Tools
3.1 Field-Theoretic Algebraic Number Theory
3.1.1 Galois Theory
In this chapter, we recall (sometimes without proof) the main definitions and results that we need from basic algebraic number theory. These can be found in many books, for example [Sam], [Bor-Sha], [Coh0], [Marc], [Fr¨o-Tay], [Ire-Ros].
ak ∈ K. But since K is perfect, there exist bk with ak = bpk, and since K has characteristic pwe have A(X) = B(X)p with B(X) =
0knbkXk, contradicting the irreducibility ofA. We thus have a contradiction, showing
that gcd(A, A) = 1, hence the proposition.
Fields of characteristic 0 are by definition perfect, as are all finite fields (Exercise 1). An important example of a field that isnotperfect isK=Fp(T):
it has characteristicp, but T ∈K is not a pth power. In fact, ifα=T1/p, the minimal polynomial ofαisA(X) =Xp−T, andA(X) = 0, soAis not separable.
Definition 3.1.2. Let K be a perfect field and let L/K be a finite field ex- tension (in other words, L is a field containing K as a subfield, and L is finite-dimensional as a K-vector space). A map σ from L to L is called a K-automorphism of L if σ is a field isomorphism that leaves K pointwise fixed. A map σfrom L toK is called a K-embeddingof L intoK ifσ is a field homomorphism(necessarily injective)that leaves K pointwise fixed.
In other words, a K-automorphism σmust be a bijection that preserves the field structure and such thatσ(a) =a for alla∈K. In particular,σ is aK-endomorphism ofL. SinceLis finite-dimensional overKit follows that the bijectivity ofσis equivalent to its injectivity or its surjectivity.
Proposition 3.1.3. Let K be a perfect field, and letL/K be an extension of degreen.
(1) Any embedding ofKintoK extends to exactlyn K-embeddings ofLinto K.
(2) There exist at mostn K-automorphisms ofL, which are theK-embeddings σ ofL intoK such thatσ(L)⊂L.
Proof. By the primitive element theorem (which is true because we have assumed K to be perfect) we can write L = K(α) for some α ∈ L. Let A(X)∈K[X] be the minimal polynomial ofα, which is therefore of degreen (aK-basis ofLis given by 1,α,α2, . . . ,αn−1). Any element ofLis therefore of the formU(α) withU(X)∈K[X], andU(X) is unique moduloA(X).
For (1), letτbe an embedding ofKintoK. To extendτto an embedding of L, for any U ∈K[X] we must define τ(U(α)) = Uτ(τ(α)), where Uτ is the polynomial obtained fromU by applyingτto all the coefficients. For this to make sense we must have 0 =τ(A(α)) =Aτ(τ(α)); hence τ(α) must be one of the roots of the polynomial Aτ, which has degree n, and sinceK is algebraically closed,Aτ has exactlynroots in K; this proves (1).
Furthermore, it is clear that if σis aK-automorphism of Lthen σis in particular aK-embedding of L into K, and conversely, such an embedding is an automorphism if and only ifσ(L) =L if and only ifσ(L) ⊂L (since
[L:K]<∞), proving (2).
Definition 3.1.4. Let α∈ K and let A(X) ∈ K[X] be its minimal monic polynomial over K. The roots of A(X) = 0 in K are called the conjugates ofαinK. In other words, two elements αandβ of K are conjugate if they have the same minimal monic polynomial overK.
Thus, if deg(A) = d, α has exactly d conjugates. The elementsσ(α) in the above proof are the conjugates ofα.
Proposition 3.1.5. LetKbe a perfect field andL/K a finite extension. The following three properties are equivalent:
(1) L is closed under conjugation over K (i.e., ifα∈L, every conjugate of αalso belongs to L).
(2) There exist exactly n= [L:K] K-automorphisms ofL.
(3) If σis aK-embedding of LintoK, thenσ(L)⊂L.
Proof. By the proof of the above proposition, ifLis closed under conju- gation then all the roots of the minimal polynomialA(X) ofαbelong to L;
hence there are indeed n K-automorphisms, so (1) implies (2). The equiva- lence of (2) and (3) is clear from Proposition 3.1.3. Assume (3), letα∈ L, letβbe a conjugate ofαoverK, and denote byA(X)∈K[X] their common minimal monic polynomial. As above, we can define a field isomorphism σ from K(α) toK(β) by the formula σ(U(α)) =U(β), and this makes sense only becauseβ is a conjugate ofα. In particular,σis an embedding ofK(α) intoK(α) =K. By Proposition 3.1.3 (1),σcan be extended to an embedding ofLintoK(in fact, to [L:K(α)] such embeddings, but we do not need more than one). By (3) it follows thatσ(α) =β∈L; hence (3) implies (1).
Definition 3.1.6. LetK be a perfect field. We say that an extensionL/K is normal or Galois if one of the three equivalent conditions of the proposition is satisfied. The set ofK-automorphisms ofLforms a group under composition, called the Galois group ofL/K and denoted byGal(L/K).
Note that this is the definition of a normal extension. A Galois extension is one that is normal and separable. Since we have assumed thatKis perfect, this last condition is unnecessary, so the two notions coincide.
We will say that an extension is Abelian (respectively cyclic) if it is Ga- lois and its Galois group is abelian (respectively cyclic). Since the simplest finite groups are the groups Z/Z with prime, clearly the simplest Galois extensions are the cyclic extensions of prime degree. In that case we use the letter for the cardinality of the Galois group so that the letter p(and p, etc.) is still available for prime numbers or places.
Proposition 3.1.7. If L=K(α1, . . . , αk)andL contains the conjugates of all theαi, then L/K is Galois.
Proof. Any element x∈L has the formx=U(α1, . . . , αk), whereU has coefficients inK. Ifσis aK-embedding ofLintoK, then the coefficients of U are fixed byσ; hence
σ(x) =U(σ(α1), . . . , σ(αk)),
and since the σ(αi) are conjugates of αi, by assumption they belong to L, and henceσ(x)∈L. It follows from Proposition 3.1.5 thatL is Galois over
K.
Corollary 3.1.8. IfL/K is a finite extension, there exists a finite extension N ofL that is Galois overK, and any suchN will also be Galois overL.
Proof. Write L = K(α), and let α1, . . . , αk be the conjugates of α in K. Then by the above proposition, N = K(α1, . . . , αk) is a finite Galois extension ofK. Furthermore, if σis anL-embedding ofN intoL=K, then it is also aK-embedding; hence it is aK-automorphism ofN, hence anL-
automorphism ofN.
From now on, we use the following standard notation. IfL is a field and H is a group of automorphisms of L, then LH denotes the fixed field of L under H, in other words, the set of elements of L that are fixed by all the elements ofH. It is clear that LH is a subfield of L.
The following proposition is the key result that we need before proving the main theorem of Galois theory.
Proposition 3.1.9. LetL/K be a Galois extension with Galois groupGand letH be a subgroup ofG. ThenLH=K if and only ifH =G.
Proof. We have clearly LH ⊃K for all H. Choose first H =G, assume that x ∈ LG, and set K1 = K(x). Then L/K1 is a field extension, and by assumption every σ ∈ G is a K1-automorphism of L. It follows from Proposition 3.1.3 thatn=|G|[L:K1][L:K] =n, so that [L:K1] =n.
In other words,K1=K, proving thatLG=K. Now letH be any subgroup, and assume thatLH=K. WriteL=K(α), and consider
A(X) =
σ∈H
(X−σ(α)).
The coefficients of the polynomialAare the elementary symmetric functions of theσ(α), hence are fixed byH, soA(X)∈K[X]. Sinceαis a root of A, it follows that|H|= deg(A)[L:K] =|G|, so thatH =Gas claimed.
We can now state and prove the fundamental theorem.
Theorem 3.1.10 (Fundamental theorem of Galois theory). LetK be a perfect field, letL/K be a finite Galois extension, and setG= Gal(L/K).
There exists a one-to-one reverse-ordering correspondence between on the one hand subfields L1 of L containing K and on the other hand, subgroups H of G. The correspondence is as follows: if H is a subgroup of G, the corresponding subfield is LH. Conversely, if L1 is a subfield of Lcontaining K, the corresponding subgroup isGal(L/L1). In other wordsGal(L/LH) =H andLGal(L/L1)=L1.
Furthermore, the extensionLH/K is Galois if and only if H is a normal subgroup of G, and in this case we have a natural isomorphism G/H Gal(LH/K).
Proof. LetL1 be a subfield ofL containingK. Since L/K is Galois, any K-embedding ofLintoKis aK-automorphism; hence anyL1-embedding of LintoL1=K is an automorphism, hence anL1-automorphism, soL/L1 is Galois by Proposition 3.1.5.
Thus, for each subextensionL1of L/K the group Gal(L/L1) exists, and we denote byL1the fixed field ofLby Gal(L/L1). Applying Proposition 3.1.9 to the Galois extensionL/L1, we obtain LGal(L/L1) =L1. Now let H be a subgroup ofGandL1=LH. By Proposition 3.1.9 once again,L1=LHif and only ifH = Gal(L/L1). Thus the two mapsH →LH andL1→Gal(L/L1) are indeed inverse maps, proving the first part of the theorem.
For the second part, letH be a subgroup ofGandL1=LHthe extension corresponding toH under the above correspondence. Clearly, for eachσ∈G the field corresponding toσHσ−1isσ(L1). Now,L1/K is Galois if and only ifσ(L1) =L1 for eachK-embeddingσofL1intoK, and such an embedding extends to aK-embedding ofL, hence to an element ofGsinceL/Kis Galois.
ThusL1/K is Galois if and only ifσ(L1) =L1 for all σ∈G, hence by the correspondence if and only if σHσ−1 = H for all σ ∈G, in other words if and only ifH is a normal subgroup ofG. Finally, if this is the case, we have a natural group homomorphism fromGto Gal(L1/K) whose kernel is equal to H. We therefore obtain an injective group homomorphism fromG/H to Gal(L1/K), and since
|G/H|=|G|/|H|= [L:K]/[L:L1] = [L1:K] = Gal(L1/K), both groups have the same order; hence the homomorphism is an isomor-
phism.
Another important result is the following.
Theorem 3.1.11. Assume thatL/K is Galois, and letM/K be any finite extension. Then the extension LM/M is also Galois, and Gal(LM/M) can be considered as a subgroup of Gal(L/K) by restriction of automorphisms.
Furthermore, we haveGal(LM/M)Gal(L/K)if and only ifM andLare linearly disjoint overK, or equivalently in the present case,M∩L=K.
Proof. WriteL=K(α) for someα∈L. Clearly LM =M(α), and since the conjugates ofαbelong toL, hence toLM, Proposition 3.1.7 implies that LM/M is Galois.
The restriction of anM-automorphism of M L to Lgives an embedding of L into K, hence a K-automorphism of L since L/K is Galois, so we have a natural map from Gal(M L/M) to Gal(L/K). Furthermore, if σ ∈ Gal(M L/M) is the identity on L, then since it is an M-automorphism it is also the identity on M, hence on LM, so our map is injective, showing that Gal(LM/M) can be considered as a subgroup of Gal(L/K). Finally, let H be the image of Gal(LM/M) in Gal(L/K). Since the fixed field of LM under Gal(LM/M) is M, the fixed field ofL under H is M ∩L (in detail:
x∈ LH iff x∈ L and σ(x) = x for allσ ∈ H iff x ∈L and x ∈M). The fundamental theorem of Galois theory that we have just proved shows that H= Gal(L/(M∩L)), so thatH = Gal(L/K) iffM∩L=K.