Question

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?

• A. \( P1 \Rightarrow P2 \)
• B. \( P2 \Rightarrow P3 \)
• C. \( P3 \Rightarrow P4 \)
• D. \( P4 \Rightarrow P1 \)

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.

Answer 1

The Correct Option is A, B, C

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 \( P3 \Rightarrow P4 \).