and σ(1 + i) = 1 − i is 2i, which is certainly not divisible by 7. Therefore, the inertia group is the trivial group {1}. The Galois group of this residue field over the subfield Z/7Z has order 2, and is generated by the image of the Frobenius element. The Frobenius is none other than σ; this means that for any integers a and b. Summary Computing the factorisation Suppose that we wish to determine the factorisation of a prime ideal P of OK into primes of OL. The following procedure (Neukirch, p. 47) solves this problem in many cases. The strategy is to select an integer θ in OL so that L is generated over K by θ (such a θ is guaranteed to exist by the primitive element theorem), and then to examine the minimal polynomial H(X) of θ over K; it is a monic polynomial with coefficients in OK. Reducing the coefficients of H(X) modulo P, we obtain a monic polynomial h(X) with coefficients in F, the (finite) residue field OK/P. Suppose that h(X) factorises in the polynomial ring F[X] as where the hj are distinct monic irreducible polynomials in F[X]. Then, as long as P is not one of finitely many exceptional primes (the precise condition is described below), the factorisation of P has the following form: where the Qj are distinct prime ideals of OL. Furthermore, the inertia degree of each Qj is equal to the degree of the corresponding polynomial hj, and there is an explicit formula for the Qj: where hj denotes here a lifting of the polynomial hj to K[X]. In the Galois case, the inertia degrees are all equal, and the ramification indices e1 = ... = en are all equal. The exceptional primes, for which the above result does not necessarily hold, are the ones not relatively prime to the conductor of the ring OK[θ]. The conductor is defined to be the ideal it measures how far the order OK[θ] is from being the whole ring of integers (maximal order) OL. A significant caveat is that there exist examples of L/K and P such that there is no available θ that satisfies the above hypotheses (see for example ). Therefore, the algorithm given above cannot be used to factor such P, and more sophisticated approaches must be used, such as that described in. An example Consider again the case