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 $2.4$