Set of Linear Transformations over Commutative Ring forms Submodule of Module of All Mappings
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Theorem
Let $R$ be a commutative ring.
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 $H$.
Then $\map {\LL_R} {G, H}$ is a submodule of the $R$-module $H^G$.
Commutative and Unitary Ring
Let $\struct {H, +_H, \circ}_R$ be a unitary module.
Then $\map {\LL_R} {G, H}$ is also a unitary module.
Proof
By definition of abelian group, the center of a commutative ring $R$ is $R$ itself.
So, we need to show that every $\phi \in \map {\LL_R} {G, H}$ fulfils the conditions to be a linear transformation:
- $(1): \quad \forall x, y \in G: \map {\paren {\lambda \circ \phi} } {x +_G y} = \lambda \circ \map \phi x +_H \lambda \circ \map \phi y$
- $(2): \quad \forall x \in G: \forall \mu \in R: \map {\paren {\lambda \circ \phi} } {\mu \circ x} = \mu \circ \map {\paren {\lambda \circ \phi} } x$
for all $\phi \in \map {\LL_R} {G, H}$.
From Linear Transformation from Center of Scalar Ring, these conditions are indeed fulfulled.
Hence the result.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations