Definition:Vector Space over Division Subring/Special Case

Theorem
Let $\left({R, +, \circ}\right)$ be a ring with unity whose unity is $1_R$.

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

Then $\left({R, +, \circ_V}\right)_S$ is an $S$-vector space, where $\circ_V$ is the restriction of $\circ$ to $S \times R$.

Proof
A vector space is by definition a unitary module over a division ring.

$S$ is by assumption a division ring, so it ramains to show that $\left({R, +, \circ}\right)_S$ is a unitary module.

That is, $\left({R, +}\right)$ must be an abelian group, and $\left({R, +, \circ}\right)_S$ must obey the unitary module axioms.

$\left({R, +}\right)$ is an abelian group by the definition of a ring.

Let us verify the unitary module axioms.

Note that:


 * $\forall a,b \in S: a \circ_S b = a \circ b$

and:


 * $\forall a \in S: \forall x \in R: a \circ_V x = a \circ x$

since $\circ_S$ and $\circ_V$ are restrictions.

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


 * $\forall a \in S: \forall x,y \in R: a \circ_V \left({x+y}\right) = a \circ_V x + a \circ_V y$

We have:

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


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

We have:

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


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

We have:

Axiom $(4)$
We need to show that:


 * $\forall x \in R: 1_R \circ_V x = x$

From Cancellable Monoid Identity of Submonoid it follows that $1_R \in S$.

Therefore, the product $1_R \circ_V x = x$ is defined.

We now have: