# UFD Question

• September 9th 2010, 03:54 PM
Samson
UFD Question
Hello all,

I am finishing up this chapter in my book with regards to UFD's, and it presented an interesting question that ties in with unique prime factorization. Here is what it says:

It can be shown directly that $Z[\sqrt{-3}]$ is not a UFD by finding an integer which factors into primes in more than one way. These two factorizations are related to the unique prime factorization in the quadratic integers in $Q[\sqrt{-3}]$

Can anyone show me this directly and explain how those factorizations are related?

Thank you!
• September 10th 2010, 01:40 AM
Opalg
Quote:

Originally Posted by Samson
Hello all,

I am finishing up this chapter in my book with regards to UFD's, and it presented an interesting question that ties in with unique prime factorization. Here is what it says:

It can be shown directly that $Z[\sqrt{-3}]$ is not a UFD by finding an integer which factors into primes in more than one way. These two factorizations are related to the unique prime factorization in the quadratic integers in $Q[\sqrt{-3}]$

Can anyone show me this directly and explain how those factorizations are related?

Look at the factorisations $4 = 2\times2 = (1 + \sqrt{-3})(1 - \sqrt{-3}).$
• September 13th 2010, 10:33 AM
Samson
Thank you Opalg! How would you describe how those factorizations are related though? Is there some property or law that describes that relationship?
• September 13th 2010, 11:29 AM
Opalg
Quote:

Originally Posted by Samson
How would you describe how those factorizations are related though? Is there some property or law that describes that relationship?

The ring $\mathbb{Z}[\sqrt{-3}]$ has a norm associated with it, namely a multiplicative function N from the ring to the positive integers. In this case, $N(a+b\sqrt{-3}) = a^2+3b^2$. If you want to factorise a number in $\mathbb{Z}[\sqrt{-3}]$ then you need to factorise its norm. For example, $N(4) = 16$. So if $a+b\sqrt{-3}$ is a factor of 4 then $N(a+b\sqrt{-3})$ must be a factor of 16. The only factor of 16 that can be expressed in the form $a^2+3b^2$ is 4, which is equal to both $2^2+3\cdot0^2$ and $1^2+3\cdot1^2$. So the only factors of 4 in $\mathbb{Z}[\sqrt{-3}]$ are $\pm2$ and $\pm1\pm\sqrt{-3}]$. These numbers all have norm 4, and there are no proper factors of 4 that can be expressed in the form $N(a+b\sqrt{-3})$. So 2 and $1\pm\sqrt{-3}$ are all irreducible in the ring $\mathbb{Z}[\sqrt{-3}]$. Thus 4 has two essentially different factorisations in $\mathbb{Z}[\sqrt{-3}]$, which shows that this ring is not a UFD.
• September 14th 2010, 10:16 AM
Samson
Thank you!