Math 129: Algebraic Number Theory Lecture 12: Galois Extensions

William Stein

Thursday, March 15, 2004

1 Galois Extensions

Suppose K ⊂ C is a number field. Then K is Galois if every field homomorphism K → C has image K, or equivalently, # Aut(K) = [K : Q]. More generally, we have the following definition.

Definition 1.1 (Galois). An extension K/L of number fields is Galois if # Aut(K/L) = [K : L], where Aut(K/L) is the group of automorphisms of K that fix L. We write Gal(K/L) = Aut(K/L).

For example, Q is Galois (over itself), any quadratic extension K/L is Galois, since it is of the form L(\sqrt{a}), for some a ∈ L, and the nontrivial embedding is induced by \sqrt{a} ↦ −\sqrt{a}, so there is always one nontrivial automorphism. If f ∈ L[x] is an irreducible cubic polynomial, and a is a root of f, then one proves in a course in Galois theory that L(a) is Galois over L if and only if the discriminant of f is a perfect square in L. Random number fields of degree bigger than 2 are rarely Galois (I will not justify this claim further in this course).

If K/Q is a number field, then the Galois closure K^{gc} of K is the field generated by all images of K under all embeddings in C (more generally, if K/L is an extension, the Galois closure of K over L is the field generated by images of embeddings K → C that are the identity map on L). If K = Q(a), then K^{gc} is generated by each of the conjugates of a, and is hence Galois over Q, since the image under an embedding of any polynomial in the conjugates of a is again a polynomial in conjugates of a.

How much bigger can the degree of K^{gc} be as compared to the degree of K = Q(a)? There is a natural embedding of Gal(K^{gc}/Q) into the group of permutations of the conjugates of a. If there are n conjugates of a, then this is an embedding Gal(K^{gc}/Q) ↪ S_{n}, where S_{n} is the symmetric group on n symbols, which has order n!. Thus the degree of the K^{gc} over Q is a divisor of n!. Also the Galois group is a transitive subgroup of S_{n}, which constrains the possibilities further. When n = 2, we recover the fact that quadratic extensions are Galois. When n = 3, we see that the Galois closure of a cubic extension is either the cubic extension or a quadratic extension of the cubic extension. It turns out that that Galois closure of a cubic extension is obtained by adjoining the square root of the discriminant. For an extension K of degree 5, it is “frequently” the case that the Galois closure has degree 120, and in fact it is a difficult and interesting problem to find examples of degree 5 extension in which the Galois closure has degree smaller than 120 (according to MAGMA: the only possibilities for the order of a transitive proper subgroup of S_{5} are 5, 10, 20, and 60; there are five transitive subgroups of S_{5} out of the total of 19 subgroups of S_{5}).

Let n be a positive integer. Consider the field K = Q(ζ_{n}), where ζ_{n} = e^{2πi/n} is a primitive nth root of unity. If σ : K → C is an embedding, then σ(ζ_{n}) is also an nth root of unity, and the group of nth roots of unity is cyclic, so σ(ζ_{n}) = ζ_{n}^{m} for some m which is invertible modulo n. Thus K is Galois and Gal(K/Q) ↪ (Z/nZ)^{∗}. However, [K : Q] = n, so this map is an isomorphism. (Side note: Taking a p-adic limit and using the maps Gal(\overline{Q}/Q) → Gal(Q(ζ_{p^{r}})/Q), we obtain a homomorphism Gal(\overline{Q}/Q) → Z_{p}^{∗}, which is called the p-adic cyclotomic character.)

Compositums of Galois extensions are Galois. For example, the biquadratic field K = Q(\sqrt{5}, \sqrt{−1}) is a Galois extension of Q of degree 4.

Fix a number field K that is Galois over a subfield L. Then the Galois group G = Gal(K/L) acts on many of the object that we have associated to K, including:

• the integers O_{K},

• the units U_{K},

• the group of nonzero fractional ideals of O_{K},

• the class group Cl(K), and

• the set S_{𝔭} of prime ideals 𝔓 lying over a given prime 𝔭 of O_{L}.

In the next section we will be concerned with the action of Gal(K/L) on S_{𝔭}, though actions on each of the other objects, especially Cl(K), will be of further interest.

2 Decomposition of Primes

Fix a prime 𝔭 ⊂ O_{K} and write 𝔭O_{K} = 𝔓_{1}^{e_{1}} ⋯ 𝔓_{g}^{e_{g}}, so S_{𝔭} = {𝔓_{1}, …, 𝔓_{g}}.

Definition 2.1 (Residue class degree). Suppose 𝔓 is a prime of O_{K} lying over 𝔭. Then the residue class degree of 𝔓 is

f_{𝔓/𝔭} = [O_{K}/𝔓 : O_{L}/𝔭],

i.e., the degree of the extension of residue class fields.

If M/K/L is a tower of field extensions and 𝔮 is a prime of M over 𝔓, then

f_{𝔮/𝔭} = [O_{M}/𝔮 : O_{L}/𝔭] = [O_{M}/𝔮 : O_{K}/𝔓] · [O_{K}/𝔓 : O_{L}/𝔭] = f_{𝔮/𝔓} · f_{𝔓/𝔭},

so the residue class degree is multiplicative in towers.

Note that if σ ∈ Gal(K/L) and 𝔓 ∈ S_{p}, then σ induces an isomorphism of finite fields O_{K}/𝔓 → O_{K}/σ(𝔓) that fixes the common subfield O_{L}/𝔭. Thus the residue class degrees of 𝔓 and σ(𝔓) are the same. In fact, much more is true.

Theorem 2.2. Suppose K/L is a Galois extension of number fields, and let 𝔭 be a prime of O_{L}. Write 𝔭O_{K} = ∏_{i=1}^{g} 𝔓_{i}^{e_{i}}, and let f_{i} = f_{𝔓_{i}/𝔭}. Then G = Gal(K/L) acts transitively on the set S_{𝔭} of primes 𝔓_{i},

e_{1} = ⋯ = e_{g}, f_{1} = ⋯ = f_{g},

and efg = [K : L], where e is the common value of the e_{i} and f is the common value of the f_{i}.

Proof. For simplicity, we will give the proof only in the case L = Q, but the proof works in general. Suppose p ∈ Z and pO_{K} = 𝔭_{1}^{e_{1}} ⋯ 𝔭_{g}^{e_{g}}, and S = {𝔭_{1}, …, 𝔭_{g}}. We will first prove that G acts transitively on S. Let 𝔭 = 𝔭_{i} for some i. Recall that we proved long ago, using the Chinese Remainder Theorem, that there exists a ∈ 𝔭 such that (a)/𝔭 is an integral ideal that is coprime to pO_{K}. The product

I = ∏_{σ∈G} σ((a)/𝔭) = ∏_{σ∈G} \frac{(σ(a))O_{K}}{σ(𝔭)} = \frac{(Norm_{K/Q}(a))O_{K}}{∏_{σ∈G} σ(𝔭)}  (2.1)

is a nonzero integral O_{K} ideal since it is a product of nonzero integral O_{K} ideals. Since a ∈ 𝔭 we have that Norm_{K/Q}(a) ∈ 𝔭 ∩ Z = pZ. Thus the numerator of the rightmost expression in (2.1) is divisible by pO_{K}. Also, because (a)/𝔭 is coprime to pO_{K}, each σ((a)/𝔭) is coprime to pO_{K} as well. Thus I is coprime to pO_{K}. Thus the denominator of the rightmost expression in (2.1) must also be divisibly by pO_{K} in order to cancel the pO_{K} in the numerator. Thus for any i we have

∏_{j=1}^{g} 𝔭_{j}^{e_{j}} = pO_{K} | ∏_{σ∈G} σ(𝔭_{i}),

which in particular implies that G acts transitively on the 𝔭_{i}.

Choose some j and suppose that k ≠ j is another index. Because G acts transitively, there exists σ ∈ G such that σ(𝔭_{k}) = 𝔭_{j}. Applying σ to the factorization pO_{K} = ∏_{i=1}^{g} 𝔭_{i}^{e_{i}}, we see that

∏_{i=1}^{g} 𝔭_{i}^{e_{i}} = ∏_{i=1}^{g} σ(𝔭_{i})^{e_{i}}.

Taking ord_{𝔭_{j}} on both sides we get e_{j} = e_{k}. Thus e_{1} = e_{2} = ⋯ = e_{g}.

As was mentioned right before the statement of the theorem, for any σ ∈ G we have O_{K}/𝔭_{i} ≅ O_{K}/σ(𝔭_{i}), so by transitivity f_{1} = f_{2} = ⋯ = f_{g}. Since O_{K} is a lattice in K, we have

[K : Q] = dim_{Z} O_{K} = dim_{F_{p}} O_{K}/pO_{K}

= dim_{F_{p}} (⨁_{i=1}^{g} O_{K}/𝔭_{i}^{e_{i}}) = ∑_{i=1}^{g} Norm_{K/Q}(𝔭_{i}^{e_{i}}) = ∑_{i=1}^{g} e_{i}f_{i} = efg,

which completes the proof.

The rest of this section illustrates the theorem for quadratic fields and a cubic field and its Galois closure.

2.1 Quadratic Extensions

Suppose K/Q is a quadratic field. Then K is Galois, so for each prime p ∈ Z we have 2 = efg. There are exactly three possibilties:

• Ramified: e = 2, f = g = 1: The prime p ramifies in O_{K}, so pO_{K} = 𝔭^{2}. There are only finitely many such primes, since if f(x) is the minimal polynomial of a generator for O_{K}, then p ramifies if and only if f(x) has a multiple root modulo p. However, f(x) has a multiple root modulo p if and only if p divides the discriminant of f(x), which is nonzero because f(x) is irreducible over Z. (This argument shows there are only finitely many ramified primes in any number field. In fact, we will later show that the ramified primes are exactly the ones that divide the discriminant.)

• Inert: e = 1, f = 2, g = 1: The prime p is inert in O_{K}, so pO_{K} = 𝔭 is prime. This happens 50% of the time, which is suggested by quadratic reciprocity (but not proved this way), as we will see illustrated below for a particular example.

• Split: e = f = 1, g = 2: The prime p splits in O_{K}, in the sense that pO_{K} = 𝔭_{1}𝔭_{2} with 𝔭_{1} ≠ 𝔭_{2}. This happens the other 50% of the time.

Suppose, in particular, that K = Q(\sqrt{5}), so O_{K} = Z[γ], where γ = (1 + \sqrt{5})/2. Then p = 5 is ramified, since pO_{K} = (\sqrt{5})^{2}. More generally, the order Z[\sqrt{5}] has index 2 in O_{K}, so for any prime p ≠ 2 we can determine the factorization of p in O_{K} by finding the factorization of the polynomial x^{2} − 5 ∈ F_{p}[x]. The polynomial x^{2} − 5 splits as a product of two distinct factors in F_{p}[x] if and only if e = f = 1 and g = 2. For p ≠ 2, 5 this is the case if and only if 5 is a square in F_{p}, i.e., if (\frac{5}{p}) = 1, where (\frac{5}{p}) is +1 if 5 is a square mod p and −1 if 5 is not. By quadratic reciprocity,

(\frac{5}{p}) = (−1)^{\frac{5−1}{2} · \frac{p−1}{2}} · (\frac{p}{5}) = (\frac{p}{5}) = \begin{cases}+1 & if p ≡ ±1 (mod 5) \\ −1 & if p ≡ ±2 (mod 5). \end{cases}

Thus whether p splits or is inert in O_{K} is determined by the residue class of p modulo 5.

2.2 The Cube Roots of Two

Suppose K/Q is not Galois. Then e_{i}, f_{i}, and g are defined for each prime p ∈ Z, but we need not have e_{1} = ⋯ = e_{g} or f_{1} = ⋯ = f_{g}. We do still have that ∑_{i=1}^{g} e_{i}f_{i} = n, by the Chinese Remainder Theorem.

For example, let K = Q(\sqrt[3]{2}). We know that O_{K} = Z[\sqrt[3]{2}]. Thus 2O_{K} = (\sqrt[3]{2})^{3}, so for 2 we have e = 3 and f = g = 1. To factor 3O_{K}, we note that working modulo 3 we have

x^{3} − 2 = (x − 2)(x^{2} + 2x + 1) = (x + 1)(x + 1)^{2} = (x + 1)^{3} ∈ F_{3}[x],

so

3O_{K} = (3, \sqrt[3]{2} + 1)^{3}.

Thus e_{1} = 3, f_{1} = 1, and g = 1. Next, working modulo 5 we have

x^{3} − 2 = (x + 2)(x^{2} + 3x + 4) ∈ F_{5}[x],

and the quadratic factor is irreducible. Thus

5O_{K} = (5, \sqrt[3]{2} + 2) · (5, \sqrt[3]{2}^{2} + 3\sqrt[3]{2} + 4).

Thus here e_{1} = e_{2} = 1, f_{1} = 1, f_{2} = 2, and g = 2.

3 The Decomposition and Inertia Groups

Fix a finite Galois extension K/Q with Galois group G = Gal(K/Q). Let 𝔭 ⊂ O_{K} be a prime lying over a prime p ∈ Z.

Definition 3.1 (Decomposition Group). The Decomposition group of 𝔭 is

D_{𝔭} = {σ ∈ G : σ(𝔭) = 𝔭} ⊂ G.

Remark 3.2. Note that D_{𝔭} is called the “splitting group” in [Swinnerton-Dyer], but everybody I know now calls it the decomposition group, so that is what we’ll call it.

Let F_{𝔭} = O_{K}/𝔭 denote the residue class field of 𝔭. In this section we will prove that there is a natural exact sequence

1 → I_{𝔭} → D_{𝔭} → Gal(F_{𝔭}/F_{p}) → 1,

where I_{𝔭} is the inertia subgroup of D_{𝔭}, and #I_{𝔭} = e. The most interesting part of the proof is showing that the natural map D_{𝔭} → Gal(F_{𝔭}/F_{p}) is surjective.

We will also discuss the structure of D_{𝔭} and introduce Frobenius elements, which play a crucial roll in understanding Galois representations.