IDA: Example

Examples

Z -> Z

Z -> Z/6Z

C -> C

Z/6Z -> Z/2Z

Q[X] -> Q[X]/(X²)

R[X]/(X² + 1) -> R + Ri

The map f : Z -> Z given by f(a) = 3a is not a morphism of rings because f(1) is not equal to 1.

The map f : Z -> Z/6Z given by f(a) = a + 6Z is a morphism of rings:

f(0) = 0 + 6Z,

f(1) = 1 + 6Z,

f(a + b) = (a + b) + 6Z = (a + 6Z) + (b + 6Z) = f(a) + f(b), and

f(ab) = (ab) + 6Z = (a + 6Z) * (b + 6Z) = f(a) * f(b).

The kernel of the map is exactly 6Z.

Complex conjugation is a morphism of rings C -> C. In fact, it is an isomorphism: It is its own inverse.

The map f : Z/6Z -> Z/2Z given by f(a + 6Z) = a + 2Z is a morphism of rings.
The kernel consists of all a + 6Z for a Z such that a = 0 mod 2. The kernel is therefore {0, 2, 4}. It is not hard to verify that f is surjective.

The morphism f : Q[X] -> Q[X]/(X²) which sends a polynomial to its class modulo X² is a morphism as is easily verified.
The kernel of this morphism consists of all polynomials that are divisible by X². The morphism is surjective, so the image is the whole ring Q[X]/(X²).

The map f : R[X]/(X² + 1) -> R + Ri, where i is the usual imaginary number (square root of -1), is defined by
g + (X² + 1) -> g(i),
for every residue class g + (X² + 1) R[X]/(X² + 1).
Observe that it is indeed well defined. This follows from the fact that if g - h is divisible by X² + 1, then g(i) = h(i). So g + (X² + 1) = h + (X² + 1) implies g(i) = h(i).
This map is in fact an isomorphism. The inverse map is given by a + bi -> a + bX + (X² + 1), as can be easily checked.