Artin symbol of the Kummer fields - Creative Math. and Inf.

64 Diana SavinProposition 1.1. ([5]). Let L/K be a Galois extension and Z K , Z l be the rings of algebraicintegers of the fields K and L.Let P ∈Spec(Z K ) such that the extension K⊂L is unramifiedin P. Let P ′ be a prime ideal in the ring Z L such that P ′ /P Z L . Then there exists a uniqueautomorphism σ∈G(L/K) such that:σ(x) ≡ x N(P) (modP ′ ).Definition 1.2. ([1]). The element σ of the Proposition 1.1. is denotedIf the extension K⊂L is Abelian, thenbut only on P = P ′ ∩ Z K and it is denoted( )L/KP ′ (L/KP( )L/K.P ′does not depend on P ′ ∈Spec(Z L ),)Definition 1.3. ([1]). Let K⊂L be a Galois extension of fields, and let P ′ be a maximalideal in the ring Z L . The setZ P′ ={τ ∈ G(L/K)/τ(P ′ ) = P ′}is a subgroup in G(L/K) and it is called the group of decomposition of P ′ in theextension K ⊂ L.Theorem 1.4. ([1]). Let K⊂L be a Galois extension of fields, and let P be a maximal idealin the ring Z K .i) For any P ′ ∈Max P (Z L ) we have [G(L/K) : Z P′] = g P , where g P is the number ofprime ideals fromZ L which divide P.ii) If(K⊂L is)a unramified in P and Abelian extension of fields and Z K /P is a finite field,then L/Kgenerate the group ZP ′ P′ and |Z P′ | = f P′Proposition 1.2. ([6]). Let l be a prime natural number l ≥ 3 and ξ be a primitive root ofunity of l- th order. A prime natural number p ≥ 3 is a prime in the ring Z [ξ] if and onlyif p is generating the group (Z ∗ l , ·)Let l be a odd prime natural number and ξ be a primitive root of unity of order l. Z [ξ]is the ring of integers of the cyclotomic field Q (ξ).Let p be a prime natural number, p≠l, and P be a prime ideal in the ring Z [ξ], P dividingthe ideal generated by p, (p), in the ring Z [ξ].Proposition 1.3. ([3]). Let α ∈ Z [ξ], α/∈P. There is an integer c, unique modulo l, suchthat α N(P)−1l ≡ ξ c (modP ).Definition 1.4. ([3]). The root of unity ξ c is called the power-character of the numberα with respect{to the prime ideal P in the ring Z [ξ]. Following Hilbert([3]), weα}denote ξ c by .PProposition 1.4. ([3]). If α,β ∈ Z [ξ], (α), (β) are not divisible with P , then:{ } αβ{ α} { } β= · .P P P

66 Diana SavinWe know that L/K is the trivial automorphism, thereforeqZ[ξ]( L/K) { }√( l p) = l√ p p = l√ p.qZ[ξ]qZ[ξ]{ } pThe case II: If ≠ 1, we obtain that f qZL = l.qZ[ξ]We denote( ) L/K √(qZ[ξ]l p) = ξc√ l p.Uging Proposition 1.1. we have:( ) L/K √(qZ l p) ≡ l√ N(qZ p L) (modqZ L ).LBut N(qZ[ξ]) = N(qZ L ),thereforeξ c l√ p ≡ l√ p N(qZ[ξ]) (modqZ L ).The last congruence implies that:l√ p(l√ p) N(qZ[ξ])−1 − ξ c) ≡ (modqZ L ).Since l √ p∈U(ZL ) and P ∈Spec(Z L ), it results that:This equality is equivalent with:√ qBut l p l−1 −1ll√ p N(qZ[ξ])−1 ≡ ξ c (modqZ L ).l√ pq l−1 −1l≡ ξ c (modqZ L ).− ξ c ∈Z[ξ] and qZ L ∩Z[ξ] = qZ[ξ], therefore we obtain:ql√ l−1 −1pl≡ ξ c (modqZ[ξ]).According to the Proposition 1.3. and Definion 1.4., we get that{ } pξ c = .qZ[ξ]From the previously proved, we obtain:( ) { }L/K √ p(qZ[ξ]l p) =qZ[ξ]l√ p.□We give now an application of the above result:

Artin symbol of the Kummer fields 67Proposition 2.9. Let p and l be odd prime distinct natural numbers, l ≡ 1(mod3)(,ǫ bea primitive root of order 3 of unity, K = Q (ǫ) be the cyclotomic field. Let L = Q ǫ; 3√ )lbe the Kummer field with the ring of integers A. If there exist x,y∈N, p does not divide xsuch that p = x 3 + ly 3 , then the Artin symbol:( ) L/K= 1 L ,P(∀) P ∈Spec(Z[ǫ]), P/pZ[ǫ].Proof. We know that: pZ[ǫ] = P 1 ...P r , where P i ∈Spec(Z[ǫ]), i = 1,r, r =ϕ(3)ord (Z ∗3 ,·)p .We obtain that: if p ≡ 1(mod3) then r = 2;if p ≡ 2(mod3) then r = 1.The case I: p ≡ 1(mod3). We obtain that: pZ[ǫ] = P 1 P 2 , where P 1 , P 2 ∈Spec(Z[ǫ]).The equality p = x 3 + ly 3 is equivalent with:p = (x + √ 3ly)(x + ǫ 3√ ly)(x + ǫ 2 3√ ly). (2.1)Passing to the ideals in the ring A, in the equality (1), we have:P 1 A · P 2 A = (x + 3√ ly)A(x + ǫ 3√ ly)A(x + ǫ 2 3√ ly)A. (2.2)N(P 1 ) = N(P 2 ) = p f , where f is the inertial degree of P 1 , in the extension of fieldsQ ⊂ Q(ǫ).From the Theorem 1.3., we have that efg = [Q(ǫ) : Q] = 2. But g = 2,e = 1,therefore f = 1 and N(P 1 ) = N(P 2 ) = p.Using the Proposition 1.3. and the Definition 1.4., we have:{ } lBut ∈ { 1,ǫ,ǫ 2} , i = 1,2. We can have:P ioror{ lP i}≡ l p−13 (modPi ),i = 1,2. (2.3){ } l= ǫ c1 ≠ 1,P 1{ } l= ǫ c2 ≠ 1P 2{ } l= 1,P 1{ } l= ǫ c ≠ 1P 2{ } { } l l= = 1.P 1 P 2

68 Diana Savin{ } { }llIf = ǫ c1 ≠ 1, = ǫ c2 ≠ 1, using the Theorem 1.5., it resultsP 1 P 2P 1 A,P 2 A∈Spec(A).{ }This implies{that}the equality (2.2) is impossible. Therefore,llcannot have = ǫ c1 ≠ 1, = ǫ c2 ≠ 1.{ }P 1 { }P 2l lIf = 1, = ǫ c ≠ 1, usind the Theorem 1.5. we obtain thatP 1 P 2P 1 A = P 1P ′ 2P ′ 3, ′ where P ′ i ∈Spec(A) and P 2A∈Spec(A).Passing to the ideals in the ring A, in the equality (2.1), we have:P ′ 1P ′ 2P ′ 3(P 2 A) = (x + 3√ ly)A(x + ǫ 3√ ly)A(x + ǫ 2 3√ ly)A. (2.4)We know that p does not divide x, l ≡ 1(mod3) and we can prove easily that 3does not divide x + y, g.c.d.(x,y) = 1. According to the Proposition 1.7., the ideals(x + 3√ ly)A, (x + ǫ 3√ ly)A, (x + ǫ 2 3√ ly)Aare comaximal ideals in pairs.It is known that the Galois group G(L/K) is a cyclic group and let σ∈G(L/K),σ:L ↦→(L, σ(ǫ) = ǫ, σ( 3√ l) = ǫ 3√ l. We consider three cases.(i) If x + y 3√ )l A∈Spec(A), using the Proposition 1.6., we obtain that((σ x + y 3√ ) ) (l A = x + yǫ 3√ )l A ∈ Spec(A)and(( σ 2 x + y 3√ ) ) (l A = x + yǫ 2 3√ )l A ∈ Spec(A).This implies that the equality (2.4) is impossible. Similarly((we obtain that theequality (2.4) is impossible, in the case (ii) (when the ideal σ x + y 3√ ) )l A is aproduct((of two distinct prime ideals in the ring A) and in the case (iii) (when theideal σ x + y 3√ ) )l A is the 2-power of a prime ideal in the ring A).{ {llIf = = 1, using the congruences (2.3) and the fact thatP 1}P 2}P 1 ,P 2 ∈Spec(Z[ǫ]), we obtain that:1 ≡ l p−13 (modpZ[ǫ]).But p,l∈N ∗ , therefore 1 ≡ l p−13 (modp).The last congruence is possible because p≡1(mod3).Since{ lP 1}={ } l= 1, using the Proposition 2.8., we obtain that:P 2( ) L/K= 1 L , (∀)i = 1,2P iThe case II: p ≡ 2(mod3). This implies that the ideal pZ[ǫ]∈Spec(Z[ǫ]).Similarly with the case I we obtain that N(pZ[ǫ]) = p 2 .

Artin symbol of the Kummer fields 69Using the Proposition 1.3. and the Definition 1.4., we have:{ } l≡ l p2 −13 (modpZ[ǫ]).pZ[ǫ]Since p and l are prime distinct natural numbers, it results that{ } lIf = ǫ c ≠ 1, hence pA∈Spec(A).pZ[ǫ]Passing to the ideals in the ring A, in the equality (1), we have:pA = (x + 3√ ly)A(x + ǫ 3√ ly)A(x + ǫ 2 3√ ly)A.The last equality is impossible. Therefore, we cannot have{ } l= ǫ c ≠ 1.pZ[ǫ]{ } lIf = 1 similar with the case I we obtain:pZ[ǫ]1 ≡ l p2 −13 (modpZ[ǫ]).But p, l p2 −13 ∈N ∗ , therefore 1 ≡ l p2 −13 (modp).From p≡2(mod3) results (p −1)/ p2 −1{ lpZ[ǫ]}= 1 implies that( L/KpZ[ǫ]{ } l≠ 0.pZ[ǫ]) 3. This implies that the last congruence is true.= 1 L . □REFERENCES[1] Albu T. and Ion D.I, Chapters of the algebraic theory of numbers (in Romanian), Ed. Academiei,Bucureşti, 1984[2] Cohen H., A Course in Computational Algebraic Number Theory, Springer-Verlag, 1993[3] Hilbert D., The theory of algebraic number fields (translated into Romanian), Ed. Corint, Bucureşti,1998[4] Ireland K. and Rosen M., A Classical Introduction to Modern Number Theory, Springer-Verlag, 1992[5] Magioladitis M., Primes of the form x 2 + ny 2 , University of Duisburg-Essen, June 2004[6] Savin D., About systems of Diophantine equations, Automation, Computers, Applied Mathematics,13 (2004), 191-196.[7] Savin D., Using the properties of Kummer fields in the proof on the diophantine equation x 4 − q 4 = py r ,Conference on Combinatorics, Automata and Number Theory, University of Liege, Belgium, May2006 (paper submitted)[8] Ştefănescu M., Galois Theory (in Romanian), Ed. Ex Ponto, Constanta, 2002[9] Stevenhagen P., Kummer Theory and Reciprocity Laws, Universiteit Leiden, 2005”OVIDIUS” UNIVERSITY OF CONSTANŢAFACULTY OF MATHEMATICS AND INFORMATICSDEPARTMENT OF MATHEMATICSBD. MAMAIA 124900527 CONSTANTA, ROMANIAE-mail address: savin.diana@univ-ovidius.roE-mail address: dianet72@yahoo.com