Understanding UFD's with Quadratic Integer & Norm Questions

[Brimley]

Hello PhysicsForums!

I was reading up on UFD's and I came up with a few quick questions.

1. Why don't the integers of Q[\sqrt{-5}] form a UFD? I was trying to tie in the quadratic integers that divide 6 to help me understand this, but I am stuck.

2. Why is Z a UFD?

3. Assuming Q[\sqrt{d}] is a UFD, \alpha is an integer in Q[\sqrt{d}] so that \alpha and \bar{\alpha} have no common factor. The norm of \alpha, N(\alpha), is a perfect square in Z. Is \alpha a perfect square in the quadratic integers in Q[\sqrt{d}]? I believe I saw a proof for this idea somewhere on the web but now I can't find it, however I do remember that it revealed that the variable was indeed a perfect square as described above, however I just forgot why. Can anyone explain this?

Thank you :-)

 

[Brimley]

Is anyone able to help with these? It seems like UFD's are a fairly common subject. I hope somebody will chime in.

Thank you, --Brim

 

[Petek]

1. In \mathbb{Q}(\sqrt{-5}), we have

6 = 2 \times 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})

and all four of the factors are non-associated primes in \mathbb{Q}(\sqrt{-5}).

2. It follows from the Fundamental Theorem of Arithmetic.

3. Something isn't right here. Consider the Gaussian integers \mathbb{Z}. They form a UFD. Let \alpha = 2 + i. Then \bar{\alpha} = 2 - i. The two numbers are non-associated primes and hence have no common factor. But N(\alpha) = (2 + i)(2 - i) = 5 isn't a perfect square in \mathbb{Z}. Also, since \alpha is a prime, it isn't a perfect square in \mathbb{Z}.

 

[Brimley]

1. In \mathbb{Q}(\sqrt{-5}), we have

6 = 2 \times 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})

and all four of the factors are non-associated primes in \mathbb{Q}(\sqrt{-5}).

2. It follows from the Fundamental Theorem of Arithmetic.

3. Something isn't right here. Consider the Gaussian integers \mathbb{Z}. They form a UFD. Let \alpha = 2 + i. Then \bar{\alpha} = 2 - i. The two numbers are non-associated primes and hence have no common factor. But N(\alpha) = (2 + i)(2 - i) = 5 isn't a perfect square in \mathbb{Z}. Also, since \alpha is a prime, it isn't a perfect square in \mathbb{Z}.

2. Okay, I found a few things that were useful from that link:
- A ring in which the Fundamental Theorem of Arithmetic holds is called a unique factorization domain.
- A ring is said to be a unique factorization domain if the Fundamental theorem of arithmetic (for non-zero elements) holds there. For example, any Euclidean domain or principal ideal domain is necessarily a unique factorization domain. Specifically, a field is trivially a unique factorization domain.

And from the UFD page I found:
- In UFDs, every irreducible element is prime. (In any integral domain, every prime element is irreducible, but the converse does not always hold.) Note that this has a partial converse: any Noetherian domain is a UFD if every irreducible element is prime.

While all of this information is useful, I can't tie it together with why Z is a UFD directly. However, I'm pretty sure with the three above statements that someone might know how to explain this.

3. So you believe this fails and that there is no general solution? Is there an example where this is indeed true and \alpha is indeed a perfect square in the quadratic integers of Q[\sqrt{d}] ?

 

[Petek]

Regarding Z being a UFD: I like to be helpful, and I don't want to come across as being impatient, but this can be found in just about any undergraduate texts on abstract algebra and number theory. It's also correctly discussed on Wikipedia. If you're not getting it, you need to sit down with a book or two and start with the basics: Read the definitions of ring, integral domain, unique factorization domain, and so on. It's unrealistic to expect to ask about UFDs without knowing the basic concepts.

On the third topic, yes there are algebraic integers that satisfy the properties you stated. For example, let \alpha = 7 + 24i \ in \ \mathbb{Z}. Try to prove that \alpha is a perfect square, that it has no factors in common with its conjugate and that its norm is a perfect square. If you need help with this, please show what you tried and where you got stuck.

 

[Brimley]

Regarding Z being a UFD: I like to be helpful, and I don't want to come across as being impatient, but this can be found in just about any undergraduate texts on abstract algebra and number theory. It's also correctly discussed on Wikipedia. If you're not getting it, you need to sit down with a book or two and start with the basics: Read the definitions of ring, integral domain, unique factorization domain, and so on. It's unrealistic to expect to ask about UFDs without knowing the basic concepts.

On the third topic, yes there are algebraic integers that satisfy the properties you stated. For example, let \alpha = 7 + 24i \ in \ \mathbb{Z}. Try to prove that \alpha is a perfect square, that it has no factors in common with its conjugate and that its norm is a perfect square. If you need help with this, please show what you tried and where you got stuck.

Well here is my attempt at both of them (Respectively):

1. Z is a UFD because the Fundamental Theorem of Arithemtic holds, and every irreducible element in Z is prime. However, I don't believe that is enough in case I wanted to really explain why, or is it? If not, could someone explain why it is?

2. So here we go:
\alpha = 7 + 24i
\bar{\alpha} = 7 - 24i

Neither of them have common factors

N(\alpha)=(7+24i)(7-24i)
= 49 -576 (i^2)
= 49+576
= 625
\sqrt{625} = 25.

Does that seem correct?

 

[robert Ihnot]

He also asks about showing the existence of an a+bi, who's square is 7+24i.

 

[Brimley]

He also asks about showing the existence of an a+bi, who's square is 7+24i.

Yes, \sqrt{(7+24i)}=4+3i. What does this change?

 

[Petek]

Re: \mathbb{Z} is a UFD. Let's write out the definitions. Perhaps that will make it more clear.

Definition: An integral domain, R, is a unique factorization domain if

(a) any nonzero element in R is either a unit or can be written as the product of a finite number of irreducible elements of R;
(b) the decomposition in part (a) is unique up to the order and association of the irreducible elements.

Definition: A commutative ring with unity is an integral domain if it has no zero-divisors.

Definition: An element a that is not a unit in a ring R will be called irreducible if whenever a = bc with both b, c in R, then one of b or c must be a unit in R.

Note: In the context of UFDs, the terms prime element and irreducible element are equivalent. Therefore, in the preceding definition you can replace irreducible with prime.

Can you now see that \mathbb{Z} is a UFD because it is an integral domain whose elements (the integers) satisfy the requirements of parts (a) and (b) above (with prime replacing irreducibles)?

Yes, \sqrt{(7+24i)}=4+3i . What does this change?

One of your requirements in the original post was that \alpha be a perfect square.

2. So here we go:
\alpha = 7 + 24i
\bar{\alpha} = 7 - 24i

Neither of them have common factors

The last statement requires proof. I made the example slightly more difficult than intended, so I'll sketch the proof. The basic idea is to calculate the prime decompositions of 7 + 24i and 7 - 24i, observe that they have no prime factors in common, and conclude that they have no common factors. I'll work with 7 + 24i. The same arguments apply to 7 - 24i.

We know that 7 + 24i = (4 + 3i)^2. Is 4 + 3i prime? No, as some quick calculations show that 4 + 3i = -i(1 + 2i)^2. Is 1 + 2i prime? Yes. Here's the proof:

Suppose that 1 + 2i = (a + bi)(c + di) with neither factor a unit in \mathbb{Z} and a, b, c, d all integers. Taking the norm of both sides of the equality, we see that

N(1 + 2i) = 5 = (a^2 + b^2)(c^2 + d^2)

It's obvious that the only solutions to this equation imply that one of the factors a + bi or c + di is a unit, contrary to assumption. Therefore, 1 + 2i is prime in \mathbb{Z}. Putting this together, we have

7 + 24i = (4 + 3i)^2 = [-i(1 + 2i)^2]^2 = -(1 + 2i)^4

and similarly

7 - 24i = -(1 - 2i)^4.

After checking that 1 + 2i and 1 - 2i aren't associates, we can conclude that \alpha and \bar \alpha have no common factors.

 

[Brimley]

Re: \mathbb{Z} is a UFD. Let's write out the definitions. Perhaps that will make it more clear.

Definition: An integral domain, R, is a unique factorization domain if

(a) any nonzero element in R is either a unit or can be written as the product of a finite number of irreducible elements of R;
(b) the decomposition in part (a) is unique up to the order and association of the irreducible elements.

Definition: A commutative ring with unity is an integral domain if it has no zero-divisors.

Definition: An element a that is not a unit in a ring R will be called irreducible if whenever a = bc with both b, c in R, then one of b or c must be a unit in R.

Note: In the context of UFDs, the terms prime element and irreducible element are equivalent. Therefore, in the preceding definition you can replace irreducible with prime.

Can you now see that \mathbb{Z} is a UFD because it is an integral domain whose elements (the integers) satisfy the requirements of parts (a) and (b) above (with prime replacing irreducibles)?



One of your requirements in the original post was that \alpha be a perfect square.



The last statement requires proof. I made the example slightly more difficult than intended, so I'll sketch the proof. The basic idea is to calculate the prime decompositions of 7 + 24i and 7 - 24i, observe that they have no prime factors in common, and conclude that they have no common factors. I'll work with 7 + 24i. The same arguments apply to 7 - 24i.

We know that 7 + 24i = (4 + 3i)^2. Is 4 + 3i prime? No, as some quick calculations show that 4 + 3i = -i(1 + 2i)^2. Is 1 + 2i prime? Yes. Here's the proof:

Suppose that 1 + 2i = (a + bi)(c + di) with neither factor a unit in \mathbb{Z} and a, b, c, d all integers. Taking the norm of both sides of the equality, we see that

N(1 + 2i) = 5 = (a^2 + b^2)(c^2 + d^2)

It's obvious that the only solutions to this equation imply that one of the factors a + bi or c + di is a unit, contrary to assumption. Therefore, 1 + 2i is prime in \mathbb{Z}. Putting this together, we have

7 + 24i = (4 + 3i)^2 = [-i(1 + 2i)^2]^2 = -(1 + 2i)^4

and similarly

7 - 24i = -(1 - 2i)^4.

After checking that 1 + 2i and 1 - 2i aren't associates, we can conclude that \alpha and \bar \alpha have no common factors.

Thank you Petek, the breakdown was very thorough and simple to follow.