Ring Homomorphisms, Isomorphisms, and Automorphisms

# Ring Homomorphisms, Isomorphisms, and Automorphisms

We have already defined Group Homomorphisms, Group Isomorphisms, and Group Automorphisms. We define analogous concepts for 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.
 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 Isomorphism 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) $\phi(1_R) = 1_S$. 4) $\phi$ is bijective. If such a ring isomorphism exists then $(R, +_1, *_1)$ and $(S, +_2, *_2)$ are said to be Isomorphic denoted $R \cong S$. In other words, a ring isomorphism from $(R, +_1, *_1)$ to $(S, +_2, *_2)$ is a bijective ring homomorphism.
 Definition: Let $(R, +, *)$ be a ring with multiplicative identity $1$. A function $\phi : R \to R$ is a Ring Automorphism if: 1) $\phi (a + b) = \phi (a) + \phi (b)$. 2) $\phi (a * b) = \phi (a) * \phi (b)$. 3) $\phi (1) = 1$. 4) $\phi$ is bijective. In other words, a ring automorphism of a ring $(R, +, *)$ is a bijective ring homomorphism from the ring onto itself.

We now state a simple and obvious result regarding ring isomorphisms:

 Theorem 1: Let $(R, +_1, *_1)$, $(S, +_2, *_2)$ and $(T, +_3, *_3)$ be rings. a) If $R \cong S$ then $S \cong R$. b) If $R \cong S$ and $S \cong T$ then $R \cong T$.

Let's look at an example. Consider the rings $(R, +, *)$ and $(\mathbb{C}, +, *)$ of real and complex numbers respectively. We will show that $\mathbb{R}$ and $\mathbb{C}$ are NOT isomorphic.

To show this, suppose the opposite, i.e., suppose that $\mathbb{R}$ is isomorphic to $\mathbb{C}$. Then there exists an isomorphism, $\phi : \mathbb{C} \to \mathbb{R}$. Consider the element $i \in \mathbb{C}$. Then:

(1)
\begin{align} \quad \phi^2 (i) = \phi(i) * \phi (i) = \phi (i * i) = \phi (i^2) = \phi (-1) = -\phi (1) = -1 \end{align}

But $\phi (i) \in \mathbb{R}$. The equation above saids that a real number squared is equal to $-1$. This is a contradiction. So the assumption that $\mathbb{R}$ was isomorphic to $\mathbb{C}$ is false! Hence:

(2)
\begin{align} \quad \mathbb{R} \not \cong \mathbb{C} \end{align}