Subring Module is Module

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Theorem

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

Let $\struct {S, +_S, \times_S}$ be a subring of $R$.

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

Let $\circ_S$ be the restriction of $\circ$ to $S \times G$.


Let $\struct {G, +_G, \circ_S}_S$ be subring module induced by $S$.

Then $\struct {G, +_G, \circ_S}_S$ is an $S$-module.


Unitary Subring

Let $\struct {R, +, \times}$ be a ring with unity.

Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module.

Let $1_R \in S$.


Then $\struct{G, +_G, \circ_S}_S$ is also unitary.


Special Case

Let $S$ be a subring of the ring $\struct {R, +, \circ}$.

Let $\circ_S$ be the restriction of $\circ$ to $S \times R$.


Then $\struct {R, +, \circ_S}_S$ is an $S$-module.


Proof

We have that:

$\forall a, b \in S: a +_S b = a + b$
$\forall a, b \in S: a \times_S b = a \times b$
$\forall a \in S: \forall x \in G = a \circ_S x = a \circ x$

as $+_S$, $\times_S$ and $\circ_S$ are restrictions.


Let us verify the module axioms.


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

We need to show that:

$\forall a \in S: \forall x, y \in G: a \circ_S \paren {x +_G y} = a \circ_S x +_G a \circ_S y$

We have:

\(\ds a \circ_S \paren {x +_G y}\) \(=\) \(\ds a \circ \paren {x +_G y}\)
\(\ds \) \(=\) \(\ds a \circ x +_G a \circ y\) Module Axiom $\text M 1$: Distributivity over Module Addition on $\struct {G, +_G, \circ}_R$
\(\ds \) \(=\) \(\ds a \circ_S x +_G a \circ_S y\)

$\Box$


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

We need to show that:

$\forall a, b \in S: \forall x \in G: \paren {a +_S b} \circ_S x = a \circ_S x +_G b \circ_S y$

We have:

\(\ds \paren {a +_S b} \circ_S x\) \(=\) \(\ds \paren {a + b} \circ x\)
\(\ds \) \(=\) \(\ds a \circ x + b \circ x\) Module Axiom $\text M 2$: Distributivity over Scalar Addition on $\struct {G, +_G, \circ}_R$
\(\ds \) \(=\) \(\ds a \circ_S x +_G b \circ_S x\)

$\Box$


Module Axiom $\text M 3$: Associativity

We need to show that:

$\forall a, b \in S: \forall x \in G: \paren {a \times_S b} \circ_S x = a \circ_S \paren {b \circ_S x}$

We have:

\(\ds \paren {a \times_S b} \circ_S x\) \(=\) \(\ds \paren {a \times b} \circ x\)
\(\ds \) \(=\) \(\ds a \circ \paren {b \circ x}\) Module Axiom $\text M 3$: Associativity on $\struct {G, +_G, \circ}_R$
\(\ds \) \(=\) \(\ds a \circ_S \paren {b \circ_S x}\)

$\Box$


Thus $\struct {G, +_G, \circ_S}_S$ is an $S$-module.

$\blacksquare$


Sources

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