Discussion
Let \(R = \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}\) and \(I = \mathbb{Z} \times \mathbb{Z} \times \{0\}\). Then which of the following statement is correct?
Answer: B
\(I\) is a prime ideal but not a maximal ideal of \(R\).
The projection \((a, b, c) \mapsto c\) is a surjective ring homomorphism from \(R\) onto \(\mathbb{Z}\) whose kernel is exactly \(I = \mathbb{Z} \times \mathbb{Z} \times \{0\}\). So \(R/I \cong \mathbb{Z}\).
An ideal is prime exactly when the quotient is an integral domain, and maximal exactly when the quotient is a field. \(\mathbb{Z}\) is an integral domain but not a field, so \(I\) is prime and not maximal. Concretely, \(I \subsetneq \mathbb{Z} \times \mathbb{Z} \times 2\mathbb{Z} \subsetneq R\), so \(I\) is not maximal. That rules out the three other options.
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