Subring Module is Module

Theorem
Let $\left({R, +, \times}\right)$ be a ring.

Let $\left({S, +_S, \times_S}\right)$ be a subring of $R$.

Let $\left({G, +_G, \circ}\right)_R$ be an $R$-module.

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

Then $\left({G, +_G, \circ_S}\right)_S$ is an $S$-module.

The module $\left({G, +_G, \circ_S}\right)_S$ is called the $S$-module obtained from $\left({G, +_G, \circ}\right)_R$ by restricting scalar multiplication. (Eugh, that's a bit of a mouthful. We gotta come up with something better than that.)

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

Proof
Note that:


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

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

Let us verify the module axioms.

Axiom $(1)$
We need to show that:


 * $\forall a \in S: \forall x,y \in G: a \circ_S \left({x +_G y}\right) = a \circ_S x +_G a \circ_S y$

We have:

Axiom $(2)$
We need to show that:


 * $\forall a,b \in S: \forall x \in G: \left({a +_S b}\right) \circ_S x = a \circ_S x +_G b \circ_S y$

We have:

Axiom $(3)$
We need to show that:


 * $\forall a,b \in S: \forall x \in G: \left({a \times_S b}\right) \circ_S x = a \circ_S \left({b \circ_S x}\right)$

We have:

Thus, $\left({G, +_G, \circ_S}\right)_S$ is an $S$-module.

Let us prove the final statement:

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

To show that $\left({G, +_G, \circ_S}\right)_S$ 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: