# Thread: UFD Question

1. ## 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!

2. 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 $4 = 2\times2 = (1 + \sqrt{-3})(1 - \sqrt{-3}).$

3. Thank you Opalg! How would you describe how those factorizations are related though? Is there some property or law that describes that relationship?

4. 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.

5. Thank you!