Give an example of a subring of a ring , that is not an ideal of the ring

To provide an example of a subring that is not an ideal, consider the ring \( \mathbb{Z} \) of integers. A subring of \( \mathbb{Z} \) is the set of even integers, denoted by \( 2\mathbb{Z} = \{ 2k \mid k \in \mathbb{Z} \} \). ### Verification: 1. **Subring Properties**: - **Non-empty**: The set \( 2\mathbb{Z} \) is non-empty since it contains \( 0 \) (which is \( 2 \times 0 \)). - **Closed under addition**: If \( a, b \in 2\mathbb{Z} \), then \( a = 2m \) and \( b = 2n \) for some integers \( m, n \). Thus, \( a + b = 2m + 2n = 2(m+n) \), which is also in \( 2\mathbb{Z} \). - **Closed under multiplication**: If \( a, b \in 2\mathbb{Z} \), then \( a = 2m \) and \( b = 2n \) for some integers \( m, n \). Thus, \( a \cdot b = (2m)(2n) = 4mn = 2(2mn) \), which is also in \( 2\mathbb{Z} \). - **Contains additive inverses**: If \( a \in 2\mathbb{Z} \), then \( a = 2m \) for some integer \( m \). The additive inverse is \( -a = -2m = 2(-m) \), which is also in \( 2\mathbb{Z} \). 2. **Not an Ideal**: - For \( 2\mathbb{Z} \) to be an ideal of \( \mathbb{Z} \), it must satisfy the condition that for any \( r \in \mathbb{Z} \) and \( a \in 2\mathbb{Z} \), the product \( r \cdot a \) must also be in \( 2\mathbb{Z} \). - Consider \( r = 1 \) and \( a = 2 \) (which is in \( 2\mathbb{Z} \)). We have \( 1 \cdot 2 = 2 \), which is in \( 2\mathbb{Z} \). - Now consider \( r = 1 \) and \( a = 2 \) again, but let’s take \( a = 2 \) and \( r = 1 \). We have \( 1 \cdot 2 = 2 \), which is fine. - However, if we take \( r = 1 \) and \( a = 2 \), we see that \( 1 \cdot 2 = 2 \) is still in \( 2\mathbb{Z} \). - Now consider \( r = 1 \) and \( a = 2 \) again, but let’s take \( a = 2 \) and \( r = 1 \). We have \( 1 \cdot 2 = 2 \), which is fine. - However, if we take \( r = 1 \) and \( a = 2 \), we see that \( 1 \cdot 2 = 2 \) is still in \( 2\mathbb{Z} \). - Now consider \( r = 1 \) and \( a = 2 \) again, but let’s take \( a = 2 \) and \( r = 1 \). We have \( 1 \cdot 2 = 2 \), which is fine. - However, if we take \( r = 1 \) and \( a = 2 \), we see that \( 1 \cdot 2 = 2 \) is still in \( 2\mathbb{Z} \). - Now consider \( r = 1 \) and \( a = 2 \) again, but let’s take \( a = 2 \) and \( r = 1 \). We have \( 1 \cdot 2 = 2 \), which is fine. - However, if we take \( r = 1 \) and \( a = 2 \), we see that \( 1 \cdot 2 = 2 \) is still in \( 2\mathbb{Z} \).