Then there exists a unit u such that a = ub. Therefore,  N(a) = N(ub) = N(u) N(b) [Product rule for norms] = N(b)  since N(u) = 1. It remains to show that if d < 0, then the norm of any integer in Q($\sqrt{d}$) is non-negative. That's a simple calculation.  I hope this makes my previous post more clear.  2. $a$ is a set, not a number, so you need to define what you mean by $\bar{a}$ and what you mean by two sets being associates. I could try to guess, but prefer that you give your definition. I'm not trying to be obstinate, but your usage isn't standard, AFAIK.
First off, thank you for explaining that 1st point to me, I really appreciate it!

2. $$\bar{a}$$ is simply the conjugate of $a$. It is to my understanding that since $a$ is a set of quadratic integers of the form $a+b*\sqrt{-3}$, that $\bar{a}$ would be a set of integers of the form $a-b*\sqrt{-3}$.

Associate: If $\alpha$ and $\beta$ are non-zero integers in $Q[\sqrt{d}]$ such that $\alpha = \beta*\epsilon$, where $\epsilon$ is a unit, then $\alpha$ is said to be an associate of $\beta$. In other words, $\alpha$ is only an associate of $\beta$ if and only if $\alpha / \beta$ is a unit.

Recognitions:      Gold Member  Is this your question?  Let a be an integer in Q($\sqrt{-3}$). Under what conditions is $\bar{a}$ an associate of a?  If so, start by describing both the integers and units of Q($\sqrt{-3}$). Hint: There are more of them than you might think.

Quote by Petek 					 				 				Is this your question?  Let a be an integer in Q($\sqrt{-3}$). Under what conditions is $\bar{a}$ an associate of a?  If so, start by describing both the integers and units of Q($\sqrt{-3}$). Hint: There are more of them than you might think.
No, I already know what an associate is. I'm simply saying this:

If $a$ is a set of quadratic integers in Q($\sqrt{-3}$) and $\bar{a}$ is the associate set of quadratic integers with respect to $a$, what would the set of quadratic integers in $a$ have to look like? I essentially would like to know a description of $a$.

Recognitions:      Gold Member  Let's fix some notation. Perhaps that will help me to understand your question.  Let K = Q($\sqrt{-3}$).  Let $O_{K}$ denote the integers of K.  Also, let's use uppercase letters to denote sets and lowercase to denote elements of sets.  Let A $\subseteq\\O_{K}$ (A is a subset of the integers of K.)  Let $$\overline{A}$$ = {$$\bar{a}$$: a $$\in$$ A} (the set of conjugates of elements in A)  Let A~ = {ua: a$$\in$$A and u is a unit in $$O_K$$} (the set of all associates of elements in A)  As I understand it, you want A to be a subset of integers in K that satisfies certain relationships with its set of conjugates and set of associates. Is that correct? If so, what is the relationship?

Quote by Petek 					 				 				Let's fix some notation. Perhaps that will help me to understand your question.  Let K = Q($\sqrt{-3}$).  Let $O_{K}$ denote the integers of K.  Also, let's use uppercase letters to denote sets and lowercase to denote elements of sets.  Let A $\subseteq\\O_{K}$ (A is a subset of the integers of K.)  Let $$\overline{A}$$ = {$$\bar{a}$$: a $$\in$$ A} (the set of conjugates of elements in A)  Let A~ = {ua: a$$\in$$A and u is a unit in $$O_K$$} (the set of all associates of elements in A)  As I understand it, you want A to be a subset of integers in K that satisfies certain relationships with its set of conjugates and set of associates. Is that correct? If so, what is the relationship?
You are correct. I don't know what you mean by asking me what is the relationship. I'm trying to determine what the set of $A$ will be like though so that the conditions are met.

Recognitions:      Gold Member  Let me rephrase your original question using the above notation:   Assume that A $$\subseteq \\ O_K$$ is such that $$\overline{A}$$ and A are associates. Can someone help describe what this set would be like when this occurs?   So the condition on the set A is that "$$\overline{A}$$ and A are associates." In a prior post, I asked for your definition of what it means for two sets to be associates. You replied with the definition of what it means for two integers in $O_K$ to be associates. So, suppose that A and B are subsets of $O_K$. Then A and B are said to be associates if ... . Please supply the rest of the definition*. I know what it means for two numbers to be associates, but not what you mean by saying that two sets are associates. Thanks.  * For example, a possible definition would be that A and B are said to be associates if for every a $\in$ A, there exists a b $\in$ B such that a and b are associates.

Quote by Petek 					 				 				 * For example, a possible definition would be that A and B are said to be associates if for every a $\in$ A, there exists a b $\in$ B such that a and b are associates.
That is exactly what I'm saying! :-)

Recognitions:
Gold Member

Quote by Brimley 					 				 				That is exactly what I'm saying! :-)
Good! So let's define A = {a $\in O_K$: a is an associate of some element in $$\overline{A}$$}

Now, try to prove the following:

1. Any rational integer (that is, any element of Z) belongs to A.

2. Any unit in $O_K$ belongs to A.

3. 1 + $\sqrt{-3} \in$ A.

Can you find any other elements of A?

Quote by Petek 					 				 				Good! So let's define A = {a $\in O_K$: a is an associate of some element in $$\overline{A}$$}  Now, try to prove the following:  1. Any rational integer (that is, any element of Z) belongs to A.  2. Any unit in $O_K$ belongs to A.  3. 1 + $\sqrt{-3} \in$ A.  Can you find any other elements of A?
+-1 (Units)
$+-1+*\sqrt{-3}$
+-2
$+-\sqrt{-3}$

I believe those all work, and I know that each shown above and its negative are associates, however $1 +*\sqrt{-3}$ and $1 - \sqrt{-3}$ are not associates.

Recognitions:      Gold Member  You need to work on your notation: That extra i isn't necessary. And, 1 + $\sqrt{-3}$ and 1 - $\sqrt{-3}$ are associates. There are 6 units in $O_K$. Two are +1 and -1. Can you find the other 4?

Quote by Petek 					 				 				You need to work on your notation: That extra i isn't necessary. And, 1 + $\sqrt{-3}$ and 1 - $\sqrt{-3}$ are associates. There are 6 units in $O_K$. Two are +1 and -1. Can you find the other 4?
My bad! I totally goofed that up, but I'm glad you knew what I meant.

Perhaps 2 of them are 0+-i ? I'm not sure what would the others be.

Recognitions:      Gold Member  No problem. I'll be mostly offline for the next few days. Take a look at the Wikipedia articles on Quadratic Integers and Eisenstein Integers. Use the article on Quadratic Integers to determine the elements of $O_K$. Figure out why the Eisenstein integers are relevant to your situation. I'll check back next week to see how you did. Perhaps someone else will add to the discussion.

Quote by Petek 					 				 				No problem. I'll be mostly offline for the next few days. Take a look at the Wikipedia articles on Quadratic Integers and Eisenstein Integers. Use the article on Quadratic Integers to determine the elements of $O_K$. Figure out why the Eisenstein integers are relevant to your situation. I'll check back next week to see how you did. Perhaps someone else will add to the discussion.
If someone could help me with this, I'd really appreciate it. I looked at those Wiki's and I was totally lost.
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