Greatest Common Divisor in a strange extension ring.

[tomtom690]

Homework Statement

I need to show that two elements in $$\textbf{Z}$$[$$\sqrt{-5}$$] have gcd = 1.
The elements are 3 and 2+$$\sqrt{-5}$$

Homework Equations

The Attempt at a Solution

My way of thinking was if I can show that both elements are irreducible, then they are both prime and hence have gcd of 1. I can show they are both irreducible, using the norm function - ie showing that if eg 3 = ab then either N(a) or N(b) is 1. This means that 3 is irreducible in this ring. I think.
Can somebody tell me if this is correct please? Like I said, I'm almost there, just need to polish it off!
Thanks in advance.

 

[mighty2000]

Yes, your approach is correct. To show that an element is irreducible in \textbf{Z}[\sqrt{-5}], we can use the norm function as you mentioned. For the element 3, we have N(3) = 9, and the only elements in \textbf{Z}[\sqrt{-5}] with norm 9 are 3 and -3. But neither of these can be written as a product of two non-units in \textbf{Z}[\sqrt{-5}], so 3 is irreducible. Similarly, for the element 2+\sqrt{-5}, we have N(2+\sqrt{-5}) = 9, and the only elements with norm 9 are 3 and -3. Since 2+\sqrt{-5} cannot be written as a product of two non-units in \textbf{Z}[\sqrt{-5}], it is also irreducible. Therefore, both 3 and 2+\sqrt{-5} are prime elements in \textbf{Z}[\sqrt{-5}], and their gcd is 1.