Ring Homomorphisms

# Ring Homomorphisms

We have already defined Group Homomorphisms. We now extend the concept to rings.

 Definition: Let $(R, +_1, *_1)$ and $(S, +_2, *_2)$ be rings with multiplicative identities $1_R$ and $1_S$ respectively. A function $\phi : R \to S$ is a Ring Homomorphism if: 1) $\phi(a +_1 b) = \phi(a) +_2 \phi(b)$ for all $a, b \in R$. 2) $\phi(a *_1 b) = \phi(a) *_2 \phi(b)$ for all $a, b \in R$. 3) $\phi(1_R) = 1_S$. If such a ring homomorphism exists then $(R, +_1, *_1)$ and $(S, +_2, *_2)$ are said to be Homomorphic.

For example, consider the ring $(R, +, *)$ and the polynomial ring $(R[x], +, *)$. We will show that $R$ is homomorphic to $R[x]$. Let $\phi : R \to R[x]$ be defined for all $r \in R$ by:

(1)
\begin{align} \quad \phi (r) = r(x) \end{align}

Where $r(x) = r$, i.e., $r(x)$ is the constant polynomial that gives the value $r$.

Let $a, b \in R$. Then:

(2)
\begin{align} \quad \phi (a + b) = (a + b)(x) = a(x) + b(x) = \phi (a) + \phi(b) \end{align}
(3)
\begin{align} \quad \phi (a * b) = (a * b)(x) = a(x) * b(x) = \phi(a) * \phi(b) \end{align}

Furthermore, if $1 \in R$ is the multiplicative identity in $R$ then $1(x)$ is the multiplicative identity in $R[x]$ and:

(4)
\begin{align} \quad \phi(1) = 1(x) \end{align}

So $\phi$ is a homomorphism from $R$ to $R[x]$. So $R$ and $R[x]$ are homomorphic.