Let's solve the problem step by step for \( p = 2 \): --- ### **Step 1: Analyze the Rings for \( p = 2 \)** We are asked whether: \[ \frac{\mathbb{Z}_2[x]}{(x^2 - 2)} \cong \frac{\mathbb{Z}_2[x]}{(x^2 - 3)} \] But in \( \mathbb{Z}_2 \), both \( 2 \) and \( 3 \) reduce modulo 2 as follows: - \( 2 \equiv 0 \pmod{2} \) - \( 3 \equiv 1 \pmod{2} \) So the quotients become: \[ \frac{\mathbb{Z}_2[x]}{(x^2 - 0)} = \frac{\mathbb{Z}_2[x]}{(x^2)} \] and \[ \frac{\mathbb{Z}_2[x]}{(x^2 - 1)} \] **Explanation:** We reduced the constants modulo 2, simplifying the polynomials. This step checks how the ideals look in \( \mathbb{Z}_2[x] \). --- ### **Step 2: Check If the Two Quotient Rings are Isomorphic** - \( \frac{\mathbb{Z}_2[x]}{(x^2)} \): Elements look like \( a + bx \), since \( x^2 = 0 \). - \( \frac{\mathbb{Z}_2[x]}{(x^2-1)} \): Elements look like \( a + bx \), but with \( x^2 = 1 \). These two rings are not isomorphic: - In the first ring, \( x^2 = 0 \) - In the second ring, \( x^2 = 1 \) Their structure is different (one is local with nilpotent \( x \), and the other is semisimple and splits as \( \mathbb{Z}_2 \times \mathbb{Z}_2 \)), so they are **not isomorphic**. **Explanation:** We compared the relations in the quotient rings and saw that they are fundamentally different in structure. --- ### **Final Answer:** For \( p = 2 \), the rings \( R_1 \) and \( R_2 \) are **not isomorphic**.