This is way over my head.
Let R and S be commutative runs and phi:R---->S be a ring homomorphism. a) Given an ideal J < S (that < means subset), define
={a (element of) R: phi(a) (element of) J] < R. (usually this is denoted by phi^-1(J)). PRove this is an ideal
b) Given an ideal I < R define
={phi(aO:a (element of) I] < S (usually denoted by phi(I)). Prove this is an ideal provided phi maps onto S.
c) In a case with b, then prove that if phi maps onto S then phi(<a>)=<phi(a)>.
Let us assume these are rings with unity.
To remind, a "principal ideal" <a> is the ideal {ra|r in R}.
We want to show,
phi[<a>] = <phi(a)>
The standard trick is to show each is a subset of eachother.
First note that,
phi[<a>] = phi [{ra| r in R}] = {phi(r)phi(a)| r in R}.
Then since phi(a) in <phi(a)> and phi(r) in R we have that,
phi(r)phi(a) in <phi(a)>
Second note that,
<phi(a)> = {s phi(a) | s in S}
Then since phi(a) in phi[<a>] and s = phi(r) for some r in R (because it is onto) we have that,
s phi(a) in <phi(a)>
Thus, they are equal.
That is easy.
Theorem: A subgroup of a cyclic group is a cylic group.
Now, if "I" is an ideal of Z_m it must be an additive subgroup of Z_m. But as an additive subgroup it must be cyclic. That is basically the meaning of what a principla ideal means.
See if you can do it from there.
Is this an undergraduate or a graduate course?