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| In [[mathematics]], the interplay between the [[Galois group]] ''G'' of a [[Galois extension]] ''L'' of a [[number field]] ''K'', and the way the [[prime ideal]]s ''P'' of the [[ring of integers]] ''O''<sub>''K''</sub> factorise as products of prime ideals of ''O''<sub>''L''</sub>, 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 covering]]s of [[Riemann surface]]s, which is simpler in that only one kind of subgroup of ''G'' need be considered, rather than two. This was certainly familiar before Hilbert.
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| == Definitions ==
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| Let ''L'' / ''K'' be a finite extension of number fields, and let ''B'' and ''A'' be the corresponding [[ring of integers]] of ''L'' and ''K'', respectively, which are defined to be the [[integral closure]] of the integers '''Z''' in the field in question.
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| : <math> \begin{array}{ccc} A & \hookrightarrow & B \\ \downarrow & & \downarrow \\ K & \hookrightarrow & L \end{array} </math>
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| Finally, let ''p'' be a non-zero prime ideal in ''A'', or equivalently, a [[maximal ideal]], so that the residue ''A''/''p'' is a [[field (mathematics)|field]].
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| From the basic theory of one-[[Krull dimension|dimensional]] rings follows the existence of a unique decomposition
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| : <math> pB = \prod_{j} P_j^{e(j)} </math>
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| <!-- :''pB'' = Π ''P''<sub>''j''</sub><sup>''e''(''j'')</sup> -->
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| of the ideal ''pB'' generated in ''B'' by ''p'' into a product of distinct maximal ideals ''P''<sub>''j''</sub>, with multiplicities ''e''(''j'').
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| The multiplicity ''e''(''j'') are called '''ramification indices''' of the extension at ''p''. If they are all equal to 1 and if in addition the field extensions ''B''/''P''<sub>''j''</sub> over ''A''/''p'' is [[separable extension|separable]], the field extension ''L''/''K'' is called '''unramified at ''p'''''.
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| If this is the case, by the [[Chinese remainder theorem]], the quotient
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| ::''B''/''pB''
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| :is a product of fields
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| ::''F''<sub>''j''</sub> = ''B''/''P''<sub>''j''</sub>.
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| ===The Galois situation===
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| In the following, the extension ''L'' / ''K'' is assumed to be a [[Galois extension]]. Then the [[Galois group]] ''G'' [[transitive group action|acts transitively]] on the ''P''<sub>''j''</sub>. That is, the prime ideal factors of ''p'' in ''L'' form a single [[orbit (group theory)|orbit]] under the [[automorphism]]s of ''L'' over ''K''. From this and the [[unique prime factorisation|unicity of prime factorisation]], it follows that ''e''(''j'') = ''e'' is independent of ''j''; something that certainly need not be the case for extensions that are not Galois.
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| The basic relation then reads
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| :''pB'' = (Π ''P''<sub>''j''</sub>)<sup>''e''</sup>
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| ===Facts===
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| *Given an extension as above, it is unramified in all but finitely many points.
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| *In the unramified case, because of the transitivity of the Galois group action, the fields ''F''<sub>''j''</sub> introduced above are all isomorphic, say to the finite field ''F ′'', containing
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| ::''F'' = ''A''/''p''
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| :A counting argument shows that
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| ::[''L'':''K'']/[''F ′'':''F'']
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| :equals the number of prime factors of ''P'' in ''B''. By the [[orbit-stabilizer formula]] this number is also equal to
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| ::|''G''|/|''D''|
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| :where by definition ''D'', the '''decomposition group''' of ''p'', is the subgroup of elements of ''G'' sending a given ''P''<sub>''j''</sub> to itself. That is, since the degree of ''L''/''K'' and the order of ''G'' are equal by basic Galois theory, the order of the decomposition group ''D'' is the degree of the '''residue field extension''' ''F ′''/''F''. The theory of the [[Frobenius element]] goes further, to identify an element of ''D'', for ''j'' given, which generates the Galois group of the finite field extension.
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| *In the ramified case, there is the further phenomenon of ''inertia'': the index ''e'' is interpreted as the extent to which elements of ''G'' are not seen in the Galois groups of any of the residue field extensions. Each decomposition group ''D'', for a given ''P''<sub>''j''</sub>, contains an '''inertia group''' ''I'' consisting of the ''g'' in ''G'' that send ''P''<sub>''j''</sub> to itself, but induce the identity automorphism on
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| ::''F''<sub>''j''</sub> = ''B''/''P''<sub>''j''</sub>.
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| In the geometric analogue, for [[complex manifold]]s 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 [[preimage]]s.
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| 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 field]]s usually are 'regulated' by a degree 6 field containing them.
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| == Example — the Gaussian integers ==
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| 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''<sub>''K''</sub> is simply '''Z''', and ''O''<sub>''L''</sub> = '''Z'''[i] is the ring of [[Gaussian integers]]. Although this case is far from representative — after all, '''Z'''[i] has [[unique factorisation]] — it exhibits many of the features of the theory.
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| 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.
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| === The prime ''p'' = 2 ===
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| The prime 2 of '''Z''' ramifies in '''Z'''[i]:
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| : (2) = (1+i)<sup>2</sup>,
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| so the ramification index here is ''e'' = 2. The residue field is
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| : ''O''<sub>''L''</sub> / (1+i)''O''<sub>''L''</sub> | |
| 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'''[i] above 2. The inertia group is also all of ''G'', since
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| : ''a'' + ''bi'' ≡ ''a'' − ''bi''
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| modulo (1+i), for any integers ''a'' and ''b''.
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| In fact, 2 is the ''only'' prime that ramifies in '''Z'''[i], since every prime that ramifies must divide the [[discriminant of an algebraic number field|discriminant]] of '''Z'''[i], which is −4.
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| === Primes ''p'' ≡ 1 mod 4 ===
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| Any prime ''p'' ≡ 1 mod 4 ''splits'' into two distinct prime ideals in '''Z'''[i]; this is a manifestation of [[Fermat's theorem on sums of two squares]]. For example,
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| : (13) = (2 + 3i)(2 − 3i).
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| The decomposition groups in this case are both the trivial group {1}; indeed the automorphism σ ''switches'' the two primes (2 + 3i) and (2 − 3i), so it cannot be in the decomposition group of either prime. The inertia group, being a subgroup of the decomposition group, is also the trivial group. There are two residue fields, one for each prime,
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| : ''O''<sub>''L''</sub> / (2 ± 3i)''O''<sub>''L''</sub>,
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| which are both isomorphic to the finite field with 13 elements. The Frobenius element is the trivial automorphism; this means that
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| : (''a'' + ''bi'')<sup>13</sup> ≡ ''a'' + ''bi''
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| modulo (2 ± 3i), for any integers ''a'' and ''b''.
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| === Primes ''p'' ≡ 3 mod 4 ===
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| Any prime ''p'' ≡ 3 mod 4 remains ''inert'' in '''Z'''[i]; that is, it does ''not'' split. For example, (7) remains prime in '''Z'''[i]. In this situation, the decomposition group is all of ''G'', again because there is only one prime factor. However, this situation differs from the ''p'' = 2 case, because now σ does ''not'' act trivially on the residue field
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| : ''O''<sub>''L''</sub> / (7)''O''<sub>''L''</sub>,
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| which is the finite field with 7<sup>2</sup> = 49 elements. For example, the difference between 1 + i 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'''/7'''Z''' has order 2, and is generated by the image of the Frobenius element. The Frobenius is none other than σ; this means that
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| : (''a'' + ''bi'')<sup>7</sup> ≡ ''a'' − ''bi''
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| modulo 7, for any integers ''a'' and ''b''.
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| === Summary ===
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| {| class="wikitable"
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| |-
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| ! Prime in '''Z'''
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| ! How it splits in '''Z'''[i]
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| ! Inertia group
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| ! Decomposition group
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| |-
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| | 2
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| | Ramifies with index 2
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| | ''G''
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| | ''G''
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| |-
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| | p ≡ 1 mod 4
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| | Splits into two distinct factors
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| | 1
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| | 1
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| |-
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| | p ≡ 3 mod 4
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| | Remains inert
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| | 1
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| | ''G''
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| |}
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| == Computing the factorisation ==
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| Suppose that we wish to determine the factorisation of a prime ideal ''P'' of ''O''<sub>''K''</sub> into primes of ''O''<sub>''L''</sub>. We will assume that the extension ''L''/''K'' is a finite ''separable'' extension; the extra hypothesis of normality in the definition of Galois extension is not necessary.
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| The following procedure (Neukirch, p47) solves this problem in many cases. The strategy is to select an integer θ in ''O''<sub>''L''</sub> 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 (field theory)|minimal polynomial]] ''H''(''X'') of θ over ''K''; it is a monic polynomial with coefficients in ''O''<sub>''K''</sub>. Reducing the coefficients of ''H''(''X'') modulo ''P'', we obtain a monic polynomial ''h''(''X'') with coefficients in ''F'', the (finite) residue field ''O''<sub>''K''</sub>/''P''. Suppose that ''h''(''X'') factorises in the polynomial ring ''F''[''X''] as
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| : <math> h(X) = h_1(X)^{e_1} \cdots h_n(X)^{e_n}, </math>
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| where the ''h''<sub>''j''</sub> 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:
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| : <math> P O_L = Q_1^{e_1} \cdots Q_n^{e_n}, </math>
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| where the ''Q''<sub>''j''</sub> are distinct prime ideals of ''O''<sub>''L''</sub>. Furthermore, the inertia degree of each ''Q''<sub>''j''</sub> is equal to the degree of the corresponding polynomial ''h''<sub>''j''</sub>, and there is an explicit formula for the ''Q''<sub>''j''</sub>:
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| : <math> Q_j = P O_L + h_j(\theta) O_L. </math>
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| In the Galois case, the inertia degrees are all equal, and the ramification indices ''e''<sub>1</sub> = ... = ''e''<sub>''n''</sub> are all equal.
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| The exceptional primes, for which the above result does not necessarily hold, are the ones not relatively prime to the [[conductor (ring theory)|conductor]] of the ring ''O''<sub>''K''</sub>[θ]. The conductor is defined to be the ideal
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| : <math> \{ y \in O_L : yO_L \subseteq O_K[\theta]\}; </math>
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| it measures how far the [[order (ring theory)|order]] ''O''<sub>''K''</sub>[θ] is from being the whole ring of integers (maximal order) ''O''<sub>''L''</sub>.
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| 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 <ref>http://modular.math.washington.edu/papers/undergrad/decomp/decomp/node4.html</ref>). Therefore the algorithm given above cannot be used to factor such ''P'', and more sophisticated approaches must be used, such as that described in.<ref>http://modular.math.washington.edu/papers/undergrad/decomp/decomp/node3.html</ref>
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| === An example ===
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| Consider again the case of the Gaussian integers. We take θ to be the imaginary unit ''i'', with minimal polynomial ''H''(''X'') = ''X''<sup>2</sup> + 1. Since '''Z'''[<math>i</math>] is the whole ring of integers of '''Q'''(<math>i</math>), the conductor is the unit ideal, so there are no exceptional primes.
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| For ''P'' = (2), we need to work in the field '''Z'''/(2)'''Z''', which amounts to factorising the polynomial ''X''<sup>2</sup> + 1 modulo 2:
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| : <math>X^2 + 1 = (X+1)^2 \pmod 2.</math>
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| Therefore there is only one prime factor, with inertia degree 1 and ramification index 2, and it is given by
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| : <math>Q = (2)\mathbf Z[i] + (i+1)\mathbf Z[i] = (1+i)\mathbf Z[i].</math>
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| The next case is for ''P'' = (''p'') for a prime ''p'' ≡ 3 mod 4. For concreteness we will take ''P'' = (7). The polynomial ''X''<sup>2</sup> + 1 is irreducible modulo 7. Therefore there is only one prime factor, with inertia degree 2 and ramification index 1, and it is given by
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| : <math>Q = (7)\mathbf Z[i] + (i^2 + 1)\mathbf Z[i] = 7\mathbf Z[i].</math>
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| The last case is ''P'' = (''p'') for a prime ''p'' ≡ 1 mod 4; we will again take ''P'' = (13). This time we have the factorisation
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| : <math>X^2 + 1 = (X + 5)(X - 5) \pmod{13}.</math>
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| Therefore there are ''two'' prime factors, both with inertia degree and ramification index 1. They are given by
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| : <math>Q_1 = (13)\mathbf Z[i] + (i + 5)\mathbf Z[i] = \cdots = (2+3i)\mathbf Z[i]</math>
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| and
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| : <math>Q_2 = (13)\mathbf Z[i] + (i - 5)\mathbf Z[i] = \cdots = (2-3i)\mathbf Z[i].</math>
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| == External links ==
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| * {{planetmath_reference|id=6818|title=Splitting and ramification in number fields and Galois extensions}}
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| * {{Citation| url=http://modular.math.washington.edu/129/ant/ | author=William Stein | title=A brief introduction to classical and adelic algebraic number theory}}
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| == References ==
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| {{Reflist}}
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| * {{Neukirch ANT}}
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| {{DEFAULTSORT:Splitting Of Prime Ideals In Galois Extensions}}
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| [[Category:Algebraic number theory]]
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| [[Category:Galois theory]]
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