Examples
We come back to several examples given with the definition of morphism.
id : R -> R
Z -> Z/6Z
C -> C
Z/6Z -> Z/2Z
Q[X] -> Q[X]/(X2)
R[X]/(X2 + 1) -> R + Ri
For any ring R,
the identity map R -> R is an isomorphism,
which is its own inverse.
The map f : Z -> Z/6Z given by
f(a) = a + 6Z is not injective as its
kernel is 6Z.
The invertible elements of Z are 1, -1;
they are mapped onto the invertible elements of Z/6Z.
Complex conjugation is an isomorphism of rings C -> C.
The morphism f : Z/6Z -> Z/2Z given by
f(a + 6Z) = a + 2Z
has
kernel {0, 2, 4}, and so is not injective.
6 = |Z/6Z| = |Z/2Z| = 2.
Since f is surjective, another way of deriving that it is not injective is by contradiction. If it were, then, f would be a bijection, and so, by part 2 of the theorem,
The morphism
f : Q[X] ->
Q[X]/(X2) is surjective, but not injective.
The residue class of 1 + X is invertible in
Q[X]/(X2) (its inverse is the class of
1 - X), but
its inverse image does not contain an invertible element in
Q[X].
The map f : R[X]/(X2 + 1)
-> R + Ri, defined by
g + (f) -> g(i),
is an isomorphism. It demonstrates that two completely different looking
rings may nevertheless carry the same ring structure.