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:

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

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.

$\blacksquare$