Which of the following sets are rings with respect to the usual operations of addition and multiplication? If the set is a ring, is it also a field?

7z

To determine whether a given set is a ring with respect to the usual operations of addition and multiplication, we need to check if it satisfies the properties of a ring. Additionally, to determine if a ring is also a field, we must check if every non-zero element has a multiplicative inverse.

Let’s break down the analysis:

### 1. **Definition of a Ring**:
A set \( R \) with two operations (typically addition and multiplication) is a **ring** if:
1. \( R \) is an **abelian group** under addition (i.e., it is closed under addition, has an additive identity, and each element has an additive inverse).
2. \( R \) is closed under multiplication.
3. Multiplication is **associative**.
4. **Distributive laws** hold: \( a \cdot (b + c) = a \cdot b + a \cdot c \) and \( (a + b) \cdot c = a \cdot c + b \cdot c \) for all \( a, b, c \in R \).

### 2. **Definition of a Field**:
A ring is a **field** if:
1. It is a **commutative ring** (i.e., multiplication is commutative).
2. Every non-zero element has a multiplicative inverse.

### Given Set: \( 7\mathbb{Z} \)
The set \( 7\mathbb{Z} \) is defined as all integer multiples of 7, i.e., \( 7\mathbb{Z} = \{ \dots, -14, -7, 0, 7, 14, \dots \} \).

#### Step-by-Step Analysis:
1. **Addition**: The set \( 7\mathbb{Z} \) is closed under addition. Adding any two elements of the form \( 7a \) and \( 7b \), where \( a, b \in \mathbb{Z} \), gives \( 7a + 7b = 7(a + b) \), which is again an element of \( 7\mathbb{Z} \). The additive identity is 0 (which is in \( 7\mathbb{Z} \)), and each element has an additive inverse (e.g., \( -(7a) = 7(-a) \)).

2. **Multiplication**: The set \( 7\mathbb{Z} \) is closed under multiplication. Multiplying any two elements of the form \( 7a \) and \( 7b \) gives \( 7a \cdot 7b = 49ab \), which is also an element of \( 7\mathbb{Z} \).

3. **Associativity of Multiplication**: Multiplication in \( 7\mathbb{Z} \) inherits the associativity from the integers \( \mathbb{Z} \), so this condition is satisfied.

4. **Distributive Property**: The distributive property also holds in \( 7\mathbb{Z} \) because it inherits this property from \( \mathbb{Z} \).

Therefore, \( 7\mathbb{Z} \) is indeed a **ring**.

#### Is \( 7\mathbb{Z} \) a Field?
For \( 7\mathbb{Z} \) to be a field, every non-zero element must have a multiplicative inverse within \( 7\mathbb{Z} \). However, for any non-zero element \( 7a \) (where \( a \in \mathbb{Z} \) and \( a \neq 0 \)), there is no element \( b \in \mathbb{Z} \) such that \( 7a \cdot b = 1 \), because the product \( 7ab \) will always be a multiple of 7 and can never equal 1.

Thus, **\( 7\mathbb{Z} \) is not a field**.

### Summary:
- **\( 7\mathbb{Z} \) is a ring.**
- **\( 7\mathbb{Z} \) is not a field.**

Would you like more details, or do you have any questions? Here are 5 related questions to expand your understanding:

1. What are some other examples of sets that are rings but not fields?
2. How do the conditions change when considering a commutative ring?
3. Can a non-commutative ring ever be a field?
4. How does the concept of an ideal relate to the set \( 7\mathbb{Z} \)?
5. What properties make \( \mathbb{Q} \) (the set of rational numbers) a field?

**Tip:** Understanding the distinction between a ring and a field is crucial when studying abstract algebra and number theory!

Explore whether 7Z is a ring and analyze its potential to be a field. Understand the properties of rings and fields in algebra, with a focus on the set of integer multiples of 7, 7Z. Detailed mathematical analysis provided.

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