Discussion
Let \(n\) be a square free natural number. Then the quotient ring \(Q[X]/\lt X^{2} - n\gt\) is :
Answer: D
The keyed answer is an integral domain.
For a square-free \(n \gt 1\), \(\sqrt{n}\) is irrational, so \(X^{2} - n\) has no rational root and, being of degree 2, is irreducible over \(Q\). The ideal it generates is then maximal, and \(Q[X]/\langle X^{2} - n \rangle \cong Q(\sqrt{n})\) is a field. Every field is an integral domain, so (D) holds, while 'not a field' and 'an integral domain which is not a field' are false.
The key assumes \(n \gt 1\): for \(n = 1\), \(X^{2} - 1 = (X - 1)(X + 1)\) and the quotient has zero divisors, which is the only case in which 'not an integral domain' would apply.
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