Homomorphisms and Factor Rings

Let $\displaystyle \phi : R \rightarrow R' $ be a ring homomorphism and let N be an ideal of R

(a) Show that $\displaystyle \phi [N] $ is an ideal of $\displaystyle \phi [R] $

(b) Give an example to show that $\displaystyle \phi [N] $ need not be an ideal of R'

(c) Let N' be an ideal either of $\displaystyle \phi [R] $ or of R'. SHow that $\displaystyle \phi^{-1} [N'] $ is an ideal of R.

Re: Homomorphisms and Factor Rings

(a) we need to show 2 things:

1) for any r,s in N, φ(r) - φ(s) is in φ(N)

2) for any r' in φ(R), and any n' in φ(N), r'n' is in φ(N).

for 1) we have φ(r) - φ(s) = φ(r - s). since N is an ideal of R, we have r - s in N, because (N,+) is a subgroup of (R,+). thus r - s is a pre-image of φ(r) - φ(s) in N under φ, so φ(r) - φ(s) is in φ(N).

2) is similar: since r' is in φ(R), r' = φ(r), for some r in R (we don't really care which one, or if there are several such r, which there probably is).

since n' is in φ(N), n' = φ(n) for some n in N (again, it doesn't matter which one, or how many such n there are. we just need to use one of them).

thus r'n' = φ(r)φ(n) = φ(rn). since N is an ideal of R, and r is in R, and n is in N, rn is in N. thus r'n' is in φ(N).

b) the only difference here is that φ is not surjective. when looking for counter-example, a good place to look is usually in Z_{n}, or a polynomial ring.

let R = Z[x], and let R' = Z[x,y], and let φ be the inclusion map. clearly (x), the set of all polynomials in x with 0 constant term is an ideal of Z[x].

let's prove φ((x)) is not an ideal of Z[x,y]. note that the polynomial f(x) = x is in φ(Z[x]), and in fact f is in (x), so that φ(f) is in φ((x)).

also g(x,y) = y is an element of Z[x,y]. but g(x,y)φ(f) is NOT in φ((x)), since there is no polynomial h(x) in Z[x], such that φ(h) = yx

(the image of (x) contains only polynomials in x with no constant terms, none of these have any y-terms). hence φ((x)) is NOT an ideal of Z[x,y] (it doesn't "absorb" multiplication by any polynomial with y-terms, including the most basic one, g(x,y) = y).

(c) a word about why it doesn't matter of we consider φ(R) or R'. if N' has elements in it outside of φ(R), these have NO pre-image in R. in other words:

φ^{-1}(N') = φ^{-1}((N'∩φ(R))U(N'-φ(R)) = φ^{-1}(N'∩φ(R)) U φ^{-1}(N'-φ(R)) = φ^{-1}(N'∩φ(R)) U Ø = φ^{-1}(N'∩φ(R)),

so either way, we wind up considering the pre-image of N'∩φ(R), which is an ideal of φ(R) (if N' is a subset of φ(R), this is just N'). convince yourself that the intersection of two ideals is an ideal.

so without any loss of generality, we can just consider N' an ideal of φ(R). again we have to show two things:

1) if r is in φ^{-1}(N'), and s in in φ^{-1}(N') , then r - s is in φ^{-1}(N')

2) if a is ANY element of R, and r is in φ^{-1}(N'), then ar is in φ^{-1}(N').

first, we tackle 1): suppose r,s are in φ^{-1}(N'). then φ(r), and φ(s) are in N'. since N' is an ideal, φ(r) - φ(s) is in N'. hence φ(r - s) = φ(r) - φ(s) is in N', so r - s is in φ^{-1}(N').

2) is again, largely similar: clearly φ(a) is in φ(R), and φ(r) is in N' (since r is in φ^{-1}(N')), hence φ(ar) = φ(a)φ(r) is in N', since N' is an ideal of φ(R). thus ar is in φ^{-1}(N').

one can also prove 1) & 2) above just by assuming N' is an ideal of R', but the proof is JUST THE SAME (with some slight modifications).

Re: Homomorphisms and Factor Rings

Thanks for the help and guidance

As soon as my day job allows me the time I will work carefully through this post

Peter