decomposed prime

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 ''O''_''K'' factorise as products of prime ideals of ''O''_''L'', provides one of the richest parts of algebraic number theory. The splitting of prime ideals in Galois extensions is sometimes attributed to David Hilbert by calling it Hilbert 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. This was certainly familiar before Hilbert.

Definitions

Let ''L''/''K'' be a finite extension of number fields, and let ''O_K'' and ''O_L'' 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.
: $\begin O_K & \hookrightarrow & O_L \\ \downarrow & & \downarrow \\ K & \hookrightarrow & L \end$
Finally, let ''p'' be a non-zero prime ideal in ''O_K'', or equivalently, a maximal ideal, so that the residue ''O_K''/''p'' is a field.

From the basic theory of one- dimensional rings follows the existence of a unique decomposition
: $pO_L = \prod_^ P_j^$

of the ideal ''pO_L'' generated in ''O_L'' by ''p'' into a product of distinct maximal ideals ''P''_''j'', with multiplicities ''e''_''j''.

The field ''F'' = ''O_K''/''p'' naturally embeds into ''F''_''j'' = ''O_L''/''P''_''j'' for every ''j'', the degree ''f''_''j'' = 'O_L''/''P''_''j'' : ''O_K''/''p''of this residue field extension is called inertia degree of ''P''_''j'' over ''p''.

The multiplicity ''e''_''j'' is called ramification index of ''P''_''j'' 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 ''O_L''/''pO_L'' is a product of fields ''F''_''j''. The extension ''L''/''K'' is ramified in exactly those primes that divide the relative discriminant, hence the extension is unramified in all but finitely many prime ideals.

Multiplicativity of ideal norm implies
: :K\sum_^ e_j f_j.

If ''f''_''j'' = ''e''_''j'' = 1 for every ''j'' (and thus ''g'' = 'L'' : ''K'', we say that ''p'' splits completely in ''L''. If ''g'' = 1 and ''f''_''1'' = 1 (and so ''e''_''1'' = 'L'' : ''K'', we say that ''p'' ramifies completely in ''L''. Finally, if ''g'' = 1 and ''e''_''1'' = 1 (and so ''f''_''1'' = 'L'' : ''K'', we say that ''p'' is inert in ''L''.

The Galois situation

In the following, the extension ''L''/''K'' is assumed to be a Galois extension. Then the Galois group $G=\operatorname(L/K)$ acts transitively on the ''P''_''j''. 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'' = ''f''_''j'' and ''e'' = ''e''_''j'' are independent of ''j''; something that certainly need not be the case for extensions that are not Galois. The basic relations then read
: $pO_L = \left(\prod_^ P_j\right)^e$.
and
: :Kefg.
The relation above shows that 'L'' : ''K''''ef'' equals the number ''g'' of prime factors of ''p'' in ''O_L''. By the orbit-stabilizer formula this number is also equal to , ''G'', /, ''D''_''P''''j'', for every ''j'', where ''D''_''P''''j'', the decomposition group of ''P''_''j'', is the subgroup of elements of ''G'' sending a given ''P''_''j'' to itself. Since the degree of ''L''/''K'' and the order of ''G'' are equal by basic Galois theory, it follows that the order of the decomposition group ''D''_''P''''j'' is ''ef'' for every ''j''.

This decomposition group contains a subgroup ''I''_''P''''j'', called inertia group of ''P''_''j'', consisting of automorphisms of ''L''/''K'' that induce the identity automorphism on ''F''_''j''. In other words, ''I''_''P''''j'' is the kernel of reduction map $D_\to\operatorname(F_j/F)$. It can be shown that this map is surjective, and it follows that $\operatorname(F_j/F)$ is isomorphic to ''D''_''P''''j''/''I''_''P''''j'' and the order of the inertia group ''I''_''P''''j'' is ''e''.

The theory of the Frobenius element goes further, to identify an element of ''D''_''P''''j''/''I''_''P''''j'' for given ''j'' which corresponds to the Frobenius automorphism in the Galois group of the finite field extension ''F''_''j'' /''F''. In the unramified case the order of ''D''_''P''''j'' is ''f'' and ''I''_''P''''j'' is trivial. Also the Frobenius element is in this case an element of ''D''_''P''''j'' (and thus also element of ''G'').

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.

The splitting of primes in extensions that are not Galois may be studied by using a splitting field initially, i.e. a Galois extension that is somewhat larger. For example, cubic fields usually are 'regulated' by a degree 6 field containing them.

Example — the Gaussian integers

This section describes the splitting of prime ideals in the field extension Q(i)/Q. That is, we take ''K'' = Q and ''L'' = Q(i), so ''O''_''K'' is simply Z, and ''O''_''L'' = Z is the ring of Gaussian integers. Although this case is far from representative — after all, Z has unique factorisation, and there aren't many 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 prime ''p'' = 2

The prime 2 of Z ramifies in Z
:$(2)=(1+i)^2$
The ramification index here is therefore ''e'' = 2. The residue field is
:$O_L / (1+i)O_L$
which is the finite field with two elements. The decomposition group must be equal to all of ''G'', since there is only one prime of Z above 2. The inertia group is also all of ''G'', since
:$a+bi\equiv a-bi\bmod1+i$
for any integers ''a'' and ''b'', as $a+bi = 2bi + a-bi =(1+i) \cdot (1-i)bi + a-bi \equiv a-bi \bmod 1+i$ .

In fact, 2 is the ''only'' prime that ramifies in Z since every prime that ramifies must divide the discriminant