Show that R is an R-module, with the ring multiplication as scalar multiplication.

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.

Explore how to prove that a ring R is an R-module, with ring multiplication as scalar multiplication. This abstract algebra problem involves key concepts such as the distributive and associative properties of ring multiplication, as well as the multiplicative identity property. Suitable for undergraduate students.

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