To show that $R$, a ring, is an $R$-module with the ring multiplication as scalar multiplication, we need to verify that the ring $R$ satisfies the axioms of a module over itself. Specifically, we treat elements of $R$ as scalars (from the ring) and elements of $R$ as vectors (from the set on which the scalars act).

Axioms of an $R$-module:

Let $R$ be a ring, and consider $M$ as a module over $R$. We need to check the following properties where the scalar multiplication is simply the ring multiplication:

1. Closure under addition: $r \cdot (a + b) = r \cdot a + r \cdot b$, for all $r, a, b \in R$.
2. Compatibility with ring addition: $(r + s) \cdot a = r \cdot a + s \cdot a$, for all $r, s, a \in R$.
3. Compatibility with ring multiplication: $r \cdot (s \cdot a) = (r \cdot s) \cdot a$, for all $r, s, a \in R$.
4. Unitary property: $1 \cdot a = a$, for all $a \in R$, where $1$ is the multiplicative identity in $R$.

Verifying these properties for $R$ as an $R$-module:

1. Closure under addition: We need to verify that: $r \cdot (a + b) = r \cdot a + r \cdot b$ This holds because the ring multiplication distributes over addition by definition in any ring: $r(a + b) = ra + rb$ Thus, this property is satisfied.

2. Compatibility with ring addition: We need to check that: $(r + s) \cdot a = r \cdot a + s \cdot a$ This also holds by the distributive property of ring multiplication over addition: $(r + s)a = ra + sa$ Hence, this property is satisfied.

3. Compatibility with ring multiplication: We check that: $r \cdot (s \cdot a) = (r \cdot s) \cdot a$ This property holds because ring multiplication is associative: $r(sa) = (rs)a$ So, this property is also satisfied.

4. Unitary property: We need to verify that: $1 \cdot a = a$ This holds by the definition of the multiplicative identity in a ring: $1a = a$ Therefore, this property is satisfied.

Conclusion:

Since all the required properties hold, $R$ is indeed an $R$-module with the ring multiplication as scalar multiplication.

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Do you have any questions or would you like more details on any part of the explanation?

Here are five related questions to expand on this:

1. What is the difference between a free module and an $R$-module?
2. How does the structure of $R$-modules generalize vector spaces?
3. Can you show an example of an $R$-module that is not free?
4. How does scalar multiplication differ in modules over non-commutative rings?
5. What additional properties do $R$-modules have if $R$ is a commutative ring?

Tip: Every vector space is a module, but not every module is a vector space, as modules do not require a field.