Existence of Greatest Common Divisors - Dummit and Foote Ch8
(Aside: I don't know if this is true or not - you can try to prove it or disprove it if you like, but, in my mental picture of this stuff, everything having a gcd, and being a UFD, are the same condition for an integral domain.)
Continuing with the example from yesterday: What about and in ?
You should double check this - I might have err'd somewhere.
ASSUME they have a gcd in R = .
1st) 3 is irreducible in R.
Suppose 3 = gh for some g, h in R, neither of which is a unit. Then N(3) = 9 and N(g), N(h) both not 1.
Then 9 = N(3) = N(gh) = N(g)N(h), so must have N(g) = N(h) = 3. But there are no elements having norm 3.
2nd) 2+sqrt(-5) is not a unit. 2-sqrt(-5) is not a unit.
N(2+sqrt(-5)) = N(2-sqrt(-5)) = 9.
3rd) 3 = r(2+sqrt(-5)) has no solution for r in R. Likewise for 2-sqrt(-5).
3 irreducible, so either r or 2+sqrt(-5) is a unit. But 2+sqrt(-5) isn't a unit.
Therefore r is a unit.
Therefore r = 1 or r = -1, since those are the only two elements of norm 1.
But clearly neither choice of r is a solution. Thus there are no solutions.
Same for 2-sqrt(-5)
4th) Every gcd is of the form gcd = 9u or gcd = 3u for some unit u in R.
3 divides both, so 3 divides any gcd. Examine d = gcd = 3q, where q in R. Then d divides 9 so ds = 9 for some s in R.
Thus 3qs = 9, so qs = 3, so (q = 3u, s=u*) or (q=u, s=3u*) for some unit u in R.
Therefore d = 9u or d = 3u for some unit u in R.
5th) 3 is a gcd
Let d be a gcd. Then since d divides 3(2+sqrt(-5)), there exists some t in R such that td = 3(2+sqrt(-5)).
If d = 9u, then 9tu = 3(2+sqrt(-5)) for some t in R, so 3tu = (2+sqrt(-5)),
so 3(2-sqrt(-5))tu = (2+sqrt(-5))(2-sqrt(-5)) = 9, and so (2-sqrt(-5))(tu) = 3.
But have established that no such (tu) in R can ever make that hold.
Therefore d = 3u for some unit u in R.
Therefore d' = du* = 3uu* = 3 is another gcd.
Therefore 3 is a gcd.
6th) 3 cannot be a gcd:
Since (2+sqrt(-5)) divides both 9=(2+sqrt(-5))(2-sqrt(-5)) and 3(2+sqrt(-5)),
have that (2+sqrt(-5)) divides every gcd.
If 3 is a gcd, the (2+sqrt(-5)) divides 3, so there exists r in R such that (2+sqrt(-5))r = 3.
But have already established that that equation has no solutions for any r in R.
Therefore 3 cannot be a gcd.
Thus the assumption that a gcd exists has led to the conclusion that 3 both is, and is not, a gcd, with is a contradiction.
Therefore, those two elements have no gcd in R.