1. Homomorphisms

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)>.

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
It is just a question whether you know your definitions or note. The definitions in Field Theory are more harder to understrand and use than in most areas in mathematics.

b) Given an ideal I < R define
={phi(aO:a (element of) I] < S (usually denoted by phi(I)).
Here.

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.

5. Ideal being principal

If I wanted to continue this problem how could I prove that every ideal in Z_m is principal.

If I wanted to continue this problem how could I prove that every ideal in Z_m is principal.
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.