# Abstract Algebra Problem involving the ideals

## Homework Statement

Let f:R→S be a homomorphism of rings. If J is an ideal in S and I={r∈R/f(r)∈J}, prove that I is an ideal in R that contains the kernal of f.

## The Attempt at a Solution

I feel like I have the problem right, but would like to have someone look over it and see if I made a assumption that I shouldn't have. Here it is below:

To show I is an ideal in R, we have two show three things:

1. We have to show that 0 is in I. f(0) = 0 since f is a homomorphism. 0 is in J since J is an ideal. This means f(0) is in J, so 0 is in I.

2. We have to show that I is closed under addition, so if we have r, s in I, then r+s is also in I.
Assume that r, s, are in I. Then by definition of I, f(r) and f(s) are in J.
If we can show that f(r+s) is in J, then by definition of I, this means that r+s is in I, so this is what we need to do.
So consider f(r+s). Since f is a homomorphism, f(r+s) = f(r) + f(s). Since f(r) is in J and f(s) is in J and J is an ideal, then f(r) + f(s) is in J. This means f(r+s) is in J.
So r+s is in I, which is what we needed to show.

3. We have to show that I is closed under multiplication by elements of R, so if we have r in R and s in I, then rs is also in I.
Assume that r is in R and s is in I. If we can show that f(rs) is in J, then by definition of I, this means that rs is in I.
So consider f(rs). Since f is a homomorphism, f(rs) = f(r)f(s). Since s is in I, then f(s) is in J. Since J is an ideal, then it is closed under multiplication by elements of R, and f(r) is in R, so f(r)f(s) is in J. Thus, f(rs) is in J.
So rs is in I, which is what we needed to show.

This means that I is an ideal in R. We still need to show that I contains the kernel of f. The kernel of f are the elements {x in R | f(x) = 0}. Consider x in ker f. Then f(x) = 0. Since J is an ideal, 0 is in J, so f(x) is in J. Then, by definition, x is in R, so R contains ker f.