Set of Linear Transformations over Commutative Ring forms Submodule of Module of All Mappings/Unitary
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Theorem
Let $R$ be a commutative ring with unity whose unity is $1_R$.
Let $\struct {G, +_G, \circ}_R$ and $\struct {H, +_H, \circ}_R$ be $R$-modules.
Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.
Let $\struct {H, +_H, \circ}_R$ be a unitary module.
Then $\map {\LL_R} {G, H}$ is also a unitary module.
Proof
From Set of Linear Transformations over Commutative Ring forms Submodule of Module of All Mappings, $\map {\LL_R} {G, H}$ is a module.
It remains to be shown that $\map {\LL_R} {G, H}$ is a unitary module, that is:
- $\forall \phi \in \map {\LL_R} {G, H}: 1_R \circ \phi = \phi$
So, let $\struct {H, +_H, \circ}_R$ be a unitary $R$-module.
Then, by Unitary Module Axiom $\text {UM} 4$: Unity of Scalar Ring:
- $\forall x \in H: 1_R \circ x = x$
Thus:
\(\ds \map {\paren {1_R \circ \phi} } x\) | \(=\) | \(\ds 1_R \circ \paren {\map \phi x}\) | Definition of Composition of Mappings | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi x\) | Unitary Module Axiom $\text {UM} 4$: Unity of Scalar Ring |
Hence the result.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations