Verify that the set of complex numbers of the form \(x + y\sqrt{2}\), where \(x\) and \(y\) are rational, is a subfield of \(\CC\).
Solution
1. Definition and identities:
1.1. Let \(F = \{x + y\sqrt{2} \mid x, y \in \mathbb{Q}\} \subset \mathbb{C}\).
definition of subset.
1.2. \(0 = 0 + 0\sqrt{2} \in F\) and \(1 = 1 + 0\sqrt{2} \in F\) since \(0, 1 \in \mathbb{Q}\).Thus \(F\) contains the additive and multiplicative identities.
\(0, 1 \in \mathbb{Q}\).
2. Closure under subtraction and multiplication:
2.1. Let \(\alpha = x_1 + y_1\sqrt{2} \in F\) and \(\beta = x_2 + y_2\sqrt{2} \in F\) with \(x_1, y_1, x_2, y_2 \in \mathbb{Q}\).
setup.
2.2. The difference is:\[ \alpha - \beta = (x_1 - x_2) + (y_1 - y_2)\sqrt{2}. \]Since \(\mathbb{Q}\) is a field, \(x_1 - x_2 \in \mathbb{Q}\) and \(y_1 - y_2 \in \mathbb{Q}\), so \(\alpha - \beta \in F\).
closure of \(\mathbb{Q}\) under subtraction.
2.3. The product is:\[ \alpha \beta = (x_1 + y_1\sqrt{2})(x_2 + y_2\sqrt{2}) = (x_1 x_2 + 2 y_1 y_2) + (x_1 y_2 + x_2 y_1)\sqrt{2}. \]Since \(x_1 x_2 + 2 y_1 y_2 \in \mathbb{Q}\) and \(x_1 y_2 + x_2 y_1 \in \mathbb{Q}\), we have \(\alpha \beta \in F\).
ring operations in \(\mathbb{Q}\) and \((\sqrt{2})^2 = 2 \in \mathbb{Q}\).
3. Existence of multiplicative inverses for non-zero elements:
3.1. Let \(\alpha = x + y\sqrt{2} \in F \setminus \{0\}\).
non-zero element setup.
3.2. We claim \(x^2 - 2y^2 \neq 0\):If \(y = 0\), then \(x \neq 0\), so \(x^2 - 2y^2 = x^2 \neq 0\).If \(y \neq 0\), then \(x^2 - 2y^2 = 0 \implies \sqrt{2} = |x/y| \in \mathbb{Q}\), contradicting the irrationality of \(\sqrt{2}\).Thus \(x^2 - 2y^2 \in \mathbb{Q} \setminus \{0\}\).
irrationality of \(\sqrt{2}\).
3.3. Multiply numerator and denominator by the algebraic conjugate \(x - y\sqrt{2}\):\[ \frac{1}{\alpha} = \frac{1}{x + y\sqrt{2}} = \frac{x - y\sqrt{2}}{(x + y\sqrt{2})(x - y\sqrt{2})} = \left(\frac{x}{x^2 - 2y^2}\right) + \left(\frac{-y}{x^2 - 2y^2}\right)\sqrt{2}. \]Since \(\frac{x}{x^2 - 2y^2} \in \mathbb{Q}\) and \(\frac{-y}{x^2 - 2y^2} \in \mathbb{Q}\), \(\alpha^{-1} \in F\).
field axioms of \(\mathbb{Q}\).
4. Conclusion:\(F\) contains 1, is closed under subtraction, multiplication, and taking inverses of non-zero elements; hence \(F = \mathbb{Q}(\sqrt{2})\) is a subfield of \(\mathbb{C}\). Q.E.D.
1 through 3 (Subfield Criterion).