5.3: Ring Homomorphisms

Ring Homomorphisms

Definition: Ring homomorphisms

Let $R, S$ be rings. A function $\phi: R \to S$ is called a ring homomorphism if 

1. $\phi(a+b)=\phi(a)+\phi (b), \forall a,b \in R,$ and 

2. $\phi(ab)=\phi(a) \phi (b), \forall a,b \in R.$

 

If $\phi$ is bijective, then $\phi$ is called isomorphism.

Example $\PageIndex{1}$

Show that  the function $\phi:\mathbb{Z}_2 \to \mathbb{Z}_2$ defined by $\phi(x)=x^2, x \in \mathbb{Z}_2$ is a ring homomorphism.

Solution

Let $a, b \in i\mathbb{Z}_2.$

Consider $\phi(a+b)=(a+b)^2=a^2+2ab+b^2=a^2+b^2=\phi(a)+\phi (b).$

Consider $\phi(ab)=(ab)^2=a^2b^2=\phi(a)\phi (b).$

Hence, $\phi$ is a ring homomorphism.

 

Example $\PageIndex{2}$

Whether  or not the function $\phi:\mathbb{Z} \to \mathbb{Z}$ defined by $\phi(x)=2x, x \in \mathbb{Z}$ is a ring homomorphism? Justify your answer.

Solution

Choose $a=2$ and $b=3$. Then   $\phi(2)=  4$ and $\phi(3)=6$,  $\phi(a)\phi (b)=(4)(6)=24$ but $\phi(ab)=\phi((2)(3))=\phi(6)=12\ne \phi(a)\phi(b)$. 

Hence, $\phi$ is  not a ring homomorphism.

Definition: Kernel of a homomorphism

The kernel of $\phi$ is denoted and defined as $Ker(\phi) = \{ r \in R | \phi(r)=0 \}.$

Example $\PageIndex{3}$

Define $\phi: \mathbb{Z} \to \mathbb{Z}_n$ by $\phi(a)=a(mod n), \forall a\in R.$

Then $\phi$ is a ring homomorphism but not an isomorphism. In this case, $Ker(\phi) = n \mathbb{Z}.$

Properties of ring homomorphisms:

Theorem $\PageIndex{1}$

Let $R, S$ be rings. Let $\phi: R \to S$ be a ring homomorphism. Then,

Then 

1. $\phi(0)=0.$

2. if $R$ is a commutative ring, then $\phi(R)$ is also a commutative ring.

3. if $R$ is a field and $\phi(R) \ne \{0\}$ then $\phi(R)$ is also a field.

4. if $R$ and $S$ have an identity. If $\phi$ is onto then $\phi(1_R)=1_S.$

Theorem $\PageIndex{2}$

Let $R, S$ be rings. Let $\phi: R \to S$ be a ring homomorphism. Then $Ker (\phi)$ is an ideal of $R$.