2) Let \( A \) be a ring with unity, and let \( I \) be an ideal of \( A \) such that \( I \) contains a unit of \( A \) Prove that \( I=A \)

To prove that \( I = A \) given that \( I \) is an ideal of a ring \( A \) and contains a unit of \( A \), we can proceed as follows: 1. **Understanding the Setup**: - Let \( u \in I \) be a unit in \( A \). By definition of a unit, there exists an element \( v \in A \) such that \( uv = 1 \), where \( 1 \) is the multiplicative identity in \( A \). 2. **Using the Properties of Ideals**: - Since \( I \) is an ideal, it is closed under multiplication by elements of \( A \). This means that for any \( a \in A \), the product \( au \) is in \( I \). 3. **Generating Elements of \( A \)**: - Now, consider any element \( a \in A \). We can express \( a \) as \( a \cdot 1 \). Since \( 1 = uv \) (where \( u \) is a unit in \( I \)), we can rewrite \( a \) as: \[ a = a \cdot 1 = a \cdot (uv) = (au)v. \] - Since \( u \in I \) and \( I \) is an ideal, \( au \in I \). Therefore, \( (au)v \in I \) because \( v \) is an element of \( A \) and \( I \) is closed under multiplication by elements of \( A \). 4. **Conclusion**: - Since \( (au)v \in I \) and \( I \) contains \( u \), we have \( a = (au)v \in I \) for any \( a \in A \). This shows that every element \( a \in A \) is also in \( I \). - Therefore, we conclude that \( I = A \). Thus, we have shown that if \( I \) is an ideal of \( A \) that contains a unit of \( A \), then \( I \) must be equal to \( A \). \[ \boxed{I = A} \]