# Quick Q-Field Problems

• Jul 27th 2010, 03:37 AM
Samson
Quick Q-Field Problems
Hello All,

I had a few quick Quadratic Field related problems I wanted to throw out there to see if someone can give me a hand with.

1 - Does anyone know of a Quadratic Integer in Q[Sqrt(-1)] that is prime but whose norm is not prime?

2 - What is a prime factorization of 6 in Q[Sqrt(-1)] ? I have been having trouble understanding these.

Thank you! -Samson
• Jul 27th 2010, 04:04 AM
tonio
Quote:

Originally Posted by Samson
Hello All,

I had a few quick Quadratic Field related problems I wanted to throw out there to see if someone can give me a hand with.

1 - Does anyone know of a Quadratic Integer in Q[Sqrt(-1)] that is prime but whose norm is not prime?

If by "quadratic integer" you meant "Gaussian integer" or just "integer", then the only primes in $\mathbb{Z}[i]$ are elements of the form $a+bi$ , with $a^2+b^2=p\,,\,\,p$ an ordinary prime in $\mathbb{Z}$ , or elements of the form $a\,\,or\,\,ai$ , with $|a|=3\!\!\pmod 4$ a rational prime, and the prime $1+i$.
It's easy to check all these primes have a prime norm, so...

2 - What is a prime factorization of 6 in Q[Sqrt(-1)] ? I have been having trouble understanding these.

$6=2\cdot 3 = -i(1+i)^2\cdot 3=-3i(1+i)^2$ , with $\pm i$ a unit, of course.

Tonio

Thank you! -Samson

.
• Jul 27th 2010, 04:09 AM
Samson
Hey Tonio, could you edit your post so that I can quote the text that you put inside of my quote?

By the way, thank you on part 2, but what I was going to quote on part 1 was:

By Quadratic Integer, I meant what's described herein : Quadratic integer - Wikipedia, the free encyclopedia

Please note that the article states:
Quote:

"Quadratic integers are a generalization of the rational integers to quadratic fields. Important examples include the Gaussian integers and the Eisenstein integers."
Can you relate this and the article to find an example to satisfy part one? I know you have the generics, but is there a discrete example you might be able to come up with?
• Aug 3rd 2010, 04:31 AM
elemental
Quote:

Originally Posted by tonio
elements of the form , with a rational prime, and the prime .
It's easy to check all these primes have a prime norm, so...

How exactly do all of them have prime norms? You said there can be prime numbers of the form a (where a is congruent to 3 mod 4). Norm(a) = a^2, and clearly this norm is not a prime. Am I missing something fundamental?
• Aug 3rd 2010, 05:00 AM
tonio
Quote:

Originally Posted by elemental
How exactly do all of them have prime norms? You said there can be prime numbers of the form a (where a is congruent to 3 mod 4). Norm(a) = a^2, and clearly this norm is not a prime. Am I missing something fundamental?

Of course you're missing something fundamental: my typos/mistakes.(Giggle)

I wrote that about the prime norms before adding the primes of the form $a\,,\,ai\,,\,\,with\,\,a=3\!\!\pmod 4$ a prime, and then edited , forgot.

You're right. Thanx

Tonio
• Aug 3rd 2010, 06:01 AM
Samson
Hey tonio, could you reply to my post above where I attempted to quote you?
• Aug 3rd 2010, 12:12 PM
tonio
Quote:

Originally Posted by Samson
Hey tonio, could you reply to my post above where I attempted to quote you?

I honestly can't understand what you want... can't you read/understand my answer to your post or what?

Tonio
• Aug 3rd 2010, 01:06 PM
Samson
Quote:

Originally Posted by tonio
I honestly can't understand what you want... can't you read/understand my answer to your post or what?

Tonio

Well you never directly answered my last post. I don't know if you missed it. Here is what I said:

Hey Tonio, could you edit your post so that I can quote the text that you put inside of my quote?

By the way, thank you on part 2, but what I was going to quote on part 1 was:

By Quadratic Integer, I meant what's described herein : Quadratic integer - Wikipedia, the free encyclopedia (See Link Above)

Please note that the article states:
"Quadratic integers are a generalization of the rational integers to quadratic fields. Important examples include the Gaussian integers and the Eisenstein integers."
Can you relate this and the article to find an example to satisfy part one? I know you have the generics, but is there a discrete example you might be able to come up with?
• Aug 4th 2010, 05:55 AM
undefined
Quote:

Originally Posted by Samson
Well you never directly answered my last post. I don't know if you missed it. Here is what I said:

Hey Tonio, could you edit your post so that I can quote the text that you put inside of my quote?

By the way, thank you on part 2, but what I was going to quote on part 1 was:

By Quadratic Integer, I meant what's described herein : Quadratic integer - Wikipedia, the free encyclopedia (See Link Above)

Please note that the article states:
"Quadratic integers are a generalization of the rational integers to quadratic fields. Important examples include the Gaussian integers and the Eisenstein integers."
Can you relate this and the article to find an example to satisfy part one? I know you have the generics, but is there a discrete example you might be able to come up with?

EDIT: When I wrote this post originally, I was not careful about the distinction between $\mathbb{Q}(\sqrt{-1})$ and $\mathbb{Z}[i]$. Here's what I wrote, slightly edited:

~~~~~~~~~~~~~~~~~~~~~~

tonio's first post (post #2 of this thread) was not "generic" but was a direct and concrete answer to your questions. To spell it out more explicitly,

(1) The answer is, there does not exist any such Gaussian prime.

Gaussian Prime -- from Wolfram MathWorld
Gaussian integer - Wikipedia, the free encyclopedia (subheading: As a unique factorization domain)

Note: I don't know why tonio mentions $1+i$ separately, as it fits the description $a+bi$ where $N(a+bi)=a^2+b^2=p$ with $\displaystyle p$ a regular prime in $\mathbb{Z}$.

(2) tonio gave you the prime factorisation of 6 in $\mathbb{Z}[i]$. Note that 2*3 is not a prime factorisation here because 2 is not a Gaussian prime.

Unique factorization domain - Wikipedia, the free encyclopedia
Unique Factorization Domain -- from Wolfram MathWorld

~~~~~~~~~~~~~~~~~~~~~~

So, the above applies to $\mathbb{Z}[i]$. Question to the OP (Samson): Are you sure you want to know about primes in $\mathbb{Q}(\sqrt{-1})$ as opposed to $\mathbb{Z}[i]$? If so, I'll have to think/research more, or someone else will have to add some info..
• Aug 4th 2010, 07:41 AM
tonio
Quote:

Originally Posted by undefined
EDIT: When I wrote this post originally, I was not careful about the distinction between $\mathbb{Q}(\sqrt{-1})$ and $\mathbb{Z}[i]$. Here's what I wrote, slightly edited:

~~~~~~~~~~~~~~~~~~~~~~

tonio's first post (post #2 of this thread) was not "generic" but was a direct and concrete answer to your questions. To spell it out more explicitly,

(1) The answer is, there does not exist any such Gaussian prime.

Gaussian Prime -- from Wolfram MathWorld
Gaussian integer - Wikipedia, the free encyclopedia (subheading: As a unique factorization domain)

Note: I don't know why tonio mentions $1+i$ separately, as it fits the description $a+bi$ where $N(a+bi)=a^2+b^2=p$ with $\displaystyle p$ a regular prime in $\mathbb{Z}$.

Correct. I was just thinking of primes which are 1 modulo 4 when I wrote $a^2+b^2=p$ , and tried to distinguish the first prime 2 which isn't 1 modulo 4 and still fulfills the condition: $1^2+1^2=2$ , but I didn't write down the distinction.
All the above was thought to convey the fact (theorem) that a prime is a sum of squares iff it is 2 or it is 1 mod 4 , and from here to deduce, together with a little algebra, that an integer is a sum of squares iff any prime equal to 3 mod 4 that appears in its prime decompostion appears there to an even power.
Thanx

Tonio

(2) tonio gave you the prime factorisation of 6 in $\mathbb{Z}[i]$. Note that 2*3 is not a prime factorisation here because 2 is not a Gaussian prime.

Unique factorization domain - Wikipedia, the free encyclopedia
Unique Factorization Domain -- from Wolfram MathWorld

~~~~~~~~~~~~~~~~~~~~~~

So, the above applies to $\mathbb{Z}[i]$. Question to the OP (Samson): Are you sure you want to know about primes in $\mathbb{Q}(\sqrt{-1})$ as opposed to $\mathbb{Z}[i]$? If so, I'll have to think/research more, or someone else will have to add some info..

.
• Aug 5th 2010, 08:21 AM
Samson
Quote:

Originally Posted by undefined
Question to the OP (Samson): Are you sure you want to know about primes in $\mathbb{Q}(\sqrt{-1})$ as opposed to $\mathbb{Z}[i]$? If so, I'll have to think/research more, or someone else will have to add some info..

Yes, I need to find a Quadratic Integer in $\mathbb{Q}(\sqrt{-1})$ that is prime but whose norm is not prime.
• Aug 5th 2010, 09:16 AM
undefined
Quote:

Originally Posted by Samson
Yes, I need to find a Quadratic Integer in $\mathbb{Q}(\sqrt{-1})$ that is prime but whose norm is not prime.

I see you are restating what you wrote previously, but do you see why this statement is not clear out of context?

A number can be prime relative to one ring and not prime relative to another ring.

For example, 2 is a prime in the ring $\mathbb{Z}$ but is not a prime in the ring $\mathbb{Z}[i]$. But, the number 2 itself belongs to both rings.

So suppose I say: "Find a rational integer in $\mathbb{Z}[i]$ that is prime and is less than 5." Now, are we talking about primality with respect to $\mathbb{Z}$ or with respect to $\mathbb{Z}[i]$? It is not clear.

But I understand from context that you want primes with respect to $\mathbb{Q}(\sqrt{-1})$, so there's no need to re-establish that; I just posted this comment to help you understand why it was not clear in the first place.
• Aug 9th 2010, 03:26 AM
Samson
Quote:

Originally Posted by undefined
I see you are restating what you wrote previously, but do you see why this statement is not clear out of context?

A number can be prime relative to one ring and not prime relative to another ring.

For example, 2 is a prime in the ring $\mathbb{Z}$ but is not a prime in the ring $\mathbb{Z}[i]$. But, the number 2 itself belongs to both rings.

So suppose I say: "Find a rational integer in $\mathbb{Z}[i]$ that is prime and is less than 5." Now, are we talking about primality with respect to $\mathbb{Z}$ or with respect to $\mathbb{Z}[i]$? It is not clear.

But I understand from context that you want primes with respect to $\mathbb{Q}(\sqrt{-1})$, so there's no need to re-establish that; I just posted this comment to help you understand why it was not clear in the first place.

Unfortunately, I cannot provide much clarification to the problem. That is the problem out of the book, word for word, and there isn't any further explanation. I assume that this can be solved using the statement as it is. Are we all over thinking this? I just wonder if thats the case because this is one of the first example problems that they give and they are usually the easiest, and they get harder as they go on...
• Sep 4th 2010, 01:00 PM
undefined
Quote:

Originally Posted by Samson
Unfortunately, I cannot provide much clarification to the problem. That is the problem out of the book, word for word, and there isn't any further explanation. I assume that this can be solved using the statement as it is. Are we all over thinking this? I just wonder if thats the case because this is one of the first example problems that they give and they are usually the easiest, and they get harder as they go on...

I think your book must be referring to what tonio addressed already (which I explained a bit more in my first post), and the other information I provided above about $\mathbb{Q}(\sqrt{-1})$ vs $\mathbb{Z}[i]$ was a bit misleading in that I missed a crucial bit of very basic number theory that I should not have if I were more versed.

In any field, every non-zero element is a unit. Therefore every non-zero element divides every other element. Thus, even though prime and irreducible elements are defined in fields with respect to those fields (every field is trivially a UFD), they are not very interesting, because they do not exist!

Someone please correct me if this is wrong, but I believe this is the conclusion for this particular issue.