UFD Question

**Samson** · September 9th 2010, 02:54 PM

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!

**Opalg** · September 10th 2010, 12:40 AM

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

**Samson** · September 13th 2010, 09:33 AM

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

**Opalg** · September 13th 2010, 10:29 AM

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.

**Samson** · September 14th 2010, 09:16 AM

Thank you!

Thread: UFD Question

LinkBack

Thread Tools

Display

UFD Question