Thread: Principal Ideals of Rationals

Originally Posted by fuzbyone

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 ?

Now, apply these two things to your problem.

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Originally Posted by Swlabr

Firstly, note that is a field.

Secondly, let be any ring, an ideal of . Then what happens if ?

Now, apply these two things to your problem.

Thanks for the reply, I can see that anything with the Unit becomes the entire field.

A new question: What is the definition of a Principal Ideal in a integral domain? is it the same thing as a Principal Ideal in a Ring? I suspect they are different definitions.

Originally Posted by fuzbyone

Thanks for the reply, I can see that anything with the Unit becomes the entire field.

A new question: What is the definition of a Principal Ideal in a integral domain? is it the same thing as a Principal Ideal in a Ring? I suspect they are different definitions.

What definition do you have for in a ring?