Bourbaki

Exercise 8, § 1

Algebra, VII, p. 50

Let K be a quadratic field, that is an extension of Q\mathbf{Q} of degree 2. a) Show that K=Q(e)K = \mathbf{Q}(e) where e2e^2 is a rational integer d≠1d \neq 1 not divisible by a square >1> 1 (in ZZ). b) If a=a+bea = a + be (a,b∈Qa, b \in \mathbf{Q}), let αˉ\bar{\alpha} denote the conjugate of a+bea + be. We say that aa is an integer of the quadratic field K if its norm ααˉ\alpha \bar{\alpha} and its trace a+αˉa + \bar{\alpha} are rational integers. Show that the integers of K form a ring A (notice that if α\alpha and β\beta are integers of K then the sum and product of the rational numbers αβ+αβ‾\alpha \beta + \overline{\alpha \beta} and αβ‾+αβ‾\alpha \overline{\beta} + \overline{\alpha \beta} are integers, hence that these rational numbers are the roots of a monic polynomial over ZZ; deduce that they are rational integers). c) Show that the integers of K=Q(e)K = \mathbf{Q}(e) form a ZZ-module with basis (1,e)(1, e) if d−1≢0(mod4)d - 1 \not\equiv 0 \pmod{4}, or (1/2(1+e),1/2(1−e))(1/2(1 + e), 1/2(1 - e)) if d−1≡0(mod4)d - 1 \equiv 0 \pmod{4} (if a+bea + be is an integer of K, check that 2a2a and 2b2b are rational integers such that (2a)2−d(2b)2≡0(4)(2a)^2 - d(2b)^2 \equiv 0(4), and investigate whether they can be odd). Show that the discriminants of these bases (III, p. 549, Def. 3) are 4d4d and dd respectively.

Solution

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Transcribed from the printed volume. Not translated.