7 Quadratic fields

We now specialise the results of the previous section to the case of extensions of $\mathbb {Q}$ of degree 2. Such a field is of the form $\mathbb {Q}$[T]/P(T) where P(T) is an irreducible polynomial of degree 2. An order in such a field is generated by 1 and a non-rational element $\alpha$ that satisfies an equation of the type P(T) = T² - bT + c. Thus every order has the form R[T]/P(T). Now, it is clear that Trace($\alpha$) = b and Nm($\alpha$) = c. Moreover, Trace($\alpha^{2}_{}$) = Trace(b$\alpha$ - c) = b² - 2c. Thus the discriminant D_R of R is the determinant of $\left(\vphantom{ \begin{smallmatrix}2 & b\\ b & b^2-2c\end{smallmatrix} }\right.$$\begin{smallmatrix}2 & b\\ b & b^2-2c\end{smallmatrix}$$\left.\vphantom{ \begin{smallmatrix}2 & b\\ b & b^2-2c\end{smallmatrix} }\right)$ which is b² - 4c (as expected). In particular, we see that D_R = b²(mod 4); i. e. the discriminant must be 0 or 1 modulo 4. In the first case, we can replace $\alpha$ by $\alpha$ + (b - 1)/2 so that we get an element with trace 1. In the second case, we can replace $\alpha$ by $\alpha$ + b/2 to get an element with trace 0. Thus we can assume that the equation takes the form T² - T + N in the first case and T² + N in the second case. An alternative normalisation is to replace $\alpha$ by $\omega_{D}^{}$ = (D_R + $\sqrt{D_R}$)/2 in both cases; this can be done since D_R + b is even in both cases. We thus have a natural basis for R. There is also a natural involution on R which sends $\sqrt{D_R}$ to - $\sqrt{D_R}$ or equivalently $\omega_{D}^{}$ to D_R - $\omega_{D}^{}$.