No, **not every ring is a field**. While every field is a ring, the reverse is not true because fields satisfy additional properties that rings do not necessarily satisfy. Let’s explore the concepts in detail to clarify this:
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### **Definitions**
1. **Ring**:
A ring is a set \( R \), equipped with two binary operations:
- **Addition (+)**: Forms an abelian group (i.e., commutative, associative, has an identity element, and every element has an additive inverse).
- **Multiplication (·)**: Associative and distributes over addition.
- Rings can be commutative (e.g., \( a·b = b·a \)) or non-commutative.
- There is no requirement that every nonzero element has a multiplicative inverse.
Examples:
- \( \mathbb{Z} \) (integers under addition and multiplication): A commutative ring with no multiplicative inverses for most nonzero elements.
- \( \mathbb{M}_2(\mathbb{R}) \) (2×2 matrices with real entries): A non-commutative ring.
2. **Field**:
A field is a commutative ring with two additional properties:
- Every **nonzero element** has a multiplicative inverse.
- Multiplication is commutative.
Examples:
- \( \mathbb{Q} \), \( \mathbb{R} \), \( \mathbb{C} \) are fields.
- \( \mathbb{Z}_p \) (integers modulo a prime \( p \)) is a field, where \( p \) is prime.
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### **Key Difference Between Rings and Fields**
The critical distinction is the existence of a multiplicative inverse for every nonzero element. Rings do not require this, but fields do.
- In \( \mathbb{Z} \), for example, the only elements with multiplicative inverses are \( 1 \) and \( -1 \) (since \( 1 \times 1 = 1 \), \( (-1) \times (-1) = 1 \)). Thus, \( \mathbb{Z} \) is not a field.
- In \( \mathbb{Q} \), every nonzero element \( a/b \) (where \( a, b \in \mathbb{Z} \), \( b \neq 0 \)) has an inverse \( b/a \). Hence, \( \mathbb{Q} \) is a field.
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### **Examples of Rings That Are Not Fields**
1. **Integers (\( \mathbb{Z} \))**:
- Addition and multiplication work as expected.
- Nonzero integers (e.g., 2, 3, etc.) do not have multiplicative inverses in \( \mathbb{Z} \) because their inverses (e.g., \( 1/2, 1/3 \)) are not integers.
- Therefore, \( \mathbb{Z} \) is a ring but not a field.
2. **Polynomial Ring (\( \mathbb{R}[x] \))**:
- Set of polynomials with real coefficients.
- No multiplicative inverses for most nonzero polynomials. For example, the inverse of \( x \) is \( 1/x \), which is not a polynomial.
- Hence, \( \mathbb{R}[x] \) is a ring but not a field.
3. **Matrix Ring (\( \mathbb{M}_n(\mathbb{R}) \))**:
- Set of \( n \times n \) matrices with real entries.
- Not all matrices have multiplicative inverses (e.g., singular matrices have no inverse).
- Also, matrix multiplication is not commutative.
- Therefore, \( \mathbb{M}_n(\mathbb{R}) \) is a ring but not a field.
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### **When Is a Ring a Field?**
A ring \( R \) is a field if and only if:
1. \( R \) is commutative (multiplication is commutative).
2. \( R \) has a multiplicative identity (denoted as \( 1 \), where \( 1 \cdot a = a \) for all \( a \in R \)).
3. Every nonzero element in \( R \) has a multiplicative inverse.
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### **Special Cases**
- **Division Rings** (or Skew Fields): These are rings where every nonzero element has a multiplicative inverse, but multiplication may not be commutative. An example is the quaternions \( \mathbb{H} \). Division rings are not fields unless they are commutative.
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### **Conclusion**
Not every ring is a field because fields impose stricter requirements (commutativity and the existence of multiplicative inverses for all nonzero elements). Common examples of rings that are not fields include \( \mathbb{Z} \), \( \mathbb{R}[x] \), and \( \mathbb{M}_n(\mathbb{R}) \). Fields are thus a special class of rings.