Subring Module is Module

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
If $\struct {G, +_G, \circ}_R$ is a unitary $R$-module and $1_R \in S$, then $\struct{G, +_G, \circ_S}_S$ is also unitary.

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.

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:

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:

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:

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

It remains to prove the final statement:

If $\struct {G, +_G, \circ}_R$ is a unitary $R$-module and $1_R \in S$, then $\struct {G, +_G, \circ_S}_S$ is also unitary.

To show that $\struct {G, +_G, \circ_S}_S$ is unitary, we must prove that:


 * $\forall x \in G: 1_R \circ_S x = x$

Since $1_R \in S$ by assumption, the product $1_R \circ_S x$ is defined.

We now have: