# Composition of Ring Homomorphisms is Ring Homomorphism

## Theorem

Let:

- $\struct {R_1, +_1, \circ_1}$
- $\struct {R_2, +_2, \circ_2}$
- $\struct {R_3, +_3, \circ_3}$

be rings.

Let:

- $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$
- $\psi: \struct {R_2, +_2, \circ_2} \to \struct {R_3, +_3, \circ_3}$

be homomorphisms.

Then the composite of $\phi$ and $\psi$ is also a homomorphism.

## Proof 1

A specific instance of Composite of Homomorphisms on Algebraic Structure is Homomorphism.

$\blacksquare$

## Proof 2

So as to alleviate possible confusion over notation, let the composite of $\phi$ and $\psi$ be denoted $\psi \bullet \phi$ instead of the more usual $\psi \circ \phi$.

Then what we are trying to prove is denoted:

- $\paren {\psi \bullet \phi}: \struct {R_1, +_1, \circ_1} \to \struct {R_3, +_3, \circ_3}$ is a homomorphism.

To prove the above is the case, we need to demonstrate that the morphism property is held by $+_1$ and $\circ_1$ under $\psi \bullet \phi$.

We take two elements $x, y \in R_1$, and put them through the following wringer with respect to $+_1$:

\(\displaystyle \map {\paren {\psi \bullet \phi} } {x +_1 y}\) | \(=\) | \(\displaystyle \map \psi {\map \phi {x +_1 y} }\) | Definition of Composition of Mappings | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \psi {\map \phi x +_2 \map \phi y}\) | Morphism Property applied to $+_1$ under $\phi$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \psi {\map \phi x} +_3 \map \psi {\map \phi y}\) | Morphism Property applied to $+_2$ under $\psi$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map {\paren {\psi \bullet \phi} } x +_3 \map {\paren {\psi \bullet \phi} } y\) | Definition of Composition of Mappings |

The same applies to $\circ_1$:

\(\displaystyle \map {\paren {\psi \bullet \phi} } {x \circ_1 y}\) | \(=\) | \(\displaystyle \map \psi {\map \phi {x \circ_1 y} }\) | Definition of Composition of Mappings | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \psi {\map \phi x \circ_2 \map \phi y}\) | Morphism Property applied to $\circ_1$ under $\phi$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \psi {\map \phi x} \circ_3 \map \psi {\map \phi y}\) | Morphism Property applied to $\circ_2$ under $\psi$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map {\paren {\psi \bullet \phi} } x \circ_3 \map {\paren {\psi \bullet \phi} } y\) | Definition of Composition of Mappings |

Disentangling the confusing and tortuous expressions above, we (eventually) see that this shows that the morphism property is indeed held by both $+_1$ and $\circ_1$ under $\psi \bullet \phi$.

Thus $\paren {\psi \bullet \phi}: \struct {R_1, +_1, \circ_1} \to \struct {R_3, +_3, \circ_3}$ is a homomorphism.

$\blacksquare$

## Sources

- 1970: B. Hartley and T.O. Hawkes:
*Rings, Modules and Linear Algebra*... (previous) ... (next): $\S 2.2$: Homomorphisms: Definition $2.4$