Question:medium

All rings considered below are assumed to be associative and commutative with \( 1 \neq 0 \). Further, all ring homomorphisms map 1 to 1.

Consider the following statements about such a ring \( R \):

P1: \( R \) is isomorphic to the product of two rings \( R_1 \) and \( R_2 \).
P2: \( \exists r_1, r_2 \in R \) such that \( r_1^2 = r_1 \neq 0 \), \( r_2^2 = 0 \), and \( r_1 + r_2 = 1 \).
P3: \( R \) has ideals \( I_1, I_2 \subset R \) with \( R \neq I_1 \), \( (0) \neq I_2 \), and \( R = I_1 + I_2 \) and \( I_1 \cap I_2 = (0) \).
P4: \( \exists a, b \in R \) with \( a \neq 0 \), \( b \neq 0 \) such that \( ab = 0 \).

Then, which of the following is/are TRUE?

Show Hint

When working with ring homomorphisms and ideals, it is important to check the properties of the ring and its decomposition before drawing conclusions about its structure.
Updated On: Feb 2, 2026
  • \( P1 \Rightarrow P2 \)
  • \( P2 \Rightarrow P3 \)
  • \( P3 \Rightarrow P4 \)
  • \( P4 \Rightarrow P1 \)
Show Solution

The Correct Option is A, B, C

Solution and Explanation

To determine which statements about the ring \( R \) are true, we will analyze the implications between the given statements \( P1, P2, P3, \) and \( P4 \). Let's evaluate each implication step by step:

  1. \( P1 \Rightarrow P2 \):

    Given that \( R \) is isomorphic to the product of two rings \( R_1 \) and \( R_2 \), it can be expressed as \( R \cong R_1 \times R_2 \). In the product ring, we can take \( r_1 = (1,0) \) and \( r_2 = (0,1) \). Then:

    • \( r_1^2 = (1,0)^2 = (1,0) \), so \( r_1^2 = r_1 \neq 0 \).
    • \( r_2^2 = (0,1)^2 = (0,0) \), so \( r_2^2 = 0 \).
    • \( r_1 + r_2 = (1,0) + (0,1) = (1,1) = 1 \).

    This satisfies the conditions of \( P2 \). Therefore, \( P1 \Rightarrow P2 \) is true.

  2. \( P2 \Rightarrow P3 \):

    Suppose there exist \( r_1, r_2 \in R \) such that \( r_1^2 = r_1 \neq 0 \), \( r_2^2 = 0 \), and \( r_1 + r_2 = 1 \). Define two ideals in \( R \):

    • Let \( I_1 = \{ x \in R \mid x = xr_1 \} \).
    • Let \( I_2 = \{ x \in R \mid x = xr_2 \} \).

    Then, we can check:

    • \( R = I_1 + I_2 \) because any element \( x \in R \) can be written as \( x = xr_1 + xr_2 \).
    • \( I_1 \cap I_2 = (0) \) because for any \( x \in I_1 \cap I_2 \), we have \( x = xr_1 = xr_2 = 0 \).

    This fulfills the conditions of \( P3 \). Therefore, \( P2 \Rightarrow P3 \) is true.

  3. \( P3 \Rightarrow P4 \):

    Suppose \( R \) has ideals \( I_1, I_2 \subset R \) such that \( R = I_1 + I_2 \) and \( I_1 \cap I_2 = (0) \). By these conditions:

    • There exist non-zero elements \( a \in I_1 \cap I_2 \) and \( b \in I_2 \) such that \( a + b = 1 \). Since \( ab \in I_1 \cap I_2 = (0) \), we have \( ab = 0 \).

    Thus, there exist \( a, b \in R \) with \( a \neq 0 \) and \( b \neq 0 \) such that \( ab = 0 \). Therefore, \( P3 \Rightarrow P4 \) is true.

From the above implications, we conclude that the true statements are: \( P1 \Rightarrow P2 \), \( P2 \Rightarrow P3 \), and \( P3 \Rightarrow P4 \).

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