R is an integral domain.
Assume that greatest common divisors exist in R. Let p be an irreducible element of R and let a be an element of R that is not divisible by p. Show that 1 is a greatest common divisor of a and p.
Thoughts...
p is irreducible, hence if p|ab then either a=1 or b=1.
incorrect! p is irreducible not prime!
If b = 1 then either a = 1, in which case g.c.d (a,p) = 1, or a
1, in which case p|a, a contradiction. Hence if b = 1, g.c.d (a,p) = 1.
So assume b
1, hence a = 1. Hence g.c.d (a,p) = 1.
This look right?