Discussion
Which of the following statements is false for \(\mathbf{Q}[\sqrt{2}]\) and \(\mathbf{Z}[\sqrt{2}]\) ?
Answer: B
The false statement is \(\mathbf{Q}[\sqrt{2}]\) is not a subfield of complex numbers.
\(\mathbf{Q}[\sqrt{2}] = \{a + b\sqrt{2} : a, b \in \mathbf{Q}\}\) is a set of real numbers closed under the field operations, with \((a + b\sqrt{2})^{-1} = (a - b\sqrt{2})/(a^{2} - 2b^{2})\). It is a subfield of \(\mathbf{R}\) and so of \(\mathbf{C}\).
The other statements are true. \(1 + \sqrt{2}\) is a unit of \(\mathbf{Z}[\sqrt{2}]\) (its inverse is \(\sqrt{2} - 1\)), and its powers are all distinct units (A). Since \(x^{2} - 2\) is irreducible over Q, evaluating at \(\sqrt{2}\) gives \(\mathbf{Q}[x]/(x^{2} - 2) \cong \mathbf{Q}[\sqrt{2}]\) (C). And \(\mathbf{Z}[\sqrt{2}]\) has infinitely many primes, for example a prime factor of each rational prime (D).
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