Homomorphic Image of R-Module is R-Module

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Theorem

Let $\struct {R, +_R, \times_R}$ be a ring.

Let $\struct {G, +_G, \circ_G}_R$ be an $R$-module.

Let $\struct {H, +_H, \circ_H}_R$ be an $R$-algebraic structure.

Let $\phi: G \to H$ be a homomorphism.


Then the homomorphic image of $\phi$ is an $R$-module.


Proof

Let us write $\phi \sqbrk G$ to denote the homomorphic image of $\phi$.

From Image of Group Homomorphism is Subgroup, $\phi \sqbrk G$ is a subgroup of $\struct {H, +_H}$.

For any $\map \phi g$ and $\map \phi {g'}$ in $\phi \sqbrk G$, we have:

\(\ds \map \phi g +_H \map \phi {g'}\) \(=\) \(\ds \map \phi {g +_G g'}\) $\phi$ is a homomorphism
\(\ds \) \(=\) \(\ds \map \phi {g' +_G g'}\) $\struct {G, +_G}$ is an abelian group
\(\ds \) \(=\) \(\ds \map \phi {g'} +_H \map \phi g\) $\phi$ is a homomorphism

hence $\phi \sqbrk G$ is an abelian group.


Now we can turn to showing that $\phi \sqbrk G$ is an $R$-module.

To do this, we take the module axioms in turn.


Module Axiom $\text M 1$: Distributivity over Module Addition

It is to be shown that for all $\lambda \in R$ and $\map \phi g, \map \phi {g'} \in \phi \sqbrk G$:

$\lambda \circ_H \paren {\map \phi g +_H \map \phi {g'} } = \paren {\lambda \circ_H \map \phi g} +_H \paren {\lambda \circ_H \map \phi {g'} }$

Compute, using that $\phi$ is a homomorphism repetitively:

\(\ds \lambda \circ_H \paren {\map \phi g +_H \map \phi {g'} }\) \(=\) \(\ds \lambda \circ_H \map \phi {g +_G g'}\)
\(\ds \) \(=\) \(\ds \map \phi {\lambda \circ_G \paren {g +_G} }\)
\(\ds \) \(=\) \(\ds \map \phi {\paren {\lambda \circ_G g} +_G \paren {\lambda \circ_G g'} }\) $G$ is an $R$-module
\(\ds \) \(=\) \(\ds \map \phi {\lambda \circ_G g} +_H \map \phi {\lambda \circ_G g'}\)
\(\ds \) \(=\) \(\ds \paren {\lambda \circ_H \map \phi g} +_H \paren {\lambda \circ_H \map \phi {g'} }\)

$\Box$


Module Axiom $\text M 2$: Distributivity over Scalar Addition

It is to be shown that for all $\lambda, \mu \in R$ and $\map \phi g \in \phi \sqbrk G$:

$\paren {\lambda +_R \mu} \circ_H \map \phi g = \paren {\lambda \circ_H \map \phi g} +_H \paren {\mu \circ_H \map \phi g}$

Compute, using that $\phi$ is a homomorphism repetitively:

\(\ds \paren {\lambda +_R \mu} \circ_H \map \phi g\) \(=\) \(\ds \map \phi {\paren {\lambda +_R \mu} \circ_G g}\)
\(\ds \) \(=\) \(\ds \map \phi {\paren {\lambda \circ_G g} +_G \paren {\mu \circ_G g} }\) $G$ is an $R$-module
\(\ds \) \(=\) \(\ds \map \phi {\lambda \circ_G g} +_H \map \phi {\mu \circ_G g}\)
\(\ds \) \(=\) \(\ds \paren {\lambda \circ_H \map \phi g} +_H \paren {\mu \circ_H \map \phi g}\)

$\Box$


Module Axiom $\text M 3$: Associativity

It is to be shown that for all $\lambda, \mu \in R$ and $\map \phi g \in \phi \sqbrk G$:

$\paren {\lambda \times_R \mu} \circ_H \map \phi g = \lambda \circ_H \paren {\mu \circ_H \map \phi g}$

Compute, using that $\phi$ is a homomorphism repetitively:

\(\ds \paren {\lambda \times_R \mu} \circ_H \map \phi g\) \(=\) \(\ds \map \phi {\paren {\lambda \times_R \mu} \circ_G g}\)
\(\ds \) \(=\) \(\ds \map \phi {\lambda \circ_G \paren {\mu \circ_G g} }\) $G$ is an $R$-module
\(\ds \) \(=\) \(\ds \lambda \circ_H \map \phi {\mu \circ_G g}\)
\(\ds \) \(=\) \(\ds \lambda \circ_H \paren {\mu \circ_H \map \phi g}\)

$\Box$


Having verified that $\phi \sqbrk G$ satisfies the three module axioms, we conclude it is an $R$-module.

$\blacksquare$


Sources

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