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We now specialise the results of the previous section to the case of extensions of of degree 2. Such a field is of the form [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 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() = b and Nm() = c. Moreover, Trace() = Trace(b - c) = b2 - 2c. Thus the discriminant DR of R is the determinant of which is b2 - 4c (as expected). In particular, we see that DR = b2(mod 4); i. e. the discriminant must be 0 or 1 modulo 4. In the first case, we can replace by + (b - 1)/2 so that we get an element with trace 1. In the second case, we can replace by + 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 by = (DR + )/2 in both cases; this can be done since DR + b is even in both cases. We thus have a natural basis for R. There is also a natural involution on R which sends to - or equivalently to DR - .

Subsections

Next: 7.1 Prime ideals Up: Some Lectures on Number Previous: 6.7 Prime ideals
Kapil Hari Paranjape 2002-10-20