Ring Homomorphism

A homomorphism from a ring R to a ring R', where (R,+,*) and (R',+,*) are rings, is a function φ from R to R' $$ φ \ : \ R \rightarrow R' $$ that satisfies the following properties for all elements a, b in R: $$ φ(a + b) = φ(a) + φ(b) $$ $$ φ(a \cdot b) = φ(a) \cdot φ(b) $$

Depending on its properties, a ring homomorphism can be classified as:

an automorphism if R = R'
a monomorphism if φ is injective
an epimorphism if φ is surjective
an isomorphism if φ is both injective and surjective (i.e., bijective)

A Concrete Example
The Kernel (Ker)
Key Observations

A Concrete Example

One of the simplest examples of a ring homomorphism is the zero homomorphism.

This is a function that maps every element of R to the zero element in R':

$$ y = φ(x) = x \cdot 0 $$

This function preserves addition:

$$ φ(a + b) = φ(a) + φ(b) \quad \forall \ a, b \in R $$

Example. Take a = 2 and b = 3 in R: $$ φ(2+3)=φ(5)=5 \cdot 0 = 0 $$ $$ φ(2) + φ(3) = 2 \cdot 0 + 3 \cdot 0 = 0 $$ As expected, both expressions yield zero.

This function also preserves multiplication:

$$ φ(a \cdot b) = φ(a) \cdot φ(b) \quad \forall \ a, b \in R $$

Example. Again, let a = 2 and b = 3 in R: $$ φ(2 \cdot 3) = φ(6) = 6 \cdot 0 = 0 $$ $$ φ(2) \cdot φ(3) = (2 \cdot 0) \cdot (3 \cdot 0) = 0 $$ The result is still zero, confirming that the function is a ring homomorphism.

Example 2

Now consider a different function: $$ y = φ(x) = 4 \cdot x $$ This function does preserve addition:

$$ φ(a + b) = φ(a) + φ(b) \quad \forall \ a, b \in R $$

Example. Take a = 2 and b = 3 in R: $$ φ(2+3) = φ(2) + φ(3) $$ $$ φ(5) = 4 \cdot 5 = 20 $$ $$ \phi(2) + \phi(3) = 4 \cdot 2 + 4 \cdot 3 = 8 + 12 = 20 $$ So the additive property is preserved.

However, it does not preserve multiplication:

$$ φ(a \cdot b) \ne φ(a) \cdot φ(b) \quad \text{in general} $$

Example. Let a = 4 and b = 5 in R: $$ φ(4 \cdot 5) = \phi(20) = 4 \cdot 20 = 80 $$ $$ \phi(4) \cdot \phi(5) = (4 \cdot 4) \cdot (4 \cdot 5) = 16 \cdot 20 = 320 $$ Clearly, $$ 80 \ne 320 $$ So the function fails to preserve multiplication, and is therefore not a ring homomorphism.

The Kernel (Ker)

The kernel of a ring homomorphism between (R,+,*) and (R',+,*) is the subset of R consisting of all elements that are mapped to zero in R'. It is denoted by Kerφ: $$ Ker \, \phi = \{ r \in R \ | \ \phi(r) = 0_{R'} \} $$

Key Observations

Here are a few important properties of ring homomorphisms:

For any ring homomorphism φ: R → R', the zero element in R always maps to the zero element in R': $$ \phi(0) = 0 $$
Example. Take the zero element of R and either add or multiply it by itself: $$ \phi(0_R + 0_R) = \phi(0_R) + \phi(0_R) = 0_{R'} + 0_{R'} = 0_{R'} $$ $$ \phi(0_R \cdot 0_R) = \phi(0_R) \cdot \phi(0_R) = 0_{R'} \cdot 0_{R'} = 0_{R'} $$

And so on.