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# If R is isomomorphic to R' then is it necessarily true that R/Kern(R) is isomomorphic to R'?

Sorry it should be R/Kern(f) where f : R - > R'

### 1 Answer

- mcbengtLv 79 years agoFavorite Answer
If f is a *surjective* homomorphism from R to R' then R/ker(f) and R' are isomorphic, regardless of any additional assumptions on R and R' (ie, whether R and R' are themselves isomorphic or not). This is a consequence of what is often called the 'first isomorphism theorem' for rings.

[This theorem says a bit more. It tells you that if f is a homomorphism from R to R', then the map from R/ker(f) to f(R) given by sending x + ker(f) to f(x) is an isomorphism--- so, in particular, R/ker(f) is isomorphic to f(R). But it is important to realize that this theorem says *more* than just "R/ker(f) is isomorphic to f(R)" because it gives a specific map from R/ker(f) to f(R) and shows it is an isomorphism. Just saying "A is isomorphic to B" is less information because it asserts only that *some* map from A to B is an isomorphism, and doesn't give an example. It is better to think of the first isomorphism theorem as saying that *a particular map* from R/ker(f) to f(R) is an isomorphism, and not just that R/ker(f) and f(R) "are isomorphic".]

If f is not surjective, anything can happen, even if the rings R and R' are the same.

Example: let Z be the ring of integers and R = R' = Z x Z, the ring of ordered pairs of integers with componentwise operations. Define f from R to R' for all (a,b) by f(a,b) = (a,a). You can check that f is a homomorphism, that f is not surjective, that ker(f) = {(0,b): b in Z}, and that R/ker(f) is isomorphic to Z. Since the additive group of Z is cyclic, but the additive group of R' = Z x Z is not, in this case, we see that there can be no isomorphism from R/ker(f) to R'.

Example: let Z be the ring of integers and let R = R' = Z[x], the ring of polynomials with integer coefficients. Define f from R to R' by sending the polynomial f(x) to the polynomial f(x^2). (For example: 1 + 2x - 3x^3 goes to 1 + 2x^2 - 3 x^6, etc.) It is easy to check that f is a homomorphism, that f is not surjective, but that ker(f) is trivial and hence that R/ker(f) is isomorphic to R. And since R = R', it follows that R/ker(f) is isomorphic to R' in this case--- even though f is not surjective. (If you apply the first isomorphism theorem to f, you learn that the subring f(R) of R', which is not all of R', is nevertheless isomorphic to R'.)

To make a long story short:

If f is any surjective homomorphism from any ring R to any other ring R', then R/ker(f) is isomorphic to R'.

If f is a homomorphism from a ring R to another ring R', and f is *not* surjective, then R/ker(f) might or might not be isomorphic to R', depending on what the rings R, R' actually are, and what f is. Further assuming that R and R' are isomorphic, or even the same ring, does not decide the matter: both possibilities can happen even then.