Finite Submodule of Function Space
Theorem
Let $\struct {G, +}$ be a group whose identity is $e$.
Let $R$ be a ring.
Let $\struct {G, +, \circ}_R$ be an $R$-module.
Let $S$ be a set.
Let $G^S$ the set of all mappings $f: S \to G$.
Let $G^{\paren S}$ be the set of all mappings $f: S \to G$ such that $\map f x = e$ for all but finitely many elements $x$ of $S$.
Then:
- $\struct {G^{\paren S}, +', \circ}_R$ is a submodule of $\struct {G^S, +, \circ}_R$
where $+'$ is the operation induced on $G^{\paren S}$ by $+$.
Proof
Let $\struct {G, +, \circ}_R$ be an $R$-module and $S$ be a set.
We need to show that $\struct {G^{\paren S}, +'}$ is a group.
Let $f, g \in G^{\paren S}$.
Let:
- $F = \set {x \in S: \map f x \ne e}$
- $G = \set {x \in S: \map g x \ne e}$
From the definition of $f$ and $g$, both $F$ and $G$ are finite.
Since $e + e = e$ by definition of identity element, it follows that if:
- $\map f x + \map g x \ne e$
then necessarily $\map f x \ne e$ or $\map g x \ne e$.
That is:
- $\map {\paren {f +' g} } x = \map f x + \map g x \ne e \implies x \in F \cup G$
But as $F$ and $G$ are both finite, it follows that $F \cup G$ is also finite.
Hence $f +' g \in G^{\paren S}$ and $\struct {G^{\paren S}, +'}$ is closed.
Now let $f \in G^{\paren S}$.
Let $f^*$ be the pointwise inverse of $f$.
Thus:
- $\map {f^*} x = -\paren {\map f x}$
Again, let $F = \set {x \in S: \map f x \ne e}$.
It follows directly that:
- $x \in S \setminus F \implies \map {f^*} x = e$
Hence:
- $\map {f^*} x \ne e \implies x \in F$
and hence:
- $f^* \in G^{\paren S}$
So by the Two-Step Subgroup Test, it follows that $\struct {G^{\paren S}, +'}$ is a subgroup of $\struct {G^S, +}$.
Hence $\struct {G^{\paren S}, +', \circ}_R$ is an $R$-module.
The result follows.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Example $27.3$