# Composition of Ring Homomorphisms is Ring Homomorphism

## Theorem

Let:

- $\struct {R_1, +_1, \odot_1}$
- $\struct {R_2, +_2, \odot_2}$
- $\struct {R_3, +_3, \odot_3}$

be rings.

Let:

- $\phi: \struct {R_1, +_1, \odot_1} \to \struct {R_2, +_2, \odot_2}$
- $\psi: \struct {R_2, +_2, \odot_2} \to \struct {R_3, +_3, \odot_3}$

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

Let $\psi \circ \phi$ denote the composite of $\phi$ and $\psi$.

Then what we are trying to prove is denoted:

- $\paren {\psi \circ \phi}: \struct {R_1, +_1, \odot_1} \to \struct {R_3, +_3, \odot_3}$ is a homomorphism.

To prove the above is the case, we need to demonstrate that the morphism property is held by $+_1$ and $\odot_1$ under $\psi \circ \phi$.

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

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

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

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

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

The same applies to $\odot_1$:

\(\ds \map {\paren {\psi \circ \phi} } {x \odot_1 y}\) | \(=\) | \(\ds \map \psi {\map \phi {x \odot_1 y} }\) | Definition of Composition of Mappings | |||||||||||

\(\ds \) | \(=\) | \(\ds \map \psi {\map \phi x \odot_2 \map \phi y}\) | Morphism Property applied to $\odot_1$ under $\phi$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \map \psi {\map \phi x} \odot_3 \map \psi {\map \phi y}\) | Morphism Property applied to $\odot_2$ under $\psi$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \map {\paren {\psi \circ \phi} } x \odot_3 \map {\paren {\psi \circ \phi} } y\) | 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 $\psi \circ \phi$.

Thus $\paren {\psi \circ \phi}: \struct {R_1, +_1, \odot_1} \to \struct {R_3, +_3, \odot_3}$ is a homomorphism.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.2$ - 1970: B. Hartley and T.O. Hawkes:
*Rings, Modules and Linear Algebra*... (previous) ... (next): $\S 2.2$: Homomorphisms: Definition $2.4$