Composition of Ring Homomorphisms is Ring Homomorphism

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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 $+_1$:

\(\displaystyle \map {\paren {\psi \bullet \phi} } {x +_1 y}\) \(=\) \(\displaystyle \map \psi {\map \phi {x +_1 y} }\) $\quad$ Definition of Composition of Mappings $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \map \psi {\map \phi x +_2 \map \phi y}\) $\quad$ Morphism Property applied to $+_1$ under $\phi$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \map \psi {\map \phi x} +_3 \map \psi {\map \phi y}\) $\quad$ Morphism Property applied to $+_2$ under $\psi$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \map {\paren {\psi \bullet \phi} } x +_3 \map {\paren {\psi \bullet \phi} } y\) $\quad$ Definition of Composition of Mappings $\quad$


The same applies to $\circ_1$:

\(\displaystyle \map {\paren {\psi \bullet \phi} } {x \circ_1 y}\) \(=\) \(\displaystyle \map \psi {\map \phi {x \circ_1 y} }\) $\quad$ Definition of Composition of Mappings $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \map \psi {\map \phi x \circ_2 \map \phi y}\) $\quad$ Morphism Property applied to $\circ_1$ under $\phi$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \map \psi {\map \phi x} \circ_3 \map \psi {\map \phi y}\) $\quad$ Morphism Property applied to $\circ_2$ under $\psi$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \map {\paren {\psi \bullet \phi} } x \circ_3 \map {\paren {\psi \bullet \phi} } y\) $\quad$ Definition of Composition of Mappings $\quad$


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