In mathematics, the interplay between the Galois group G of a Galois extension L of a number field K, and the way the prime ideals P of the ring of integers OK factorise as products of prime ideals of OL, provides one of the richest parts of algebraic number theory.
There is a geometric analogue, for ramified coverings of Riemann surfaces, which is simpler in that only one kind of subgroup of G need be considered, rather than two.
Let L/K be a finite extension of number fields, and let OK and OL be the corresponding ring of integers of K and L, respectively, which are defined to be the integral closure of the integers Z in the field in question.
From the basic theory of one-dimensional rings follows the existence of a unique decomposition of the ideal pOL generated in OL by p into a product of distinct maximal ideals Pj, with multiplicities ej.
The field F = OK/p naturally embeds into Fj = OL/Pj for every j, the degree fj = [OL/Pj : OK/p] of this residue field extension is called inertia degree of Pj over p. The multiplicity ej is called ramification index of Pj over p. If it is bigger than 1 for some j, the field extension L/K is called ramified at p (or we say that p ramifies in L, or that it is ramified in L).
Otherwise, L/K is called unramified at p. If this is the case then by the Chinese remainder theorem the quotient OL/pOL is a product of fields Fj.
Multiplicativity of ideal norm implies If fj = ej = 1 for every j (and thus g = [L : K]), we say that p splits completely in L. If g = 1 and f1 = 1 (and so e1 = [L : K]), we say that p ramifies completely in L. Finally, if g = 1 and e1 = 1 (and so f1 = [L : K]), we say that p is inert in L. In the following, the extension L/K is assumed to be a Galois extension.
Then the prime avoidance lemma can be used to show the Galois group
That is, the prime ideal factors of p in L form a single orbit under the automorphisms of L over K. From this and the unique factorisation theorem, it follows that f = fj and e = ej are independent of j; something that certainly need not be the case for extensions that are not Galois.
This decomposition group contains a subgroup IPj, called inertia group of Pj, consisting of automorphisms of L/K that induce the identity automorphism on Fj.
is isomorphic to DPj/IPj and the order of the inertia group IPj is e. The theory of the Frobenius element goes further, to identify an element of DPj/IPj for given j which corresponds to the Frobenius automorphism in the Galois group of the finite field extension Fj /F.
In the unramified case the order of DPj is f and IPj is trivial, so the Frobenius element is in this case an element of DPj, and thus also an element of G. For varying j, the groups DPj are conjugate subgroups inside G: Recalling that G acts transitively on the Pj, one checks that if
Therefore, if G is an abelian group, the Frobenius element of an unramified prime P does not depend on which Pj we take.
Furthermore, in the abelian case, associating an unramified prime of K to its Frobenius and extending multiplicatively defines a homomorphism from the group of unramified ideals of K into G. This map, known as the Artin map, is a crucial ingredient of class field theory, which studies the finite abelian extensions of a given number field K.[1] In the geometric analogue, for complex manifolds or algebraic geometry over an algebraically closed field, the concepts of decomposition group and inertia group coincide.
There, given a Galois ramified cover, all but finitely many points have the same number of preimages.
This section describes the splitting of prime ideals in the field extension Q(i)/Q.
Although this case is far from representative — after all, Z[i] has unique factorisation, and there aren't many imaginary quadratic fields with unique factorization — it exhibits many of the features of the theory.
Writing G for the Galois group of Q(i)/Q, and σ for the complex conjugation automorphism in G, there are three cases to consider.
The Frobenius element is the trivial automorphism; this means that for any integers a and b.
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 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.
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.
), the conductor is the unit ideal, so there are no exceptional primes.
For P = (2), we need to work in the field Z/(2)Z, which amounts to factorising the polynomial X2 + 1 modulo 2: Therefore, there is only one prime factor, with inertia degree 1 and ramification index 2, and it is given by The next case is for P = (p) for a prime p ≡ 3 mod 4.
This time we have the factorisation Therefore, there are two prime factors, both with inertia degree and ramification index 1.