So, I know that an ideal of the rational numbers must include all denominator values:
a/b in I = Ideal
b/d in Q = rationals, for any integer d
(a/b)(b/d) = (a/d) in I
Therefore the denominator of I can be any integer d.
Near the end the numerator must be an ideal of Z = set of all integers. Some element from nZ, a multiple of n.
OK, I get stuck when I try to show it's principal: the trouble is when I need to show that there is a element from this ideal that generates the entire set I. I don't understand how you can express the generator of the principal Ideal with the lowest term, because the lowest term has a infinite denominator. If the denominator is infinite then it doesn't exactly exist. If a smallest positive element doesn't exist then there cannot exist a principal ideal of the rationals.
Firstly, note that is a field.
Secondly, let be any ring, an ideal of . Then what happens if ?