Kernel (Ker) of a Ring Homomorphism
The kernel of a ring homomorphism between two rings (R, +, *) and (R′, +, *) is the subset of R consisting of all elements that are mapped to zero in R′. This subset is referred to as the Ker φ: $$ Ker \ \phi = \{ r \in R \ | \ \phi(r) = 0_{R'} \} $$
The kernel is always a subring of R.
Furthermore, if you multiply any element of R by an element of the kernel - on either side - the result remains in the kernel:
$$ \phi(k \cdot r) = \phi(k) \cdot \phi(r) = 0 \cdot \phi(r) = 0 $$
Here, 0 denotes the zero element of R′, and since $$ 0 \in \ Ker \ \phi $$ the product is also in the kernel.
A Concrete Example
Consider the rings (ℤ6, +, *) and (ℤ3, +, *),
where ℤn denotes the ring of integers modulo n.
$$ \mathbb{Z}_6 = \{ 0, 1, 2, 3, 4, 5 \} $$
$$ \mathbb{Z}_3 = \{ 0, 1, 2 \} $$
A homomorphism from ℤ6 to ℤ3 can be defined by the natural projection:
$$ \phi(x) = x \mod 3 $$
The kernel of this homomorphism is the set {0, 3} in ℤ6, since both elements are mapped to the additive identity in ℤ3, i.e., $$ \phi(x) = 0_{\mathbb{Z}_3} $$ for all x in {0, 3}.
And so on.
