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.

proof:

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.

Proof:

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).

Proof:

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.

proof:

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:

proof:

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.