Ring Theory: intersection of ideals?

Question

If Z*= set of all integers except 0, and both a, b are elements of Z*, aZ = { az : z is an element of Z}, and bZ = { bz : z is an element of Z}, how do i show that

aZ intersection bZ = lcm(a,b)Z

where lcm(a,b) is the least common multiple of a and b? Any help would be greatly appreciated

We proved that aZ + bZ = dZ in lectures, using the formula that if u,v are primes, then there exists an a, b such that au + bv = 1. Then we proved that dZ was a subset of aZ + bZ, then the other way, therefore they are equal. How do i do this in this context?

Answer 1

This is one of those many cases in which you prove set equality by proving each is contained in the other.

Or if you like -- and this is just another way to say the same thing -- you prove an element is in one if and only if it is in the other.

Note that aZ is just the set of multiples of a -- i.e., the principal ideal generated by a. Similarly for b and lcm(a,b).

One way, this is trivial. lcm (a,b) is a multiple of each of a and b. And therefore so is any multiple of it. And so any multiple of lcm(a,b) is in both aZ and bZ, and therefore in theiir intersection.

The other way you're trying to prove:

If N is a multiple of both a and b, then it is also a multiple of lcm(a,b).

It's hard to know what to say here, because I'm guessing you already know that fact, or else it wouldn't be called the "lcm". If you don't, please say what seemingly relevant facts you do know, and I or somebody else can point you towards a proof.