Discussion
Consider the following conditions on two proper non-zero ideals \(J_{1}\) and \(J_{2}\) of a non-zero commutative ring R.
P: For any \(r_{1}, r_{2} \in R\), there exists a unique \(r \in R\) such that \(r - r_{1} \in J_{1}\) and \(r - r_{2} \in J_{2}\).
Q: \(J_{1} + J_{2} = R\)
Then, which of the following statements is TRUE?
Answer: A
The answer is P implies Q but Q does not imply P.
P implies Q: take \(r_{1} = 1\) and \(r_{2} = 0\). P gives an r with \(r - 1 \in J_{1}\) and \(r \in J_{2}\). Then \(1 = r - (r - 1) \in J_{2} + J_{1}\), so \(J_{1} + J_{2} = R\).
Q does not imply P: Q gives existence through the Chinese remainder theorem, but not uniqueness. In \(\mathbb{Z}\) with \(J_{1} = 2\mathbb{Z}\) and \(J_{2} = 3\mathbb{Z}\), the ideals are comaximal, yet if r solves the two congruences so does r + 6. Uniqueness in P actually forces \(J_{1} \cap J_{2} = 0\), which Q does not give.
So the options saying Q implies P, or that neither implies the other, are wrong.
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