Splitting of prime ideals in Galois extensions

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.

$\begin{array}{ccc} O_K & \hookrightarrow & O_L \\ \downarrow & & \downarrow \\ K & \hookrightarrow & L \end{array}$

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

$pO_L = \prod_{j=1}^{g} P_j^{e_j}$

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's 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

$[L:K]=\sum_{j=1}^{g} e_j f_j.$

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 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.

The Galois situation

In the following, the extension L/K is assumed to be a Galois extension. Then the Galois group $G=\operatorname{Gal}(L/K)$ 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

$pO_L = \left(\prod_{j=1}^{g} P_j\right)^e$.

and

$[L:K]=efg.$

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_Pj| for every j, where D_Pj, 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_Pj is ef for every j.

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

The theory of the Frobenius element goes further, to identify an element of D_Pj/I_Pj 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_Pj is f and I_Pj is trivial. Also the Frobenius element is in this case an element of D_Pj (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[i] is the ring of Gaussian integers. Although this case is far from representative — after all, Z[i] 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[i]:

$(2) = (1+i)^2\ ,$

so the ramification index here is 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[i] above 2. The inertia group is also all of G, since

$a + bi \equiv a - bi \textrm{\ modulo\ }(1+i)\ ,$

for any integers a and b.

In fact, 2 is the only prime that ramifies in Z[i], since every prime that ramifies must divide the discriminant of Z[i], which is −4.

Primes p ≡ 1 mod 4

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,

$(13) = (2 + 3i)(2 - 3i)\ .$

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,

$O_L / (2 \pm 3i)O_L\ ,$

which are both isomorphic to the finite field with 13 elements. The Frobenius element is the trivial automorphism; this means that

$(a + bi)^{13} \equiv a + bi \textrm{\ modulo\ }(2 \pm 3i)$

for any integers a and b.

Primes p ≡ 3 mod 4

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

$O_L / (7)O_L\ ,$

which is the finite field with 7² = 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/7Z has order 2, and is generated by the image of the Frobenius element. The Frobenius is none other than σ; this means that

$(a + bi)^7 \equiv a - bi\textrm{\ modulo\ } 7\ ,$

for any integers a and b.

Summary

Prime in Z	How it splits in Z[i]	Inertia group	Decomposition group
2	Ramifies with index 2	G	G
p ≡ 1 mod 4	Splits into two distinct factors	1	1
p ≡ 3 mod 4	Remains inert	1	G

Computing the factorisation

Suppose that we wish to determine the factorisation of a prime ideal P of O_K into primes of O_L. The following procedure (Neukirch, p47) solves this problem in many cases. The strategy is to select an integer θ in O_L 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 O_K. Reducing the coefficients of H(X) modulo P, we obtain a monic polynomial h(X) with coefficients in F, the (finite) residue field O_K/P. Suppose that h(X) factorises in the polynomial ring F[X] as

$h(X) = h_1(X)^{e_1} \cdots h_n(X)^{e_n},$

where the h_j 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:

$P O_L = Q_1^{e_1} \cdots Q_n^{e_n},$

where the Q_j are distinct prime ideals of O_L. Furthermore, the inertia degree of each Q_j is equal to the degree of the corresponding polynomial h_j, and there is an explicit formula for the Q_j:

$Q_j = P O_L + h_j(\theta) O_L,$

where h_j denotes here a lifting of the polynomial h_j to K[X].

In the Galois case, the inertia degrees are all equal, and the ramification indices e₁ = ... = e_n 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 O_K[θ]. The conductor is defined to be the ideal

$\{ y \in O_L : yO_L \subseteq O_K[\theta]\};$

it measures how far the order O_K[θ] is from being the whole ring of integers (maximal order) O_L.

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 of the Gaussian integers. We take θ to be the imaginary unit i, with minimal polynomial H(X) = X² + 1. Since Z[$i$] is the whole ring of integers of Q($i$), 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:

$X^2 + 1 = (X+1)^2 \pmod 2.$

Therefore, there is only one prime factor, with inertia degree 1 and ramification index 2, and it is given by

$Q = (2)\mathbf Z[i] + (i+1)\mathbf Z[i] = (1+i)\mathbf Z[i].$

The next case is for P = (p) for a prime p ≡ 3 mod 4. For concreteness we will take P = (7). The polynomial X² + 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

$Q = (7)\mathbf Z[i] + (i^2 + 1)\mathbf Z[i] = 7\mathbf Z[i].$

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

$X^2 + 1 = (X + 5)(X - 5) \pmod{13}.$

Therefore, there are two prime factors, both with inertia degree and ramification index 1. They are given by

$Q_1 = (13)\mathbf Z[i] + (i + 5)\mathbf Z[i] = \cdots = (2+3i)\mathbf Z[i]$

and

$Q_2 = (13)\mathbf Z[i] + (i - 5)\mathbf Z[i] = \cdots = (2-3i)\mathbf Z[i].$

External links

• Splitting and ramification in number fields and Galois extensions at PlanetMath.org.