Factoring of Prime Ideals in Galois Extensions

Transcription

1 Chater 8 Factoring of Prime Ideals in Galois Extensions 8.1 Decomosition and Inertia Grous We return to the general AKLB setu: A is a Dedekind domain with fraction field K, L is a finite searable extension of K, and B is the integral closure of A in L.But now we add the condition that the extension is normal, hence Galois.We will see shortly that the Galois assumtion imoses a severe constraint on the numbers e i and f i in the ram-rel identity (4.1.6). Throughout this chater, G will denote the Galois grou Gal() Proosition If σ G, then σ(b) =B.If is a rime ideal of B, then so is σ().moreover, if lies above the nonzero rime ideal P of A, then so does σ().thus G acts on the set of rime ideals lying above P. Proof.If x B, then σ(x) B (aly σ to an equation of integral deendence).thus σ(b) B.But σ 1 (B) is also contained in B, hence B = σσ 1 (B) σ(b).if PB = e i i, then aly σ to get PB = σ( i ) ei.the σ( i ) must be rime ideals because σ reserves all algebraic relations.note also that σ is a K-automorhism, hence fixes every element of A (and of P ).Therefore A = P σ() A = P. We now show that the action of G is transitive Theorem Let and 1 be rime ideals lying above P.Then for some σ G we have σ() = 1. Proof.If the assertion is false, then for each σ, the ideals 1 and σ() are maximal and distinct, so 1 σ().by the rime avoidance lemma (Section 3.1, exercises), there is an element x 1 belonging to none of the σ().comuting the norm of x relative to, we have N(x) = σ G σ(x) by (2.1.6). But one of the σ s is the identity, 1 is an ideal, and by (8.1.1) σ(x) B for all σ.consequently, N(x) 1.But N(x) A by 1

2 2 CHAPTER 8. FACTORING OF PRIME IDEALS IN GALOIS EXTENSIONS (2.2.2), so N(x) 1 A = P = A.Thus N(x) belongs to the rime ideal, and therefore some σ 1 (x) belongs to as well.this gives x σ(), a contradiction Corollary In the factorization PB = g i=1 P ei i of the nonzero rime ideal P, the ramification indices e i are the same for all i, as are the relative degrees f i.thus the ram-rel identity simlifies to efg = n, where n =L : K = G. Proof. This follows from (8.1.2), along with the observation that an automorhism σ reserves all algebraic relations. Since we have a grou G acting on the rime factors of PB, it is natural to consider the stabilizer subgrou of each rime factor Definitions and Comments We say that the rime ideals σ(),σ G, are the conjugates of. Thus (8.1.2) says that all rime factors of PB are conjugate.the decomosition grou of is the subgrou D of G consisting of those σ G such that σ() =.(This does not mean that σ fixes every element of.) By the orbit-stabilizer theorem, the size of the orbit of is the index of the stabilizer subgrou D.Since there is only one orbit, of size g, g =G : D = G / D, hence D = n/g = efg/g = ef, indeendent of.note also that distinct conjugates of determine distinct cosets of D. For if σ 1 D = σ 2 D, then σ2 1 σ 1 D, soσ 1 () =σ 2 (). There is a articular subgrou of D that will be of interest. By (8.1.1), σ(b) =B for every σ G.If σ D, then σ() =.It follows that σ induces an automorhism σ of B/.(Note that x y mod iff σx σy mod.) Since σ is a K-automorhism, σ is an A/P -automorhism.the maing σ σ is a grou homomorhism from D to the grou of A/P -automorhisms of B/ Definition The kernel I of the above homomorhism, that is, the set of all σ D such that σ is trivial, is called the inertia grou of Remarks The inertia grou is a normal subgrou of the decomosition grou, as it is the kernel of a homomorhism.it is given exlicitly by I = {σ D : σ(x)+ = x + x B} = {σ D : σ(x) x x B}.

3 8.1. DECOMPOSITION AND INERTIA GROUPS 3 We now introduce an intermediate field and ring into the basic AKLB setu, as follows. L B K D A D K A Take K D to be the fixed field of D, and let A D = B K D be the integral closure of A in K D.Let P D be the rime ideal A D.Note that is the only rime factor of P D B. This is because all rimes in the factorization are conjugate, and σ() = for all σ D, by definition of D Lemma Let P D B = e and f = B/ : A D /P D.Then e = e and f = f.moreover, A/P = A D /P D. Proof. First, observe that by the ram-rel identity see (8.1.3), e f =L : K D, which is D by the fundamental theorem of Galois theory.but D = ef by (8.1.4), so e f = ef. Now as in (4.1.3)-(4.1.5), A/P A D /P D B/, sof f.also, PA D P D,soP D divides PA D, hence P D B divides PA D B = PB.Consequently, e e, and this forces e = e and f = f.thus the dimension of B/ over A D /P D is the same as the dimension of B/ over A/P.Since A/P can be regarded as a subfield of A D /P D, the roof is comlete Theorem Assume (B/)/(A/P ) searable.the homomorhism σ σ of D to Gal(B/)/(A/P ) introduced in (8.1.4) is surjective with kernel I.Therefore Gal(B/)/(A/P ) = D/I. Proof.Let x be a rimitive element of B/ over A/P.Let x B be a reresentative of x.let h(x) =X r + a r 1 X r a 0 be the minimal olynomial of x over K D ; the coefficients a i belong to A D by (2.2.2). The roots of h are all of the form σ(x),σ D. (We are working in the extension D, with Galois grou D.) By (8.1.7), if we reduce the coefficients of h mod P D, the resulting olynomial h(x) has coefficients in A/P.The roots of h are of the form σ(x),σ D (because x is a rimitive element).since σ D means that σ() =, all conjugates of x over A/P lie in B/.By the basic theory of slitting fields, B/ is a Galois extension of A/P. To summarize, since every conjugate of x over A/P is of the form σ(x), every A/P - automorhism of B/ (necessarily determined by its action on x), is of the form σ where σ D.Since σ is trivial iff σ I, it follows that the ma σ σ is surjective and has kernel I.

4 4 CHAPTER 8. FACTORING OF PRIME IDEALS IN GALOIS EXTENSIONS Corollary The order of I is e.thus the rime ideal P does not ramify if and only if the inertia grou of every rime ideal lying over P is trivial. Proof.By definition of relative degree, the order of Gal(B/)/(A/P ) is f. By (8.1.4), the order of D is ef. Thus by (8.1.8), the order of I must be e. Problems For Section Let D() be the decomosition grou of the rime ideal.it follows from the definition of stabilizer subgrou that D(σ()) = σd()σ 1 for every σ G.Show that the inertia subgrou also behaves in this manner, that is, I(σ()) = σi()σ 1. 2.If is an abelian extension (the Galois grou G = Gal() is abelian), show that the grous D(σ()),σ G, are all equal, as are the I(σ()),σ G.Show also that the grous deend only on the rime ideal P of A. 8.2 The Frobenius Automorhism In the basic AKLB setu, with a Galois extension, we now assume that K and L are number fields Definitions and Comments Let P be a rime ideal of A that does not ramify in B, and let be a rime lying over P. By (8.1.9), the inertia grou I() is trivial, so by (8.1.8), Gal(B/)/(A/P ) is isomorhic to the decomosition grou D().But B/ is a finite extension of the finite field A/P see (4.1.3), so the Galois grou is cyclic. Moreover, there is a canonical generator given by x+ x q +, x B, where q = A/P.Thus we have identified a distinguished element σ D(), called the Frobenius automorhism, or simly the Frobenius, of, relative to the extension.the Frobenius automorhism is determined by the requirement that for every x B, σ(x) x q mod. We use the notation for the Frobenius automorhism.the behavior of the Frobenius under conjugation is similar to the behavior of the decomosition grou as a whole (see the exercises in Section 8.1) Proosition If τ G, then = τ τ() Proof.If x B, then τ 1. τ 1 x (τ 1 x) q = τ 1 x q conclude that τ τ 1 satisfies the defining equation for is determined by its defining equation, the result follows. mod.aly τ to both sides to.since the Frobenius τ()

5 8.2. THE FROBENIUS AUTOMORPHISM Corollary If is abelian, then deends only on P, and we write the Frobenius automorhism as ( P ), and sometimes call it the Artin symbol. Proof. By (8.2.2), the Frobenius is the same for all conjugate ideals τ(),τ G, hence by (8.1.2), for all rime ideals lying over P Intermediate Fields We now introduce an intermediate field between K and L, call it F.We can then lift P to the ring of algebraic integers in F, namely B F.A rime ideal lying over P has the form F, where is a rime ideal of PB.We will comare decomosition grous with resect to the fields L and F, with the aid of the identity B/ : A/P =B/ :(B F )/( F )(B F )/( F ):A/P. The term on the left is the order of the decomosition grou of over P, denoted by D(, P ).(We are assuming that P does not ramify, so e = 1.) The first term on the right is the order of the decomosition grou of over F.The second term on the right is the relative degree of F over P, call it f.thus D(, F ) = D(, P ) /f Since D = D(, P ) is cyclic and is generated by the Frobenius automorhism σ, the unique subgrou of D with order D /f is generated by σ f.note that D(, F )is a subgrou of D(, P ), because Gal(L/F ) is a subgrou of Gal().It is natural to exect that the Frobenius automorhism of, relative to the extension L/F,isσ f Proosition = f. L/F Proof.Let σ =.Then σ D, soσ() =; also σ(x) x q q = A/P.Thus σ f () = and σ f (x) x qf.since q f (B F )/( F ), the result follows Proosition If the extension F/K is Galois, then the restriction of σ = mod, x B, where is the cardinality of the field to F is F/K F. Proof.Let σ 1 be the restriction of σ to F.Since σ() =, it follows that σ 1 ( F )= F.(Note that F/K is normal, so σ 1 is an automorhism of F. ) Thus σ 1 belongs to D( F, P).Since σ(x) x q mod, wehaveσ 1 (x) x q mod ( F ), where q = A/P.Consequently, σ 1 = F/K F.

6 6 CHAPTER 8. FACTORING OF PRIME IDEALS IN GALOIS EXTENSIONS Definitions and Comments We may view the lifting from the base field K to the extension field L as occurring in three distinct stes.let F D be the decomosition field of the extension, that is, the fixed field of the decomosition grou D, and let F I be the inertia field, the fixed field of the inertia grou I.We have the following diagram: L e= I F I f= D /e F D K g=n/ef All ramification takes lace at the to (call it level 3), and all slitting at the bottom (level 1).There is inertia in the middle (level 2).Alternatively, the results can be exressed in tabular form: e f g Level g 2 1 f 1 3 e 1 1 As we move u the diagram, we multily the ramification indices and relative degrees. This is often exressed by saying that e and f are multilicative in towers.the basic oint is that if = e1 1 and 1 = e2 2, then = e1e2 2.The multilicativity of f follows because f is a vector sace dimension. 8.3Alications Cyclotomic Fields Let ζ be a rimitive m th root of unity, and let L = (ζ) be the corresonding cyclotomic field.(we are in the AKLB setu with A = Z and K =.) Assume that is a rational rime that does not divide m. Then by (7.2.5) and the exercises for Section 4.2, is unramified.thus () factors in B as 1 g, where the i are distinct rime ideals. Moreover, the relative degree f is the same for all i, because the extension L/ is Galois.In order to say more about f, we find the Frobenius automorhism σ exlicitly. The defining equation is σ(x) x mod i for all i, and consequently σ(ζ) =ζ.

7 8.3. APPLICATIONS 7 (The idea is that the roots of unity remain distinct when reduced mod i, because the olynomial X n 1 is searable over F.) Now the order of σ is the size of the decomosition grou D, which is f.thus f is the smallest ositive integer such that σ f (ζ) =ζ.since ζ is a rimitive m th root of unity, we conclude that f is the smallest ositive integer such that f 1 mod m. Once we know f, we can find the number of rime factors g = n/f, where n = ϕ(m). (We already know that e = 1 because is unramified.) When divides m, the analysis is more comlicated, and we will only state the result. Say m = a m 1, where does not divide m 1.Then f is the smallest ositive integer such that f 1 mod m 1.The factorization is () =( 1 g ) e, with e = ϕ( a ).The i are distinct rime ideals, each with relative degree f.the number of distinct rime factors is g = ϕ(m 1 )/f. We will now give a roof of Gauss law of quadratic recirocity Proosition Let q be an odd rime, and let L = (ζ q ) be the cyclotomic field generated by a rimitive q th root of unity.then L has a unique quadratic subfield F.Exlicitly, if q 1 mod 4, then the quadratic subfield is ( q), and if q 3 mod 4, it is ( q).more comactly, F = ( q ), where q =( 1) q 1)/2 q. Proof.The Galois grou of the extension is cyclic of even order q 1, hence has a unique subgrou of index 2.Therefore L has a unique quadratic subfield. By (7.1.7) and the exercises to Section 7.1, the field discriminant is d =( 1) (q 1)/2 q q 2.But d/, because d has an odd number of factors of q.if q 1 mod 4, then the sign of d is ositive and ( d)=( q).similarly, if q 3 mod 4, then the sign of d is negative and ( d)=( q).note that the roots of the cyclotomic olynomial belong to L, hence so does d; see (2.3.5) Remarks ( ) Let σ be the Frobenius automorhism F/, where F is the unique quadratic subfield of L, and is an odd rime unequal to q. By (4.3.2), case (a1), if q is a quadratic residue mod, then slits, so g = 2 and therefore f = 1.Thus the decomosition grou D is trivial, and since σ generates D, σ is the identity.if q is not a quadratic residue mod, then by (4.3.2), case (a2), is inert, so g =1, f = 2, and σ is nontrivial.since the Galois grou of F/ has only two elements, it may be identified with {1, 1} under multilication, and we may write (using the standard Legendre symbol) σ =( q ) ).On the other hand, σ is the restriction of σ = to F, by (8.2.6). Thus σ is the identity ( L/ on F iff σ belongs to H, the unique subgrou of Gal(L/) of index 2.This will haen iff σ is a square.now the Frobenius may be viewed as a lifting of the ma x x mod q. As in (8.3.1), σ(ζ q )=ζ q.thusσ will belong to H iff is a quadratic residue mod q.in other words, σ =( q ).

8 8 CHAPTER 8. FACTORING OF PRIME IDEALS IN GALOIS EXTENSIONS uadratic Recirocity If and q are distinct odd rimes, then ( ) ( ) q =( 1) ( 1)(q 1)/4. q Proof. By (8.3.3), ( ) ( ) ( q ( 1) (q 1)/2 = = q )( ) q = ( 1 ) (q 1)/2 ( ) q. But by elementary number theory, or by the discussion in the introduction to Chater 1, ( ) 1 =( 1) ( 1)/2, and the result follows Remark Let L = (ζ), where ζ is a rimitive th root of unity, rime.as usual, B is the ring of algebraic integers of L.In this case, we can factor () in B exlicitly. By (7.1.3) and (7.1.5), () =(1 ζ) 1. Thus the ramification index e = 1 coincides with the degree of the extension.we say that is totally ramified.