19.
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 \(r_{1}^{2} = r_{1} \neq 0 \neq r_{2} = r_{2}^{2}\), \(r_{1}r_{2} = 0\) and \(r_{1} + r_{2} = 1\).
P3: \(\exists\) ideals \(I_{1}, I_{2} \subseteq R\) with \(R \neq I_{1} \neq (0) \neq I_{2} \neq R\) such that \(R = I_{1} + I_{2}\) and \(I_{1} \cap I_{2} = (0)\).
P4: \(\exists\, a, b \in R\) with \(a \neq 0 \neq b\) such that \(ab = 0\).
Then, which of the following is/are TRUE?
More than one option may be correct
All 3 correct options found.
Exam Year(s)
GATE Mathematics - 2025,
Answer: A, B, C
Options A, B and C are correct; the key accepts all three.
A: if \(R \cong R_{1} \times R_{2}\) with both factors non-zero rings, then \(r_{1} = (1, 0)\) and \(r_{2} = (0, 1)\) satisfy \(r_{i}^{2} = r_{i} \ne 0\), \(r_{1}r_{2} = 0\) and \(r_{1} + r_{2} = 1\). So P1 implies P2.
B: given such idempotents, put \(I_{1} = r_{1}R\) and \(I_{2} = r_{2}R\). They sum to \(R\) since \(1 = r_{1} + r_{2}\). If \(x = r_{1}a = r_{2}b\), then \(x = r_{1}x\) (as \(r_{1}^{2} = r_{1}\)) \(= r_{1}r_{2}b = 0\), so \(I_{1} \cap I_{2} = (0)\). Both are non-zero (they contain \(r_{1}\), \(r_{2}\)) and proper (if \(I_{1} = R\) then \(r_{1}a = 1\) for some \(a\), so \(r_{1}\) is a unit, and \(r_{1}r_{2} = 0\) would force \(r_{2} = 0\); the same argument works for \(I_{2}\)). So P2 implies P3.
C: take non-zero \(a \in I_{1}\) and \(b \in I_{2}\). Then \(ab \in I_{1} \cap I_{2} = (0)\), so \(a\) and \(b\) are zero divisors. So P3 implies P4.
D: false. In \(\mathbb{Z}/4\), \(2 \cdot 2 = 0\) with \(2 \ne 0\), so P4 holds, but the ring has a unique maximal ideal \((2)\), is local, and has no non-trivial idempotent, so it is not a product of two non-zero rings.
