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• July 21st 2010, 05:04 AM
Samson
Hello all,

I was examining a case described in my book about primes in Quadratic Fields. The case describe examines positive prime numbers named "d" that are in Z and meet the condition that the quadratic integers in Q[Sqrt(d)] are in the form of a+b*Sqrt(d) , where of course a and b must be rational.

I need a place to start with this before I can continue. Can someone give me the first 3 or four of such primes? (Call these solutions - A)

What if they were required to be in the form of (a+b*Sqrt(d))/2 and 'a' and 'b' had to be both even or both odd? Do you just divide what we called Solutions-A by 2 as long as the numbers are still rational? I don't think that would work because I don't know of many primes that function in this way. What would the first 3 or 4 such primes of this form be? (Call these solutions - B).

Lastly, how would this change if we replaced Sqrt(d) with Sqrt(-d) for all mentioning of d above? As in, how would solutions A and solutions B differ? What would they be?

Thank you all, I really appreciate the help!
• July 21st 2010, 05:23 AM
undefined
Quote:

Originally Posted by Samson
I was examining a case described in my book about primes in Quadratic Fields. The case describe examines positive prime numbers named "d" that are in Z and meet the condition that the quadratic integers in Q[Sqrt(d)] are in the form of a+b*Sqrt(d) , where of course a and b must be rational.

I find it hard to believe your book uses the same letter "d" to describe two things at once. This can lead to confusion.

Check out Prime number - Prime elements in rings - Wikipedia.

The wording of 'positive prime numbers named "d" that are in Z and meet the condition that the quadratic integers in Q[Sqrt(d)] are in the form of a+b*Sqrt(d) , where of course a and b must be rational' doesn't make much sense. The quadratic integers in Q[sqrt(d)] will be a subset of Q[sqrt(d)], and wouldn't be characterized by saying a and b rational, which characterizes Q[sqrt(d)]. Also, your sentence has the setup "A meets the condition that B" where B is essentially unrelated to A. Please check the wording.

Quote:

Originally Posted by Samson
I need a place to start with this before I can continue. Can someone give me the first 3 or four of such primes? (Call these solutions - A)

What if they were required to be in the form of (a+b*Sqrt(d))/2 and 'a' and 'b' had to be both even or both odd? Do you just divide what we called Solutions-A by 2 as long as the numbers are still rational? I don't think that would work because I don't know of many primes that function in this way. What would the first 3 or 4 such primes of this form be? (Call these solutions - B).

Lastly, how would this change if we replaced Sqrt(d) with Sqrt(-d) for all mentioning of d above? As in, how would solutions A and solutions B differ? What would they be?

I'm a little confused. Are we looking for positive integers, or numbers of the form a + b*sqrt(d) with a and b rational? (Or something else?)

Of course if 'a' and 'b' are both even then we can get rid of that 2 in the denominator and proceed from there. What book are you using, by the way?
• July 21st 2010, 05:34 AM
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• July 21st 2010, 05:35 AM
Samson
For Solutions-A, we are looking for the first 3 or 4 positive prime numbers 'd' that meet the condition above (that they are in Z and that the quadratic integers in Q[Sqrt[d]] are in the form of a+b*Sqrt(d) where 'a' and b' are both rational integers.

For solutions-B, we do the exact same thing as Solutions-A except that the 'a' and 'b' must not only be rational, but also must be either both even or both odd and the quadratic integers in Q[Sqrt(d)] are in the form of (a+b*Sqrt(d))/2.

I don't know if that helps much, but that really is all the clarity I can provide.
• July 21st 2010, 05:41 AM
undefined
Quote:

Originally Posted by Samson
For Solutions-A, we are looking for the first 3 or 4 positive prime numbers 'd' that meet the condition above (that they are in Z and that the quadratic integers in Q[Sqrt[d]] are in the form of a+b*Sqrt(d) where 'a' and b' are both rational integers.

For solutions-B, we do the exact same thing as Solutions-A except that the 'a' and 'b' must not only be rational, but also must be either both even or both odd and the quadratic integers in Q[Sqrt(d)] are in the form of (a+b*Sqrt(d))/2.

I don't know if that helps much, but that really is all the clarity I can provide.

Okay, given a ring, we can use the standard definition for what a prime in that ring is. But we need to know what ring we're dealing with. You mention quadratic integers but don't say how it relates to what we're doing. You don't specify d, where d is the thing inside the square root (see why we should use different letters???). Does that mean you think there are primes that hold for any choice of d? Or do you want us to choose a d? Does d need to be positive? There are many many questions and that are not clear in what you ask, at least not to me. Maybe give some examples from your book. And please tell me what book it is.
• July 21st 2010, 06:06 AM
Samson
Quote:

Originally Posted by undefined
Okay, given a ring, we can use the standard definition for what a prime in that ring is. But we need to know what ring we're dealing with. You mention quadratic integers but don't say how it relates to what we're doing. You don't specify d, where d is the thing inside the square root (see why we should use different letters???). Does that mean you think there are primes that hold for any choice of d? Or do you want us to choose a d? Does d need to be positive? There are many many questions and that are not clear in what you ask, at least not to me. Maybe give some examples from your book. And please tell me what book it is.

I'm using Funktionalanalysis: Theorie und Anwendung. Lehrbuch (its a German textbook). I am however very comfortable in my translating skills.

I would assume that we can choose a d and find a finite example of such Quadratic Integers, and I would assume that d should be positive considering that the last part asks to consider if d would be negative. By quadratic integers, the book just states "Quadratic Integers of the form a+b*Sqrt(d)."
• July 21st 2010, 06:40 AM
undefined
Quote:

Originally Posted by Samson
I'm using Funktionalanalysis: Theorie und Anwendung. Lehrbuch (its a German textbook). I am however very comfortable in my translating skills.

I would assume that we can choose a d and find a finite example of such Quadratic Integers, and I would assume that d should be positive considering that the last part asks to consider if d would be negative. By quadratic integers, the book just states "Quadratic Integers of the form a+b*Sqrt(d)."

Okay, I can't speak or read German, but thanks for the info.

The last part does not ask us to consider d negative, not the way you wrote it. Rather it asks us to consider what happens when d is replaced by its additive inverse, -d.

The thing is, you ask for primes in Z over a quadratic field Q[sqrt(d)], then you make it sound like we're trying to find primes in Z over a ring Z[omega] of quadratic integers, which is a different ring, for explanation of notation see

Quadratic integer - Wikipedia, the free encyclopedia

Based on the Wikipedia article, your solutions A and B seem to correspond with whether d is congruent to 2,3 or whether congruent to 1 (mod 4), respectively -- but this is guess work on my part because it was not clear in your post.

Primes in one ring (any field is also a ring) will not necessarily be primes in another ring, although perhaps one can imply the other depending on how the rings are related, I'm not knowledgeable on that.

There is a distinction between prime element and irreducible element when working with general rings, and I'm not sure if they are the same here. We tend to think of primes in the integers the way irreducible elements are defined for an arbitrary ring. See

Prime number - Prime elements in rings - Wikipedia

IF prime and irreducible mean the same thing in these particular rings (I would have to read up on it), then for example 2 and 5 are not primes in the Gaussian integers, because they can be factored (but 3 cannot), see here

Problem 153 - Project Euler

I've mentioned some things from ring theory here that I don't know all the main theorems for, and possibly I would be able to answer better if I knew those theorems, but the fact is this could all be cleared up if you simply specified more clearly what you're after. Knowing more ring theory would just allow me to guess better what you meant. But if you were specific, then I or someone else could look up or recall the relevant theorems and hopefully give you a good answer.

Edit: There seems to be an error in Wikipedia at the time of this writing. In this article, it is stated that:

"Prime numbers give rise to two more general concepts that apply to elements of any ring R, an algebraic structure where addition, subtraction and multiplication are defined: prime elements and irreducible elements."

However, in this referenced article, the ring is required to be an integral domain, and not just any ring. I don't know which is the correct usage off hand, but I mention this for completeness and for reference.
• July 21st 2010, 07:16 AM
Samson
Quote:

Originally Posted by undefined

Based on the Wikipedia article, your solutions A and B seem to correspond with whether d is congruent to 2,3 or whether congruent to 1 (mod 4), respectively -- but this is guess work on my part because it was not clear in your post.

I would see if you could solve it predicated that d is congruent to 2. I just would like to understand an example of this. If 3 is easier, try 3. This is just for me so I can understand. I really appreciate the help thus far!
• July 21st 2010, 07:57 AM
undefined
Quote:

Originally Posted by Samson
I would see if you could solve it predicated that d is congruent to 2. I just would like to understand an example of this. If 3 is easier, try 3. This is just for me so I can understand. I really appreciate the help thus far!

I'll do a little letter re-assigning and assume that you're after prime elements $\displaystyle p_i$ in quadratic integer rings $\mathbb{Z}[\omega]$ corresponding to quadratic fields $\mathbb{Q}(\sqrt{D})$, where we restrict $\displaystyle p_i \in \mathbb{Z}, p_i > 0$ and thus there is a natural ordering of the primes.

I think it will be easiest to characterise primes when the rings happen to be unique factorisation domains (UFD), namely D = -1, -2, -3, -7, or -11. So when D = -1, it is congruent to 3 (mod 4) and we have the Gaussian integers. Because it is a UFD, we have that primes and irreducibles are equivalent, so no need to worry about that. So the first prime integer in the Guassian integers is 3. The next ones are 7, 11, 19, and 23. You can see some info here

Gaussian Primes

and here

id:A002145 - OEIS Search Results

A nice result is that these are precisely the primes over the integers that are not the sum of two squares. More info here

Math Forum - Ask Dr. Math

For choosing D other than -1, -2, -3, -7, or -11, we have to be careful about the distinction between prime and irreducible. Every prime is irreducible, but the converse is not necessarily true. Pay special attention to the example given here

Irreducible element - Wikipedia, the free encyclopedia

The example I mean is: "In the quadratic integer ring $\mathbb{Z}[\sqrt{-5}]$, the number 3 is irreducible but is not a prime since 9 can be written as $(2+\sqrt{-5})(2-\sqrt{-5})$ and 3(3)."

Another well-studied ring is the Eisenstein integers obtained by letting D = -3. More info here

Eisenstein integer - Wikipedia, the free encyclopedia

Eisenstein prime - Wikipedia, the free encyclopedia
• July 21st 2010, 08:18 AM
Samson
Quote:

Originally Posted by undefined
I'll do a little letter re-assigning and assume that you're after prime elements $\displaystyle p_i$ in quadratic integer rings $\mathbb{Z}[\omega]$ corresponding to quadratic fields $\mathbb{Q}[\sqrt{D}]$, where we restrict $\displaystyle p_i \in \mathbb{Z}, p_i > 0$ and thus there is a natural ordering of the primes.

I think it will be easiest to characterise primes when the rings happen to be unique factorisation domains (UFD), namely D = -1, -2, -3, -7, or -11. So when D = -1, it is congruent to 3 (mod 4) and we have the Gaussian integers. Because it is a UFD, we have that primes and irreducibles are equivalent, so no need to worry about that. So the first prime integer in the Guassian integers is 3. The next ones are 7, 11, 19, and 23. You can see some info here

Gaussian Primes

and here

id:A002145 - OEIS Search Results

A nice result is that these are precisely the primes over the integers that are not the sum of two squares. More info here

Math Forum - Ask Dr. Math

For choosing D other than -1, -2, -3, -7, or -11, we have to be careful about the distinction between prime and irreducible. Every prime is irreducible, but the converse is not necessarily true. Pay special attention to the example given here

Irreducible element - Wikipedia, the free encyclopedia

The example I mean is: "In the quadratic integer ring $\mathbb{Z}[\sqrt{-5}]$, the number 3 is irreducible but is not a prime since 9 can be written as $(2+\sqrt{-5})(2-\sqrt{-5})$ and 3(3)."

Another well-studied ring is the Eisenstein integers obtained by letting D = -3. More info here

Eisenstein integer - Wikipedia, the free encyclopedia

Eisenstein prime - Wikipedia, the free encyclopedia

Wow, that is a lot but it makes a lot of sense. I see how you addressed a negative D, but can you address a positive one? Also, what about the expression that has a and b halved also with them being either both even or both odd?
• July 21st 2010, 08:48 AM
undefined
Quote:

Originally Posted by Samson
Wow, that is a lot but it makes a lot of sense. I see how you addressed a negative D, but can you address a positive one? Also, what about the expression that has a and b halved also with them being either both even or both odd?

The part about them being both odd or both even I don't really follow. Are we defining a new ring altogether? Have we proven this is a ring? If they are both even then the 2 in the denominator can be gotten rid of, and for D congruent to 2,3 (mod 4) we will have just the expected quadratic integer ring. But if they are both odd, or if D is congruent to 1 (mod 4), I'm not sure what to make of it.

Well I already gave you some examples, so see what you can apply towards D > 0. It's trickier because we don't have any UFDs, so you have to be careful between primes and irreducibles. Whenever you choose a different D the resulting ring will have different properties. Off hand I don't know of any clear relation between choosing D and then choosing -D.
• July 21st 2010, 09:20 AM
Samson
Quote:

Originally Posted by undefined
The part about them being both odd or both even I don't really follow. Are we defining a new ring altogether? Have we proven this is a ring? If they are both even then the 2 in the denominator can be gotten rid of, and for D congruent to 2,3 (mod 4) we will have just the expected quadratic integer ring. But if they are both odd, or if D is congruent to 1 (mod 4), I'm not sure what to make of it.

Okay, so assuming that D = 2, what are the first 3 or 4 primes that meet the condition that the Quadratic Integers in Q[Sqrt(d)] are in the form a+b*Sqrt(d) where (a,b) are both even or both odd?
• July 21st 2010, 09:38 AM
undefined
Quote:

Originally Posted by Samson
Okay, so assuming that D = 2, what are the first 3 or 4 primes that meet the condition that the Quadratic Integers in Q[Sqrt(d)] are in the form a+b*Sqrt(d) where (a,b) are both even or both odd?

You seem not to have understood what I wrote. Quadratic integer rings are already defined; you can't just define them however you feel like.

Quadratic integer - Wikipedia, the free encyclopedia

If you change the form they must be in, then it seems you want to define a new ring, which would then have to be proven to be a ring, and would not be called a quadratic integer ring. I don't know how else to interpret what you're saying.
• July 21st 2010, 09:55 AM
Samson
Quote:

Originally Posted by undefined
You seem not to have understood what I wrote. Quadratic integer rings are already defined; you can't just define them however you feel like.

Quadratic integer - Wikipedia, the free encyclopedia

If you change the form they must be in, then it seems you want to define a new ring, which would then have to be proven to be a ring, and would not be called a quadratic integer ring. I don't know how else to interpret what you're saying.

So, is there any quadratic ring that is positive that has quadratic integers a+b*Sqrt(d) where (a,b) are either both positive or both negative and d>0 ?
• July 21st 2010, 10:36 AM
undefined
Quote:

Originally Posted by Samson
So, is there any quadratic ring that is positive that has quadratic integers a+b*Sqrt(d) where (a,b) are either both positive or both negative and d>0 ?

Hmm, I may have jumped the gun, in terms of, it looks like we can "finesse" your requirement to match the definition.

Wikipedia (and MathWorld) gives:

$\alpha = a + b\omega \in \mathbb{Z}[\omega]$

where $a, b \in \mathbb{Z}$

$\omega=\begin{cases}\sqrt{D}&\text{if}\ D\equiv2,3\pmod{4}\\\frac{1+\sqrt{D}}{2}&\text{if} \ D\equiv1\pmod{4}\end{cases}$

So as noted, if we look at $\frac{a+b\sqrt{D}}{2}$ with a and b both even, then this reduces to the case where D is congruent to 2,3 (mod 4). If we restrict a and b such that they are congruent to each other (mod 2) then I think we have an equivalence for when D is congruent to 1 (mod 4).

Using Wikipedia definition when D congruent to 1 (mod 4),

$a + b\omega = a + \frac{1+\sqrt{D}}{2}\cdot b = \cdots = \frac{(2a+b)+b\sqrt{D}}{2}$. Now letting a' = 2a+b and b'=b, we see that if we fix b even then (2a + b) is even and if we fix b odd then (2a + b) is odd, and we can vary a in order to get whatever a' we want, so we have the equivalent formulation

$a, b \in \mathbb{Z},\begin{Bmatrix}a\equiv b\equiv 0\pmod{2}&\text{if}\ D\equiv2,3\pmod{4} \\a\equiv b \pmod{2}&\text{if}\ D\equiv1\pmod{4} \end{Bmatrix}, \alpha = \frac{a + b\sqrt{D}}{2} \in \mathcal{O}_{\mathbb{Q}(\sqrt{D})}$

So the parity of a and b (whether they are even or odd) is restricted based on D. So to answer your question, choose D congruent to 1 (mod 4).
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