佈於局部體之循環代數的不變量

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(1)ON THE COMPUTATION OF THE INVARIANTS OF SOME CYCLIC ALGEBRAS OVER LOCAL FIELDS YANG, MING-WEN. In this master thesis, our goal is to compute the invariants of some cyclic algebras over local fields. We follow the theory developed in K. Nguyen, s Ph.D. thesis ([N]). Then, we produce some examples of our own. Through section 1 to 6, we recall some basic theory of central simple algebras, Brauer groups and Galois cohomology, as well as cyclic algebras over local and global fields. Come to the heart of this thesis, in section 7, we discuss the theory of computing the invariants of cyclic algebras in unramified and tamely ramified extensions of prime degrees over finite extension of QP . The main theme is its relation with the discrete logarithm problem in finite fields. In section 8, we provide some examples to demonstrate the theory in section 7.. 1.

(2) 2. YANG, MING-WEN. 1. algebra over fields An algebra over a field K is a ring A, with a K-vector space structure (λa)b = a(λb) = λ(ab), for λ ∈ K and a, b ∈ A. A field is a commutative division ring s.t.R∗ = R× . A noncommutative division ring is a skew field. Definition 1.1. A K-algebra is called ”simple ring” if A is a right artinian A-module which has no proper two-sided ideals. i.e. A does not contain any ideals except (0) and A. Theorem 1.2. Let A be a simple ring. Then the following statements are equivalent; (a) A is a skew field; (b) A has no zero-divisors 6= 0; (c) A has no idempotents 6= 0, 1; (d) A has no nilpotent elements 6= 0 Theorem 1.3. (Wedderburn’s Main Theorem) A is a simple ring if and only if one has A ' Mn (D) with a skew field D (unique up to isomorphism) and a suitable (unique) n. More precisely : if A is a simple ring and 0 6= e ∈ A an idempotent (e.g. e = 1), then: (i) All minimal right ideals of A are isomorphic; (ii) eAe ' Mm (D) , where D := EndA (η), η is any minimal right ideal of A; (iii) D according to (ii) is a skew field such that Z(D) ' Z(A). In particular, Z(A) is a commutative field; (iv) Mn (D) ' Mm (E) with skew fields D and E implies m = n and D ' E. Definition 1.4. Let A be a ring, define the ”inverse”(or ”opposite”) ring Aop to be the additive group A equipped with the new multiplication ”·” such that a · b = ba. The following can be checked easily: (1)(Aop )op = A. (2)Aop = A if and only if A is commutative. (3)Z(Aop ) = Z(A). (4)A is a simple ring if and only if Aop is a simple ring..

(3) ON THE COMPUTATION OF THE INVARIANTS OF SOME CYCLIC ALGEBRAS OVER LOCAL FIELDS3. Definition 1.5. Let R be a commutative ring. A and B be R-algebras, f : A → B an R-algebra homomorphism and Ωf : A ⊗ R B op → EndR (B). Then we denote by Bf the R-algebra B viewed as a left (A ⊗ R B op )-module via xb := Ωf (x)(b).(x ∈ A ⊗ R B op , b ∈ B) Theorem 1.6. Let A, B be R-algebras and let f , g : A → B be R-algebra homomorphisms. Then there is a left (A ⊗ R B op )-module isomorphism Bf ' Bg if and only if there exists a unit b ∈ B ∗ such that g(a) = bf (a)b−1 for all a ∈ A. Proof. ” ⇒ ” Call φ : Bf → Bg the given isomorphism of (A ⊗ R B op )-modules. Set b := φ(1), then φ(x) = φ(Ωf (1 ⊗ x)(1)) = Ωg (1 ⊗ x)(φ(1)) = g(1)φ(1) · x = bx, for all x ∈ B. Since φ is an isomorphism necessarily b ∈ B ∗ , Finally bf (a) = φ(f (a)) = φ(Ωf (a ⊗ 1)(1)) = Ωg (a ⊗ 1)(φ(1)) = g(a)φ(1) · 1 = g(a) · b, ∀ a ∈ A ” ⇐ ” Given b ∈ B ∗ , define φ : Bf → Bg , by x → bx Obviously φ is then a Z-module isomorphism. Going through the above calculations backwards below shows that φ is even an isomorphism of (A ⊗ R B op )-modules, where φ(1) = b · 1 = b, since g(a) = bf (a)b−1 . Hence bf (a) = g(a)b = g(a)φ(1) · 1 = Ωg (a ⊗ 1)(φ(1)) = φ(Ωf (a ⊗ 1(1)) = φ(f (a)1 · 1) = φ(f (a)) i.e. φ(f (a)) = bf (a), hence φ(x) = bx.. . Definition 1.7. Let A be a ring, B ⊆ A a subring and M ⊆ A a subset. Define the ”centralizer of M in B” as the set ZB (M ) := {b ∈ B | bm = mb for all m ∈ M }. In this context the following can be derived easily: (1)ZB (M ) is a subring of B and hence also of A; (2)ZB (M ) = ZB (< M >A ) where < M >A denotes the subring of A which is generated by the set M ; (3)ZB (M ) = B ∩ ZA (M ); (4)Z(A) = ZA (A) denote the centralizer of A in A, this is also called the center of A. (5)ZA (Z(A)) = A; (6)M ⊆ ZA (ZA (M )); (7)M ⊇ N implies ZB (M ) ⊆ ZB (N ); (8)C ⊆ B ⊆ A implies ZC (M ) ⊆ ZB (M );.

(4) 4. YANG, MING-WEN. (9)B ⊆ ZA (B) if and only if B is commutative; (10)B = ZA (B) if and only if B is maximal commutative in A; 0. (11)If f : A → A is an isomorphism, then f (ZA (B)) = ZA0 (f (B)). Proof. of(10) ” ⇒” Let B ⊆ C ( A where C is commutative. By (8) B = ZB (B) ⊆ ZC (B) ⊂ ZA (B) = B. Since ZC (B) = B and B ⊆ C = ZC (C) ⊆ ZC (B) = B. i.e. B = C. Hence B is maximal commutative in A. ” ⇐ ” By (9) Since B is commutative. Hence B ⊆ ZA (B). Let x ∈ ZA (B) = {x ∈ A | xb = bx(∗), ∀B ∈ B} , but x ∈ / B. Consider < x, B >⊂ A. By (∗) < x, B > is commutative →← (Since B is maximal). Hence ZA (B) = B.. . Theorem 1.8. (Skolem-Noether Theorem) Let A, B be simple rings, K := Z(B) ⊆ Z(A) and |A : K| finite. If f , g : A → B are K-algebra homomorphisms, then there exists a unit b ∈ B ∗ such that g(a) = bf (a)b−1 for all a ∈ A. Proof. By the above theorem, it suffices to show Bf ' Bg as left (A ⊗k B op )-module. Since A, B op are simple ring. Then A⊗k B op ' Mn (D) with unique n and some skew field D (which m L is unique up to isomorphism). Hence - if l is a minimal left ideal of A ⊗k B op , Bf ' l as well as Bg '. r L. i=1. l (Since A ⊗ k B op has no proper two sided ideal, and l is a minimal left. j=1 op. ideal of A ⊗ k B , and Bf view as an artinian left A ⊗k B op module. Hence Bf '. m L. l). Since. i=1. both Bf and Bg are left D- vector spaces of the same finite dimension, necessarily m = r and therefore Bf ' Bg as left (A ⊗k B op )- modules.. . Given two K-algebras A and B, A ⊗k B is a K-algebra with unity element 1 ⊗ 1. Multiplication is defined via (a1 ⊗ b1 )(a2 ⊗ b2 ) = a1 a2 ⊗ b1 b2 for a1 , a2 in A and b1 , b2 in B. 0. 0. Theorem 1.9. Let K be a commutative field and A, A , B, B are K-algebras such that 0. 0. 0. 0. 0. 0. A ⊆ A and B ⊆ B. Then ZA⊗k B (A ⊗K B ) = ZA (A ) ⊗K ZB (B ) in A ⊗K B. 0. 0. 0. 0. Proof. A ⊗K B resp. ZA (A ) ⊗K ZB (B ) are being regarded as K-subalgebras of the K-algebra A ⊗K B..

(5) ON THE COMPUTATION OF THE INVARIANTS OF SOME CYCLIC ALGEBRAS OVER LOCAL FIELDS5 0. 0. 0. 0. 0. 0. 0. 0. ” ⊆ ” Let a ⊗ b ∈ ZA (A ) ⊗ ZB (B ) i.e. a ∈ A, a a = aa , ∀a ∈ A ; b ∈ B, bb = bb , ∀b ∈ B Since a ∈ A, b ∈ B =⇒ a ⊗ b ∈ A ⊗K B. 0. 0. 0. 0. 0. 0. 0. 0. (a ⊗ b)(a ⊗ b ) = aa ⊗ bb = a a ⊗ b b = (a ⊗ b )(a ⊗ b). 0. 0. Hence a ⊗ b ∈ ZA⊗K B (A ⊗K B ). ” ⊆ ” We select bases {ei }i∈I resp. {fj }j∈J of A resp. B over K and choose 0. 0. x ∈ ZA⊗K B (A ⊗K B ). It follows that there exists bi ∈ B resp. aj ∈ A (unique determined by x and = 0 for all P P but finitely many indices i resp. j) such that ei ⊗ bi = x = aj ⊗ fj . i∈I. j∈J. Hence by our assumptions on x, P P P 0 0 0 0 0 (ei ⊗ bi b ) = (ei ⊗ bi )(1 ⊗ b ) = x(1 ⊗ b ) = (1 ⊗ b )x = ei ⊗ b bi , i∈I. 0. 0. i∈I. 0. 0. 0. ∀b ∈ B .. i∈I. 0. ∀b ∈ B .. Hence bi b = b bi , 0. 0. i.e. bi ∈ ZB (B ) for all i ∈ I. Similarly, aj ∈ ZA (A ) for all j ∈ J. 0. 0. 0. 0. Therefore x ∈ ZA (A ) ⊗K B ∩ A ⊗K ZB (B ) = ZA (A ) ⊗K ZB (B ), the last equality L L being a consequence of B ⊗K ( Ai ) ' (B ⊗K Ai ) together with the fact that a basis 0. i∈I. i∈I. of ZA (A ) over K can be extended to a basis of A over K.. . Corollary 1.10. Let K be a field and A, B are K-algebras. Then Z(A ⊗K B) = Z(A) ⊗K Z(B). Proof. ZA⊗K B (A ⊗ B) = ZA (A) ⊗K ZB (B) = Z(A) ⊗K Z(B).. . Corollary 1.11. A and B are central simple K-algebras if and only if A ⊗K B is a central simple K-algebra. Proof. As mentioned above A and B are simple rings ⇐⇒ A ⊗K B is a simple ring. And |A ⊗K B : Z(A ⊗K B)| = |A ⊗K B : Z(A) ⊗K Z(B)| = |A ⊗K B : K ⊗K K| = |A ⊗K B : K|, where |A : Z(A)| and |B : Z(B)| are finite. Since |A ⊗K B : K| is finite =⇒ A ⊗K B is a central simple K-algebra.. . In the following, we shall be considering central simple algebras over a field K. Definition 1.12. Two central simple K-algebras A, B are called ”similar” (write A ∼ B) if there are s, t ∈ N such that A ⊗K Ms (K) ' B ⊗K Mt (K)..

(6) 6. YANG, MING-WEN. Lemma 1.13. Let A, B be central simple K-algebras with corresponding skew field D, E. Then (i)A ∼ B if and only if D ' E, (ii)A ∼ B and |A : K| = |B : K| if and only if A ' B, (iii)” ∼ ” is an equivalent relation.. Proof. (i)Since A ∼ B, ∃s, t ∈ N , 3 A ⊗K Ms (K) ' B ⊗K Mt (E). A, B are simple rings. By 0. Wedderburn s Main theorem, we have A ' Mn (D), B ' Mm (E) . Hence Mn (D)⊗K Ms (K) ' Mm (E) ⊗K Mt (K) =⇒ Mns (D) ' Mmt (E). By the Main theorem again, we have D ' E and ns = mt. (ii)Since A ∼ B and |A : K| = |B : K| = n, Hence A ⊗K Ms (t) ' B ⊗K Mt (K) and s = t ⇐⇒ A ' B. (iii) 10 By (ii) Since A ' A, hence A ∼ A. 20 If A ∼ B, i.e. ∃s, t ∈ N , 3 A ⊗K Ms (K) ' B ⊗K Mt (K) =⇒ B ∼ A. 30 If A ∼ B, hence A ⊗K Ms (K) ' B ⊗K Mt (K); and B ∼ C, hence B ⊗K Mt (K) ' C ⊗K Mr (K). So A ⊗K Ms (K) ' C ⊗K Mr (K), hence A ∼ C.. . Theorem 1.14. The Brauer group of a field K is defined as the set of equivalence classes [A] = {B|B finite, central, simple, B ∼ A}. Implying Br(K) = {[A], A finite, central, simple }. Here multiplication is defined as follows : [A] × [B] = [A ⊗K B]. This is well-defined. Properties of the tensor product imply that multiplication is associative and commutative, the unit elemenet is given by 1Br(K) = [K]. Let Aop be the oppositional algebra of A, meaning that we have Aop = A as K-vectorspaces with multiplication in Aop given by Aop ×Aop → Aop , a × b 7→ ba, where the product on the right hand side is the one in A. Then [A][Aop ] = [K] = 1Br(K) . Hence Br(K) is indeed a group.. From now on, we may replace ”A is a central simple K-algebra ” by ”[A] ∈ Br(K).”. Theorem 1.15. Let L/K be a (not necessarily finite ) field extension, then the assignment A → A ⊗K L induces a Z-homomorphism rL/K : Br(K) → Br(L) which is functional in L (i.e. rM/K = rM/L rL/K if K ⊆ L ⊆ M )..

(7) ON THE COMPUTATION OF THE INVARIANTS OF SOME CYCLIC ALGEBRAS OVER LOCAL FIELDS7. Definition 1.16. Br(L/K) := Ker(rL/K ) is called the ”relative Brauer group of L/K”. If [A] ∈ Br(K), Then [A] ∈ Br(L/K) if and only if A ⊗K L ' Mn (L) ∼ L for some n. We say that L splits A, or is a splitting field of A. (or of [A]). There is an exact sequence of groups rL/K. 1 −→ Br(L/K) ,→ Br(K) −→ Br(L), where K ⊂ L. Theorem 1.17. (Centralizer Theorem) Let B be a simple subring of a simple ring A. If K := Z(A) ⊆ Z(B) and n := |B : K| is finite. Then : (i)ZA (B) ⊗K Mn (K) ' A ⊗K B op , (ii)ZA (B) is a simple ring, (iii) Z(ZA (B)) = Z(B), (iv)ZA (ZA (B)) = B, (v) If L = Z(B) and r = |L : K|, then A ⊗K L ' Mr (B ⊗L ZA (B)), (vi)A is a free left(right) ZA (B)-module of unique rank n, (vii)If, in addition to the above assumptions, m = |A : K| is also finite, then A is a free left(right) B-module of unique rank m n = |ZA (B) : K|. Theorem 1.18. Let [A] ∈ Br(K), [B] ∈ Br(L) and B ⊆ A, and let L/K be a finite field extension. Then we have in Br(L), [ZA (B)] = [A ⊗K L] − [B] = rL/K ([A]) − [B]. Corollary 1.19. Let L/K be a finite extension and [A] ∈ Br (K) such that L ⊆ A. Then [ZA (L)] = [A ⊗K L] = rL/K ([A]). Now we can prove the theorem below. Theorem 1.20. Let L/K be a finite extension and [A] ∈ Br(K). Then [A] ∈ Br(L/K)(i.e. 0. 0. 0. L is a splitting field of A) if and only if there exists an A such that [A ] = [A], L ⊆ A and 0. |L : K|2 = |A : K|. 0. |A : K| Proof. ” ⇐= ” By Centralizer theorem (vii), take B = L |ZA0 (L) : K| = = |L : K| 0 |L : K|2 = |L : K|. By statement in page 3, since L ⊆ A , L is commutative hence |L : K| 0 L ⊆ ZA0 (L). For reasons of degrees, we have ZA0 (L) = L. Hence rL/K ([A]) = rL/K ([A ]) = [ZA0 (L)] = [L] = 0. i.e. A ⊗K L ' Mn (L) ∼ L . Hence [A] ∈ Br(L/K)..

(8) 8. YANG, MING-WEN. ” =⇒ ” Set n2 = |A : K| = |A ⊗K L : L| and m = |L : K|; by assumption, we have A ⊗K L ' Mn (L) (Since L is a splitting field of [A], and −[A] = [Aop ]). Now consider the injective K-algebra homomorphism f , g (note that L and Aop have no proper two-sided ideals ) defined in the diagram below : x. L. ↓. ↓. 1 ⊗ x Aop ⊗K L ' Mn (L) ' EndL (Ln ) ⊆ Mmn (K) =: B ↑. ↑. a. Aop 0. Embed L, Aop ⊆ B(as above ) and define A = ZB (Aop ). By construction, it follows 0. L ⊆ A . Since L, Aop has no proper two-sided ideal, hence f , g are 1-1. Let x ∈ L, y ∈ Aop , i : Aop ⊗K L −→ B x −→ 1 ⊗ x −→ i(1 ⊗ x) = g(x) y −→ y ⊗ 1 −→ i(y ⊗ 1) = f (x) and f (x)g(x) = i(1 ⊗ x)i(y ⊗ 1) = i(y ⊗ x) = i(y ⊗ 1)i(1 ⊗ x) = g(x)f (x). Since by theorem 0. 1.19 [A ] = [ZB (Aop )] = [B ⊗K L]−[Aop ] = [B]−[Aop ] = [Mmn (K)]−[Aop ] = 0−(−[A]) = [A]. 0 |Mmn (K) : K| |B : K| = = The centralizer theorem (vii) gives |A : K| = |ZB (Aop ) : K| = op |A : K| n2 m2 n2 = m2 = |L : K|2 . n2 Corollary 1.21. Any maximal commutative subfield L of a K-skew field D is a splitting field of D and dimK (D) = |L : K|2 . Theorem 1.22. The following statements are equivalent : (i) A is a central simple K-algebra, (ii)A ⊗K L ' Mn (L) for some finite separable extension L/K, (iii)A ⊗K N ' Mn (N ) for some finite Galois extension N/K. (iii) may be restated as follows : Theorem 1.23. Br(K) =. S. Br(L/K). L/Kf inite Galois. Thus, the Brauer group Br(K) can be described in terms of relative Brauer groups. Hence one can study Br(L/K) with L is finite and Galois over K..

(9) ON THE COMPUTATION OF THE INVARIANTS OF SOME CYCLIC ALGEBRAS OVER LOCAL FIELDS9. Theorem 1.24. {Frobenius Theorem(1878)} Let D be an R-skew field 6= R, then necessarily D ' H, i.e. Br(R) ' Z/2Z. 0. Theorem 1.25. {Wedderburn s Theorem (1905)} All finite skew fields are commutative. i.e. Br(K) = 1 if K has only finitely many elements..

(10) 10. YANG, MING-WEN. 2. Brauer Group and Galois Cohomology In the following, we recall some basic theory of Galois cohomology. In what follows, let Γ be a group (written multiplicatively, not necessarily finite, though finite in all of our applications) and M a left Γ-module (usually written additively, although written multiplicatively in many of our applications). Definition 2.1. A discrete Γ-module is an abelian group M with an action of Γ in such a way that for all a, b ∈ M , and σ, τ ∈ Γ, we have (1)1a = a, ∀a ∈ M , (2)σ(a + b) = σa + σb, ∀σ ∈ Γ, a, b ∈ M , (3)(στ )a = σ(τ a) ∀σ, τ ∈ Γ, a ∈ M . If M , N are left Γ-module and f : M −→ N a Z-module homomorphism, then f is called a ”left Γ-module” homomorphism if f (σ m) =σ f (m) for all σ ∈ Γ and m ∈ M . Definition 2.2. As usual, we call M Γ = {m ∈ M |σ m = m for all σ∈Γ} the fixed module σ σ P of M . If Γ is a finite group, then NΓ : M −→ M , m −→ m is a left Γ-module σ∈Γ. homomorphism, which was called ”norm”. Definition 2.3. Let Γ be a finite group and M a left Γ-module. Then H 0 (Γ, M ) = M Γ /NΓ (M ) is called the ” 0-th Cohomology Group of M ”. Example: Let L/K be finite Galois, Γ = Gal(L/K) . Then L∗ and L+ are Γ-module by Γ. Γ. the definition of σ x = σ(x). We have L∗ = K ∗ and L+ = K + as well as NΓ = NL/K and NΓ = T rL/K , hence H 0 (Γ, L∗ ) = K ∗ /NL/K (L+ ) = {1}. 0. We have to show that K + = T rL/K (L+ ) . By the Dedekind s Lemma, we take an element a a x ∈ L such that T rL/K 6= 0, then T rL/k ( x) = Tr (x) = a. i.e. T rL/K (x) T rL/K (x) L/K ∃y ∈ L+ 3 T rL/K (y) = a (thanks to the k-linearity of the trace). Let M be a left M -module. Then C 1 (Γ, M ) = {x : Γ −→ M |x(1) = 0, and x(στ ) = x(σ) +σ x(τ )} is called the set of 1-cocycles. The 1-coboundaries is B 1 (Γ, M ) = {x : Γ −→ M |x(σ) =σ m − m for some m ∈ M } form a Z-submodule of C 1 (Γ, M ). Definition 2.4. H 1 (Γ, M ) = C 1 (Γ, M )/B 1 (Γ, M ) is called the ” 1st cohomology group of M ”..

(11) ON THE COMPUTATION OF THE INVARIANTS OF SOME CYCLIC ALGEBRAS OVER LOCAL FIELDS 11. The following result is classical and of great importance. 0. Norther s Equations :H 1 (Γ, L∗ ) = {1}. H 1 (Γ, L+ ) = {0}. Let M be a left Γ-module. C 2 (Γ, M ) = {x : Γ × Γ −→ M |x(σ, τ ) + x(στ, ρ) =σ x(τ, ρ) + x(σ, τ ρ) and x(σ, 1) = 0 = x(1, τ )} is called the set of 2-cocycles. C 2 (Γ, M ) carries the structure of a Z-module. B 2 (Γ, M ) = {x : Γ × Γ −→ M |x(σ, τ ) = z(σ) − z(στ ) +σ z(τ ) where z : Γ −→ M with z(1) = 0} is called the set of 2-coboundaries. B 2 (Γ, M ) is a Z-submodule of C 2 (Γ, M ). Definition 2.5. H 2 (Γ, M ) := C 2 (Γ, M )/B 2 (Γ, M ) is called the ”2nd cohomology group of M ”. Theorem 2.6. Let L/K be finite Galois, Γ = Gal(L/K), [A] ∈ Br(L/K), L ⊆ A and |L : K|2 = |A : K|. Then there exist elements eσ ∈ A∗ (σ ∈ Γ) such that L (i)A = Leσ , eid = 1, eσ x =σ xeσ (x ∈ L, σ ∈ Γ), and σ∈Γ. (ii)eσ eτ = x(σ, ρ)eστ (σ, τ ∈ Γ , x(σ, τ ) ∈ L∗ ), here (iii)x(σ, τ )x(στ, ρ) =σ x(τ, ρ)x(σ, τ ρ) and x(σ, id) = 1 = x(id, τ ). Moreover, if there are elements fσ satisfying (i) just as the elements eσ do, and if those fσ define elements y(σ, τ ) according to (ii), then (iv)eσ = z(σ)fσ (σ ∈ Γ), where z(σ) ∈ L∗ , z(id) = 1, z(σ)σ z(τ ) x(σ, τ ) = (σ, τ ∈ Γ). (v) y(σ, τ ) z(στ ) Proof. (i)is clear from theorem (Let A be a simple ring, L a commutative subfield of A such that K=Z(A)⊆ L ⊆ A, n := |L : K| and B:=ZA (L); then L/K is a Galois extension n L if and only if there exist elements ei ∈ A∗ such that Bei = ei B (i=1,....n) and Bei = A i=1. (e1 = 1)). Thank to ZA (L) = L.(cf. P.K.Draxl :Skew fields ,chapter 7 Theorem 3 ). Now we −1 −1 −1 have eστ xeστ =στ x =σ (τ x) = eσ eτ xe−1 τ eσ for all x ∈ L, hence eστ eσ eτ ∈ ZA (L) = L and. therefore (ii) with x(σ, τ ) =στ (e−1 στ eσ eτ ). (iii)Merely reflects the law of associativity in A, namely x(σ, τ )x(στ, ρ)eστ ρ = x(σ, τ )eστ eρ = (eσ eτ )eρ = eσ (eτ eρ = eσ x(τ, ρ)eτ ρ =σ x(τ, ρ)eσ eτ ρ =σ x(τ, ρ)x(σ, τ ρ)eστ ρ , hence (iii) since the elements eστ ρ are invertible (the rest of (iii) is obvious because of (ii) and eid = 1 ). σ −1 Now (iv): again eσ xe−1 for all x ∈ L, hence fσ−1 eσ ∈ ZA (L) = L and σ = x = fσ xfσ. therefore (iv) with z(σ) := (fσ−1 eσ ) ∈ L. Since z(σ)fσ = fσ−1 eσ fσ = fσ (fσ−1 eσ ) = eσ . i.e. fσ eσ = eσ fσ ..

(12) 12. YANG, MING-WEN. x(σ, τ )eστ fστ x(σ, τ ) = y(ρ, τ ) y(ρ, τ )eστ fστ z(σ)σ z(τ ) = . z(στ ). Finally (v) results from a simple calculation using (i)(ii) and (iv) =. z f z(τ )fτ fστ z(σ)σ z(τ )fσ fτ fστ eσ eτ fστ = σ σ = y(ρ, τ )eστ fστ fσ fτ z(στ )fστ fσ fτ z(στ )fστ. Theorem 2.7. Let L/K be finite Galois, Γ = Gal(L/K), n = |L : K| = |Γ| and x ∈ L C 2 (Γ, L∗ ). Define on the n2 -dimensional K-vector space (x, L/K) = Leσ a multiplication σ∈Γ P P by the following formula eσ x =σ xeσ (x ∈ L), eσ eτ = x(σ, τ )eστ and ( xσ eσ )( yτ eτ ) σ∈Γ τ ∈Γ P = xσ eσ yτ eτ . Then the elements eσ are invertible, eid = 1, and (x, L/K) is a central σ,τ ∈Γ. simple K-algebra with splitting field L, i.e. [x, L/K] := [(x, L/K)] ∈ Br(L/K). Proof. Write A := (x, L/K); the associativity of the multiplication defined above. By Theorem 3.1(iii) ∀x ∈ C 2 (Γ, L∗ ) x(σ, τ )x(στ, %) =σ x(τ, ρ)x(σ, τ ρ), so is eid = 1, thanks −1. −1 −1 σ x(σ, σ −1 ))−1 eσ−1 Now to x(σ, id) = 1 = x(id, τ ) and hence (1)e−1 σ = x(σ , σ) eσ −1 = ( P P P embed L ⊆ A via x −→ xeid = x, and assume 0 = x( xσ eσ ) − ( xσ eσ )x = xxσ eσ − σ∈Γ σ∈Γ σ∈Γ P P σ (x −σ x)xσ eσ Since xσ , σ x ∈ L, L is field. Hence commutative. This implies xσ xeσ = σ∈Γ. σ∈Γ. that ZA (L) = L(choose x ∈ L such that σ x 6= x for all σ 6= id). Hence Z(A) = ZA (A) ⊆ ZA (L) = L and therefore ZA = FixL (Γ) = K. Now we must show that A has no proper P two-sided ideal. Let y 6= 0 ∈ A. Then y = xeσ . If {0} 6= a ⊆ A is a two-sided ideal of σ∈Γ. A and if |Γ| = 1. For some y ∈ A, it follows y = xeσ for some σ .1 = x−1 ye−1 σ ∈ a, hence a = A. If on the other hand |Γ| ≥ 2, select x ∈ L and σ, τ ∈ Γ 3 σ j (x) 6= σ r (x) and consider P P −1 0 0 a ∈ y − σ j x−1 yx = (x − σ j (x)−1 σ i xx)eσ =: x eσ =: y Because |Γ | < |Γ|, we may σ∈Γ0. σ∈Γ. conclude a = A by induction. Hence A has no proper two-sided ideals. Hence A is a central 0. 0. simple K-algebra. Then by theorem 1.21. Since[A] ∈ Br(L/K) ⇔ ∃A , 3 [A ] = [A], L ⊆ A 0. 0. and |L : K|2 = |A : K| whereas A = A.. 0. . Definition 2.8. An algebra (x, L/K) as in theorem 2.7 is called a ”crossed product ”. Lemma 2.9. In the situation of Theorem 2.7. Let x, y ∈ C 2 (Γ, L∗ ) be such that x ≡ y mod B 2 (Γ, L∗ ). Then (x, L/K) ' (y, L/K) Proof. By assumption we have. x(σ, τ ) z(σ)σ z(τ ) = for some function z : Γ → L∗ y(σ, τ ) z(στ ). where z(id) = 1 Consider A := (x, L/K) =. L. Leσ , B = (y, L/k) =. σ∈Γ. σ∈Γ z(σ)σ xfσ =σ xz(σ)fσ =σ. Then we find eσ0 = z(σ)fσ x =. L. Lfσ and call eσ0 = z(σ)fσ ∈ B. eσ0 (x ∈ L).

(13) ON THE COMPUTATION OF THE INVARIANTS OF SOME CYCLIC ALGEBRAS OVER LOCAL FIELDS 13. as well as eσ0 eτ 0 = z(σ)fσ z(τ )fτ = z(σ)σ z(τ )fσ fτ = z(σ)σ z(τ )fσ = z(σ)σ z(τ )y(σ, τ )fστ = x(σ, τ )z(στ )fστ = x(σ, τ )eστ 0 P P hence f : A → B : xσ eσ → xσ eσ0 is a K-algebra homomorphism σ∈Γ. σ∈Γ. and thus an isomorphism.. . Theorem 2.10. Let L/K be finite Galois, Γ = Gal(L/K), any x, y ∈ C 2 (Γ, L∗ ). Then (x, L/K) ⊗K (y, L/K) ∼ (xy, L/K). Note that we write the Z-module C 2 (Γ, L∗ ) multiplicatively. Proof. Set A := (x, L/K) =. L. Leσ , B = (y, L/K) =. σ∈Γ. L. Lfσ , C = (xy, L/K) =. σ∈Γ. L. Lgσ ,. σ∈Γ. and consider the commutative K-algebra L ⊗K L ⊆ A ⊗K B. Now write L = K(θ) and n P consider the (monic) minimal polynomial fθ ∈ K[T ] of θ over K. Then fθ (T ) = ai T i = i=0 Q Q θ ⊗ 1 − 1 ⊗ σθ (T −σ θ) ∈ L[T ](an = 1, n = |L : K|). Consider the element e := ∈ θ ⊗ 1 −σ θ ⊗ 1 σ∈Γ id6=σ∈Γ L ⊗K L ⊆ A ⊗K B, And regard fθ as an element K ⊗KQ K[T ] rather than in K[T ]. Then e(θ ⊗ (θ ⊗ 1 − 1 ⊗σ θ) Q θ ⊗ 1 − 1 ⊗ σθ fθ (θ ⊗ 1) Q 1 − 1 ⊗ θ) = (θ ⊗ 1 − 1 ⊗ θ) = σ∈Γ = = σ σ θ ⊗ 1 − θ ⊗ 1 demoninator (θ ⊗ 1 − θ ⊗ 1) id6=σ∈Γ id6=σ∈Γ. Q fθ (θ) ⊗ 1 = 0, when σ = id, fθ (θ) = (θ −σ θ) = 0. Hence e(θ ⊗ 1) = (1 ⊗ θ)e, and demoninator σ∈Γ Ln−1 i Kθi ) therefore (by induction). e(θ ⊗ 1) = (1 ⊗ θi )e, consequently (because of L = i=0 (?) Since e(x ⊗ 1) = (1 ⊗ x)e for all x ∈ L. Q τ θ ⊗ 1 − 1 ⊗τ σ θ Now fix τ ∈ Γ and define e(τ ) := ∈ L ⊗K L ⊆ A ⊗K B. τ θ ⊗ 1 −τ σ θ ⊗ 1 id6=σ∈Γ Using (?) (n − 1)-times and taking into account that L ⊗K L is commutative gives Q τ θ ⊗ 1 − 1 ⊗τ σ θ = e. In particular(in the case τ = id) e(τ θ ⊗ 1 − 1 ⊗τ σ (??) ee(τ ) = τ τσ θ ⊗ 1 − θ ⊗ 1 id6=σ∈Γ θ) = e(τ θ ⊗ 1) − e(1 ⊗τ σ θ) = (τ θ ⊗ 1)e − (τ σ θ ⊗ 1)e = (τ θ ⊗ 1 −τ σ θ ⊗ 1)e. When σ = id, e(τ θ ⊗ 1 − 1 ⊗τ θ) = 0. Hence e2 = e. By our construction the numerator of e is a polynomial 1. of degree n − 1 in θ ⊗ 1 over K ⊗K K with construct term ±1⊗ θ NL/K (θ) 6= 0. Therefore n−1 n−1 L Ln−1 i L -because of L ⊗K L = Kθi ⊗K L = i=1 (θ ⊗ 1)(K ⊗K L) = (θ ⊗ 1)i (K ⊗K L)i=0. i=0. We get e 6= 0, hence e ∈ L ⊗K L ⊆ A ⊗K B is an idempotent 6= 0. Let us now consider 0. the ring e(A ⊗K B)e =: C with unit element e. Fix σ, τ ∈ Γ, then Q θ ⊗ 1 − 1 ⊗ρ θ Q σ θ ⊗ 1 − 1 ⊗τ ρ θ e(eσ ⊗ fτ )eτ = e(eσ ⊗ fτ ) =e (e ⊗ fτ ) ρ σ θ⊗1− θ⊗1 θ ⊗ 1 −σρ θ ⊗ 1 σ id6=ρ∈Γ id6=ρ∈Γ = ee(σ) (eσ ⊗ fσ ) = e(eσ ⊗ fσ ). f or. σ=τ. =0. f or. σ 6= τ.

(14) 14. YANG, MING-WEN 0. 0. Define gσ = e(eσ ⊗ fσ )e ∈ C ; from the last calculation and from (?) and (??) it follows P P P P immediately e[( xσ eσ ) ⊗ ( yτ fτ )]e = e(xσ ⊗ 1)(1 ⊗ yτ )(eσ ⊗ fτ )e = (xσ ⊗ 1)(yσ ⊗ σ∈Γ τ ∈Γ σ;τ ∈Γ σ∈Γ P 1)e(eσ ⊗ fσ )e = (xσ yσ ⊗ 1)gσ0 (xσ , yσ ∈ L). gσ0 gτ 0 = e(eσ ⊗ fσ e2 (eτ ⊗ fτ )e = e(eσ ⊗ fσ )(eτ ⊗ σ∈Γ. fτ )e = e(x(σ, τ ) ⊗ y(σ, τ ))(eστ ⊗ fστ )e= (x(σ, τ ) ⊗ yσ,τ )gστ 0 and gσ0 (x ⊗ 1) = e(eσ x ⊗ fσ )e = P 0 e(σ xeσ ⊗ fσ )e = (σ x ⊗ 1)gσ0 (x ∈ L, σ ∈ Γ). Consequently f : C → C , xσ gσ → σ∈Γ P (xσ ⊗ 1)gσ0 is a surjective K-algebra homomorphism, hence an isomorphism. Now by the σ∈Γ. 0. Wedderburn s Main Theorem (ii) take A = Mr (D), D = skew field with center K. After 0. a suitable change of basis, assume e = diag(1, . . . 1, 0, . . . 0). Where s 1 s occur. Hence 0. eAe ' Ms (D) ∼ D ∼ A. i.e. C ∼ A ⊗K B, therefore C ∼ A ⊗K B.. . We are now in a position to relative crossed-product algebras with the Brauer group. We have denote by Br(L/K) the subgroup of the Brauer group Br(K) consisting of all classes of central K-algebras splits by L. Theorem 2.11. (Crossed Product Theorem) Let L/K be finite Galois, Γ := Gal(L/K). Then the assignment x → [x, L/K], (x ∈ C 2 (Γ, L∗ )) induces an isomorphism ΩL/K : H 2 (Γ, L∗ ) ' Br(L/K) Proof. By Theorem 2.7 ΩL/K is a well defined map. Theorem 2.10 says that this map is in fact a homomorphism which is injective thanks to the second part of Theorem 2.6 (see (v) ibid). Finally ΩL/K is also surjective, this follows from 0. 0. the first part of Theorem 2.7 together with Theorem 1.20 ∀[A] ∈ Br (K). ∃[A ] 3 [A ] = [A], 0. 0. L ⊆ A and |L : K|2 = |A : K|. By theorem 2.6(i)(ii)(iii) together with theorem 1.20.. .

(15) ON THE COMPUTATION OF THE INVARIANTS OF SOME CYCLIC ALGEBRAS OVER LOCAL FIELDS 15. 3. Cyclic Algebra Definition 3.1. Let L/K be a cyclic extension, n = |L : K|, Γ = Gal(L/K) =< σ > , n−1 L i and a ∈ K ∗ . Then one calls the ring (a, L/K, σ) := Le with multiplication en = a, i=0. ex =σ xe(x ∈ L) is a ”cyclic algebra ”.. Theorem 1.23 shows that the relative Brauer group Br(L/K) can be described completely in terms of the ground field K. But we have to prove some Lemmas below. Lemma 3.2. (a, L/K, σ) ' (b, L/K, σ) provided ab ∈ NL/K (L∗ ). Lemma 3.3. Let [A] ∈ Br(K) and L/K is a cyclic extension of degree n with generating automorphism σ such that L ⊆ A and n2 = |A : K|. Then there exists an e ∈ A∗ such n−1 L i that a := en ∈ K ∗ , ex =σ xe(x ∈ L) and A = Le . Moreover, if f ∈ A∗ such that i=0. b = f n ∈ K ∗ , f x =σ xf (x ∈ L) and A =. n−1 L i=0. Lf i , then ab ∈ NL/K (L∗ ).. Lemma 3.4. For cyclic algebras we have the rule (n = |L : K|). (a, L/K, σ) ⊗K (b, L/K, σ) ' Mn [(ab, L/K, σ)]. Theorem 3.5. Let L/K be finite cyclic with generating automorphism σ. Then the assignment a → [a, L/K, σ] (a ∈ K ∗ ) induces an isomorphism θσ : K ∗ /NL/K (L∗ ) ' Br(L/K). Note that in view of example in page 10, the latter isomorphism may be related in the form H 2 (Γ, L∗ ) = K ∗ /NL/K (L∗ ) ' Br(L/K). Proof. If ab ∈ NL/K (L∗ ) by lemma 3.2 ⇒ (a, L/K, σ) ' (b, L/K, σ). And [a, L/K, σ] = [(a, L/K, σ)] = [(b, L/K, σ)] = [b, L/K, σ] ∈ Br(K). i.e. θσ (a) = θσ (b), so the map θσ is well-defined. Since θσ (a) ⊗K θσ (b) = [a, L/K, σ] ⊗K [b, L/K, σ] ' [Mn (ab, L/K, σ)] = [ab, L/K, σ] = θσ (ab). Lemma 3.4 says that θσ is even a homomorphism. The injectivity of n−1 L i n−1 L θσ follows from Lemma 3.3 A = Le = Lf i when en = a, f n = b, then ab ∈ NL/K (L∗ ). i=0. i=0. i.e. [a] = [b]( mod NL/K (L∗ )). Whereas the surjectivity is a consequence of Lemma 3.3 and the theorem 1.20.. . The following we will consider the case that L/K is a cyclic Galois extension of degree |L : K| = n. In this case we can restrict ourselves to the following simple type of 2-cocycles: Let G=Gal(L/K)=< σ >. For a ∈ K × we consider the map fσ,a : G × G → L× ,.

(16) 16. YANG, MING-WEN.   a : i+j ≥n given by fσ,a (σ i , σ j ) =  1 : i+j <n Obviously fσ,a is a normalized 2-cocycle. Let (L, σ, a) be the crossed product (L, G, fσ,a ) n−1 L and set u = uσ . Then obviously ui = uσi for i = 1, ...., n − 1 whence (L, σ, a) = Lui , i=0. with un = a, and ux = σ(x)u, ∀x ∈ L. It turns out that every crossed product is isomorphic to an algebra of the form (L, σ, a). Theorem 3.6. Let f be a normalized 2-cocycle . Then (L/K, f ) ' (a, L/K, σ) with a = n−1 Q f (σ m , σ) ∈ K ∗ . m=0. Proof. We have (L/K, f ) =. n−1 L. Lvσi with v1 = 1 and vσi x = σ i (x)vσi for x ∈ L.. i=0 i j. Furthermore, vσi vσj = f (σ , σ )vσi+j for 0 ≤ i, j ≤ n − 1. Now vσ2 = vσ vσ = f (σ, σ)vσ2 also vσ3 = vσ2 vσ = f (σ, σ)vσ2 vσ = f (σ, σ)f (σ 2 , σ)vσ3 i−1 Q and in general vσi = ( f (σ j , σ))vσi for i = 2, · · · , n − 1. j=1. we obtain vσn = f (σ, σ)f (σ 2 , σ), · · · , f (σ n−1 , σ)vσn = avσn = a. n−1 n−1 L n Q Therefore (L/K, f ) = Lvσ since f (σ j , σ) ∈ L for i = 2, · · · , n − 1. Considering. vσn. i=0. j=0. Also we have vσn = a and vσ x = σ(x)vσ for all x ∈ L. Hence (L/K, f ) ' (a, L/K, σ). Since a = vσn lies in the center of (L/K, f ), we have a ∈ K.. .

(17) ON THE COMPUTATION OF THE INVARIANTS OF SOME CYCLIC ALGEBRAS OVER LOCAL FIELDS 17. 4. Extensions Of Complete Fields and Ramification Index , Inertial Degree Definition 4.1. Let K be a field complete with respect to a valuation ϕ, either archimedean ˆ be an algebraic closure of K. Then we may extend ϕ to a valuation ϕˆ or not, and let K ˆ as follows : each a ∈ K ˆ lies in some field L with K ⊂ L ⊂ K, ˆ |L : K| finite. Set on K, 1. ϕ(a) ˆ = {ϕ(NL/K a)} |L:K| . Then we find that the vales ϕ(a) ˆ is independent of the choice of L, and that every finite ˆ is complete with respect to the valuation ϕ. extension of K contained in K ˆ When ϕ is archimedean, there are only two possibilities : ˆ ϕ = ϕ. (i) K=C = K, ˆ ˆ = C, ϕˆ extends ϕ, (ii) K=R, K where ϕ and ϕˆ are usual absolute vales on R or C. If ϕ is non-archimedean, so is ϕ. ˆ However, ϕˆ need not be a discrete valuation, even if ϕ is discrete. If ϕ is a discrete valuation on K, denote by OK its valuation ring, and by PK the maximal ¯ K = OK /PK be the residue class field, and let PK = πK · OK , so πK ideal of OK . Let O is a prime element of OK . Let vK be the exponential valuation on K, defined by setting v (a). aR = PKK. , a ∈ K, a 6= 0 and vK (0) = +∞.. Definition 4.2. We define the ”ramification index” e = e(L/K) and ”residue class degree ¯L : O ¯K ) = f . f = f (L/K)” by the formula, vL (πK ) = e, (O 0. Theorem 4.3. (Hensel s Lemma) ¯ K [x] there is a factorization f¯(x) = u(x)v(x), u(x) Let f (x) ∈ OK [x] and suppose that in O monic, where u(x) and v(x) are relatively prime. Then there is a factorization in OK [x], ¯ = v. f (x) = g(x)h(x), g(x) monic, such that g¯ = u, h ¯ L is a separable Definition 4.4. L is an ”unramified extension” of K if e(L/K) = 1” and O ¯ K . L is a ”completely ramified extension of K” if O ¯L = O ¯ K , that is, f (L/K) = extension of O 1. The study of unramified extensions reduce at once to the study of separable extensions of ¯ K by virtue of the following basic result : the residues class field O.

(18) 18. YANG, MING-WEN. Theorem 4.5. (i)There is a one-to-one inclusion-preserving correspondence between the set ˆ L= finite unramified extension of K, and the set of fields k fields L such that K ⊂ L ⊂ K, ¯ ¯L finite separable over O(K). This correspondence assigns to each such field k given by k = O (ii)If L = K(a), where a is a zero of a monic f (x) ∈ OK [x] such that a ¯ is a simple zero ¯L = O ¯ K (¯ ¯L : O ¯ K |. of f¯(x) , then L is unramified over K, and OL = OK [a], O a), |L : K| = |O Conversely, every unramified extension L of K is of this form. ¯ K (ζ), let f (x) ∈ OK [x] be any monic polynpmial such that f¯(x) = (iii)Given k = O ˜ such that L is unramified over K, and min.polO¯ ζ. Then there exists a field L = K(a) ⊂ K K. ¯ L = k, and further f (x) = min.pol.K a, a ¯L. O ¯ = ζ in O ¯ L /O ¯ K is a galois extension. If G is the galois (iv) L/K is a galois extension if and only if O ¯ of O ¯ L /O ¯ K . The group of L/K, then each σ ∈ G induces an element σ ¯ in the galois group G ¯ map σ → σ ¯ gives an isomorphism G ' G. Corollary 4.6. Let |E : K| < ∞. There is a unique field W which is maximal with respect to the properties K ⊂ W ⊂ E, W is unramified over K. This field W is the ”inertia field” ¯ W is of the extension E/K. It is characterized by the facts that W/K is unramified , and O ¯ K in O ¯ L , that is, O ¯ W = {x ∈ O ¯ L : x is separable over O ¯ K }. the separable closure of O ¯ K , and let Theorem 4.7. Suppose that the complete field K has finite residue class field O ¯ K . Then for each positive integer f , there is a unique unramified extension W of q = card O ¯W : O ¯ K | = f , namely , W = k(w) for w is a primitive (q f − 1)-th K such that |W : K| = |O ¯W = O ¯ K (w). root of 1 over K. Furthermore, OW = OK [w], O ¯ ¯ W /O ¯ K are galois extensions, Corollary 4.8. Keep the notation above. Then W/K and O ¯ cyclic of order f . The group G has generator σ : w → wq , which G ¯ with galois groups G, G is generated by σ ¯ : w¯ → w¯ q . Call σ the Frobenius automorphism of the extension W/K. Let D be a skewfield of finite dimension over a complete P -adic field K contained in the center of D, and let ∆ be the unique maximal R-order in D. Let v be the exponential valuation on K, and let w be the extension of v to the skewfield D. We have seen the value group of v is Z, while the value group of w is a subgroup of m−1 Z, where m = |D : K|. Hence the value group of w equals e−1 Z, for some positive integer e dividing m. We write e = e(D/K), and call e the ramification index of D over K. Thus, the least positive number in the set {w(a) : a ∈ 4} is precisely 1/e..

(19) ON THE COMPUTATION OF THE INVARIANTS OF SOME CYCLIC ALGEBRAS OVER LOCAL FIELDS 19. Theorem 4.9. Let πD be a prime element of ∆, and set p = πD ∆. Then every nonzero one-sided ideal of ∆ is a two-sided ideal, and is a power of p. The residue class ring ∆/p is ¯ and p ∩ R = P . a skewfield over the field R, Proof. For L be any nonzero left ideal in ∆, let l = min{vD (a) : a ∈ L}, and choose an element x ∈ L such that vD (x) = l. Then for each y ∈ L we have yx−1 ∈ ∆ (Since vD (yx−1 ) = vD (y) − vD (x) = vD (y) − l ≥ 0), so y ∈ ∆x ⊂ L . And ∀y ∈ L, y = (yx−1 )x, hence L ⊂ ∆x. This proves that L = ∆x. Furthermore, ∆x = x∆ since vD (x−1 ax) = vD (a), −l l l a ∈ D, and clearly L = πD ∆ = ∆πD = pl , since x · πD ∈ ∆.. The above shows that p is a maximal left ideal of 4. Thus L is a two-sided ideal of ∆. ¯ = ∆/p therefore has no left ideals except 0 and ∆, ¯ and thus ∆ ¯ is a skewfield. The ring ∆ Finally, p ∩ R is a prime ideal of R, and is nonzero since p is a full R-lattice in D. Thus ¯ ¯ = 0, so ∆ ¯ is a skewfield over R. p ∩ R = p, whence p · ∆ ¯ : R). ¯ Note that We now define the ”inertial degree” of D over K as f = f (D/K) = (∆ ¯ = ∆/p is a vector space over R, ¯ since P ⊂ p. ∆ Theorem 4.10. Let D be a skewfield of finite dimension m over a complete P -adic field K contained in the center D. Let e = e(D/K) be the ramification index of D over K, and f = f (D/K) the inertial degree of D over K. Both e and f are finite, and ef = |D : K| = m. ¯ a¯i , and let b0 , · · · , be−1 ∈ ∆ be such that vD (bj ) = j, ¯ = Σ· R Let a1 , · · · , af ∈ ∆ be such that ∆ 0 ≤ j ≤ e − 1. Then the ef products {ai bj } form a free R-basis for ∆. ¯ are linearly indeProof. First, let {ai : 1 ≤ i ≤ s} be elements in ∆ whose images {a¯i } in ∆ ¯ and let {bj : 1 ≤ j ≤ t} be elements of K such that the integers {vD (bj )} pendently over R, are incongruent mod e. We claim that the st products {ai bj } are linearly independent over K. If not, there is a relation. (?) (Σα1i ai )b1 + · · · +(Σαti ai )bi = 0, αji ∈ K. Where we may assume that for each bj , at least one of the coefficients αj1 , · · · , αjs is a nonzero. Write nj = min{v(αji ) : 1 ≤ i ≤ s}, αji = π nj βji . Then each βji ∈ R, and for each j there is an index k for which βjk ∈ / P . We may rewrite (?) as (Σβ1i ai )π n1 b1 + · · · +(Σβti ai )π nt bt = 0. Since vD (π) = e, replacing each bj by π nj bj does not affect the hypotheses or the conclusion. 0. 0. After making this replacement, and renumbering the a s and b s if need be. We obtain a relation (Σr1i ai )b1 + · · · +(Σrti ai )bt = 0, where each rji ∈ R, and vD (b1 ) < · · · <vD (bt ), and PP r11 ∈ / P . Therefore ( rji ai )bj /b1 = (Σr1i ai ) + · · · +(Σrti ai )bt /b1 = 0 and bj /b1 ∈ p for j. i.

(20) 20. YANG, MING-WEN. ¯ we obtain the linear independent of the {a¯i } 2 ≤ j ≤ t. Passing to the residue class ring ∆, ¯ We have therefore established that the st products {ai bj } are elements of D linearly over R. independent over K, so st ≤ m. This proves that f is finite, and shows also that ef ≤ m. Now let a1 , · · · , af , b0 , · · · , be−1 ∈ ∆ be elements satisfying the hypotheses of the theorem. The ef elements {ai bj } are linearly independent over K, by the first part of the proof, and P we need only show that ∆ = Rai bj . Let x ∈ ∆, x 6= 0, and let vD (x) = ke + j, where 0 ≤ j ≤ e − 1. Then x =. i,j k (π bj )u,. ¯ ¯ we u ∈ u(∆). Since the {a¯i } from an R-basis for ∆,. may find elements {ri } in R such that u = r1 a1 + · · · +rf af + z, with z ∈ p. Hence we have P P x = (π k bj )u = π k bj ( ri ai + Z) = ( ri π k ai )bj + x1 , where vD (x1 ) > vD (x). If x1 6= 0, we i. i. may repeat this procedure with x1 , and continue if need be with x2 , x3 , · · · . Note that k is j the greatest integer in vD (x)/e, vD (x)/e = ke + j/e = k + e , 0 ≤ j ≤ e − 1. Hence after f e−1 P P (n) (2) (1) ne steps, we obtains an equality (??)x = ( rij π k + rij π k+1 + · · · +rij π k+n−1 )ai bj + yn . j=0 i=1 (1). (1). (2). Where vD (yn ) ≥ vD (x) + ne. For fixed i, j, the sequence rij π k , rij π k + rij π k+1 , · · · , is a Cauchy sequence from R, relative to the P -adic valuation on R. Since R is assumed comP 0 plete, this sequence has a limit Sij ∈ R. If we set x =: Sij ai bj ∈ ∆, then for each n we i,j 0. 0. 0. have from (??) vD (x−x ) ≥ min(vD (yn ), k + n − 1). Thus vD (x−x ) = ∞ , hence x−x = 0 0. so x = x . Therefore x ∈ ΣRai bj , and the theorem is proved.. . Theorem 4.11. Let K be the center of the skewfield D, and set |D : K| = n2 , e = e(D/K), f = f (D/K). Then ef = n2 , e|n, n|f . Proof. Let w be the extension of v to D defined w(a) of e = vD (πK ). We have w(πD ) = 1e . Since w(πD ) = e v(N vD (πD ) = 1 = m D/K πD )). Then K(πD ) is a subfield. = 12 v(ND/K (a)). By definition n 1 v(N 1m 1 D/K πD ) = m e = e (By m of D, and has ramification index. e. Since ef = |K(πD ) : K|, we have e | |K(πD ) : K|. But K(πD ) lies in some maximal subfield E of D, whence |K(πD ) : K| | |E : K|. Since |E : K| = n. Then |D : K| = n2 and E ⊗K D ' Mn (E), the result follows e|n, ∴ n = ke. n2 = (ke)2 = k 2 e2 = ef . f = ek 2 = nk. Hence n|f .. . ¯ is a finite field with q elements. Let D be a skewfield In the following we assume that R with center K, and let |D : K| = n2 . We call n the index of D. Denote by v the valuation vK on K, and by vD that on D. Let ∆ be the unique maximal R-order in D. We shall see.

(21) ON THE COMPUTATION OF THE INVARIANTS OF SOME CYCLIC ALGEBRAS OVER LOCAL FIELDS 21. the structure of D and ∆ can be described in this case, and depend only on the index n and some integer r such that 1 ≤ r ≤ n, (r, n) = 1 ¯ is finite, there is a unique unramified extension W of K with |W : K| = f . Since R Given by W = K(w), where w is a primitive (q f − 1)th root of 1. If OW denotes the ¯ W the residue class field of OW , then OW = R[w], O ¯ w = R[ ¯ w], valuation ring of W , and O ¯ ¯ W : R| ¯ = f . Further, W/K is a galois extension with cyclic galois group of |W : K| = |O ¯ is a galois ¯ W /R order f , generated by the Frobenius automorphism σ defined by w → wq . O extension, with galois group cyclic of order f , generated by the automorphism σ ¯ which maps w¯ onto w ¯q . Now we are ready to consider skewfields. The first important consequence of the hypothesis ¯ be finite is as follows : that R. Theorem 4.12. If |D : K| = n2 , then e(D/K) = f (D/K) = n.. ¯ = R(ξ), ¯ By the theorem above, we know that π∆ = pn , ∆ where p is the unique maximal ¯ ideal of ∆, π is a prime element of R, and ξ is a primitive (q n − 1)th root of 1 over R. We wish to prove that the skewfield D comes from an unramified extension, followed by a completely ramified extension. To begin with, choose any element w ∈ ∆ such that w¯ = ξ, and set W = K(w). Then W is a subfield of D, and is an unramified extension of K such ¯ : R| ¯ = n, and hence W is a maximal subfield of D. Further, W is a that |W : K| = |∆ cyclotomic extension of K containing all of the (q n − 1)th roots of 1. Hence we could have chosen w to be a primitive (q n − 1)th root of 1 over K. Call W an inertia field of D. n ∆, the field K(πD ) Let πD be a prime element of ∆.i.e. vD (πD ) = 1. Since π∆ = πD. is a completely ramified extension of K of degree n, and is a maximal subfield of D.i.e. K ⊂ K(πD ) ⊂ E, e(K(πD )/K) = 1, f (E/K(πD )) = 1 , f (K(πD )/K) = n = f (E/K). We n−1 P· j have ∆ = Rwi πD = R[w, πD ], D = K[w, πD ]. i,j=0. The following theorem is to improve our choice of the prime element πD . Thus D is obtained by adjoining the element πD to any of its inertia fields K(w), or equivalently, by adjoining w to the field K(πD ). Note that w and πD do not commute, unless n=1. The inertia field K(w) is uniquely determined up to K-isomorphic by the index n, and our next task is to improve our choice of the prime element πD ..

(22) 22. YANG, MING-WEN. Theorem 4.13. Let w ∈ D be a primitive (q n − 1)th root of 1, and let π be any prime r. −1 n element of R. Then there exists a prime element πD ∈ ∆ such that πD = π, πD wπD = wq ,. where r is a positive integer such that 1 ≤ r ≤ n, (r, n) = 1. The integer r is uniquely determined by D, and does not depend upon the choice of w or π.. i. Proof. ∵ wq is conjugate to w in D, 0 ≤ i ≤ f − 1. There exists a nonzero α ∈ D such that αwα−1 = wq . Replacing α by π l α if need be, we may assume that α ∈ ∆. Now that vD (α) = k 0 0 ε , ε ∈ u(∆) ⇒ k, and put h = n . Then αh = ε · π b , ε ∈ u(∆), b ∈ Z, (Since α = πD (k, n) n k k 0h (k, n) 0h h (k, α = πD ε = π n) ε = π b ε) and h is the least positive integer for which such an expression for αh is possible. Then αh wα−h = αh−1 (αwα−1 )α−(h−1) = αh−1 wq α−(h−1) = h. h. αh−2 (αwα−1 )q α−(h−2) = · · · = wq . Hence wq = αh wα−h = εwε−1 ≡ w( mod p). Since ¯ = ∆ is a field. But w¯ is a primitive (q n − 1)th root of 1 in ∆, ¯ whence n = h, and so ∆ p (k, n) = 1. Now choose r, t ∈ Z, with kr − nt = 1, 1 ≤ r ≤ n, so surely (r, n) = 1. Setting πD = π −t αr . We have vD (πD ) = vD (π −t αr ) = rvD (α) − tvD (π) = rk − tn = 1. So πD is r. −1 indeed a prime a prime element of ∆. Furthermore πD · w · πD = αr · wα−r = wq . Therefore n commutes with w, and certainly also commutes with πD . πD n lies in the center of D, and is then obviously a prime element of R. We do not Thus πD n know at the point that πD equals a preassigned prime element π of R. We do not know at n equals a preassigned prime element π of R, and we show next that we can this point that πD. modify the above πD so as to accomplish this. For any λ ∈ (OW ) = {λ ∈ OW , vD (λ) = 0} = {λ ∈ OW , Nλ ∈ u(R)}, λπD is also a prime element of ∆ since (vD (λπD ) = vD (λ) + vD (πD ) = r. r. 0 + 1 = 1) and (λπD )w(λπP )−1 = λwq λ−1 = wq , since λ lies in the inertia field K(w). It remains to prove that λ may be chosen so that (λπD )n = π. Now the galois group Gal(W/K) is cyclic of order n, generated by the automorphism r. σ : w → wq . Since (r, n) = 1, the automorphism ρ : w → wq also generates this galois group. (n−1). −1 −2 n−1 −1 2 , x ∈ W . Therefore NW/K x = x(πD xπD )(πD xπD ) · · · (πD xπD But ρ(x) = πD xπD. )=. −n n , x ∈ W . This shows that for each λ ∈ u(OW ), (λπD )n = (Nw/k λ)πD . (xπD )(xπD ) · · · (xπD )n πD n Now any preassigned prime element π of R is expressible as π = βπD for some β ∈ u(R).. We may choose λ ∈ u(OW ) such that NW/K λ = β is solvable for λ ⇔ f | v(β) = 0.Therefore n n (λπD )n = π, as descried ((λπD )n = (NW/K λ)πD = βπD = π)..

(23) ON THE COMPUTATION OF THE INVARIANTS OF SOME CYCLIC ALGEBRAS OVER LOCAL FIELDS 23 0. Finally, we prove the uniqueness of r. To begin with. If w is another primitive (q n − 1)th 0. root of 1 in D, then there exist a nonzero β ∈ D such that w = βwt β −1 for some t. Let 0. 0. 0. πD = βπD β −1 , another prime element of ∆. Then we have (πD )n = (βπD β −1 )n = βπDn β −1 = 0. 0. 0. r. −1 −1 βπβ −1 = π, πD w πD−1 = (βπD β −1 )(βwt β −1 )(βπD β−1 )−1 = βπD wt πD β = β(wt )q β −1 = 0. r. (w )q . So the choice of r does not depend on which root of unity is used. We must still 0. show that r does not depend on the choice of the prime element πD . Let π be any prime 0. 0. s. element of ∆, and suppose that π wπ −1 = wq for some s between 1 and n. We may write 0 s r ¯ is finite π = rπD with r ∈ u(∆), and then wq = rπD · w · (rπD )−1 = r · wq · r−1 . Since (∆ ¯ we obtain the equality w¯ qs = w¯ qr . Therefore s = r, field, hence commutative ) Passing to ∆ since w¯ is a primitive (q n − 1)th root of unity.. . The above shows that once the complete field K is given, the skewfield D is completely determined by its index n, and by the integer r. Indeed, we first form the field K(w), with w any primitive (q n − 1)th root of 1. Then we pick any prime π ∈ R, and adjoin to the field r. −1 n = π, πD wπD W an element πD satisfying the conditions πD = wq . This determines the r the ” Hasse invariant skewfield D = K(w, πD ) up to K-isomorphism. We call the fraction n of D”.. Theorem 4.14. Let 1 ≤ r ≤ n, (r, n) = 1. Given the complete field K, there exists a r. skewfield D with center K, index n, and Hasse invariant n Proof. Let W = K(w), where w is a primitive (q n − 1)th root of 1. If there is such a skewfield D with the desired properties, then W must be a maximal subfield of D, and hence W splits D. Therefore W ⊗K D ' Mn (W ). And each element d ∈ D is representable by a matrix d∗ ∈ Mn (W ). We still therefore try to find a set of matrices in Mn (W ) which constitute a skewfield having the desired properties. r. Let θ be the automorphism of W for which θ(w) = wq , and let π be a prime element of R. For α ∈ W , define  α 0 0 0   0 θ(α) 0 0   α∗ =  0 θ2 (α) 0  0  . .. .. ..  .. . . .  0 0 0 .... .... 0. . . .... 0. ... .. .. 0 .. ..         .     ∗ πD =    . 0. θn−1 (α). 0 1 0 0 ... 0 0 .. .. 0 .. .. 1 .. .. 0 ... .. .. . .. 0 0 0 0. 0. π 0 0 0. 0. .        1   0 0 .. ..

(24) 24. YANG, MING-WEN. The map α → α∗ , α ∈ W , gives a K-isomorphism of w onto the field W ∗ = K(w∗ ) ⊂ Mn (K), where we identify each λ ∈ K with the scalar matrix λIn ∈ Mn (k). It is checked ∗ n below that (πD )   0   0 ∗ 2 ∵ (πD ) =  ..  .  π. ∗ ∗ −1 = πIn , πD · w∗ · (πD ) = (w∗ )q  1 0 ... 0  0 1 0 ...  0 1 ... 0  0 0 1 ...  .. .. ..  . . . . . . . 1   .. .. .. ..  0 0 ... 0 π 0 0 .... r. . . . 0      0   =   1      0. 0 0 1 0 ... 0 0 0 0 .. .. .. . . ..        ... 0   ... 0   0 −w  . −πθ(w) 0. 1 ... .. .. . .. π 0 0 0. . 0 .. .. 0 π 0 0 the others can be    checked.    0 −1 0 1 0 1 w 0 ∗ ∗ ∗ −1   −1   = −1  πD w (πD ) = π π π 0 0 θ(w) −π 0 π 0   θ(w) 0  = (θ(w))∗ = θ(w∗ ) = (w∗ )qr . = 0 w ∗ ] = Just show for n=2, the other can be checked easily. We now set D = K[w∗ , πD n−1 P ∗ j ) , a K-subalgebra of Mn (W ). We shall verify that D is the desired skewfield. K(w∗ )i (πD. i,j=0 ∗ j ) }. Each element a ∈ D is expressible as a K-linear combination of the n2 element {(w∗ )i ·(πD n−1 P ∗ ∗ j Since (w∗ )i = (wi )∗ . It follows that a is expressible in the form a = αj (πD ) , αj ∈ W . j=0   0 In−j ∗ j , we obtain Since (πD ) = πIj 0   α0 α1 α2 ... αn−1     πθ(α ) θ(α ) θ(α ) . . . θ(α ) n−1 0 1 n−2     2 2   πθ2 (α ) πθ2 (α ) θ (α ) . . . θ (α ) n−2 n−1 0 n−3   a=  .. .. .. .. ..   . . . . .      πθn−2 (α2 ) πθn−2 (α3 ) πθn−2 (α4 ) . . . θn−2 (α1 )    n−1 n−1 n−1 n−1 πθ (α1 ) πθ (α2 ) πθ (α3 ) . . . θ (α0 ). Hence if a = 0, then each αi = 0. This shows that D is a vector space over W with basis ∗ j {(πD ) : 0 ≤ j ≤ n − 1}, and therefore |D : K| = n2 .. Next suppose that a ∈ D is a zero divisor, but that a 6= 0. Let a be given as above, after replacing a by π l a with suitable l, we may assume that each αj ∈ Ow , and that at least one αj is a unit in OW . (This depends on the observation that π is also a prime element for OW ). Now det a ≡ NW/K α0 mod πOW . But det a = 0, since a is a zero divisor in D. Therefore.

(25) ON THE COMPUTATION OF THE INVARIANTS OF SOME CYCLIC ALGEBRAS OVER LOCAL FIELDS 25 0. NW/K α0 ∈ (πOW )∩R, whence α0 isn t a unit in OW . Hence we may write α0 = πβ0 for some β0 ∈ OW . But then det a = πNW/K α1 mod π 2 OW , so also α1 = πβ1 for some β1 ∈ OW . Continuing in this way , it follows that each αj is a multiple of π, which is contradiction that at least one αj is a unit. This proves that D has no zero divisors except 0. Hence for each nonzero a ∈ D, the K-homomorphism x → ax, x ∈ D, is monic. Since |D : K| is finite, this shows that D = aD = Da for each nonzero a ∈ D, and hence D is a skewfield. Observe next that W ∗ is a subfield of D, and that |W ∗ : K| = n. If a ∈ D commutes with w∗ , then since the diagonal entries w, θ(w), · · · , θn−1 (w) of w∗ are pairwise distinct. It follows that a is a diagonal matrix. i.e. . 0  α0   0 θ(α0 ) a=   ... ...  0 0. 0 0 .. .. 0 0 .... 0 θn−1 (α0 ).      = α0∗   . Hence we have a ∈ W ∗ , and so we have shown that W ∗ is a maximal subfield of D.(Since ∗ dimW ∗ = n) On the other hand, if a also commutes with πD , then writing a = α0∗ with. α0 ∈ W , it follows    at once that θ(α0 ) = α0 .   0 α0 0 0 1 . . . 0 α 0 0 0     0   .. .. . . ..     . .  0 θ(α0 ) 0 . .   0 0 θ(α0 ) 0 ∗ ∗ =   Since α0 πD =  . . .. . .     ... . .. .. .. ..  0 0 ··· 1    .. .     πθn−1 (α0 ) 0 ··· π 0 ··· 0 0 0 0 θn−1 (α0 )     0 0 0 0 ··· 0  0 1 . . . 0   α0   0 θ(α0 )  .. .. . . ..      . .   0 . .   0 θ(α0 ) 0 0 0 θ2 (α0 ) · · · 0 ∗ ∗      πD α0 =  .. .. .. ..  .  =  .. .. .. . .  0 0 · · · 1   ..   . . . . .     π 0 ··· 0 0 0 0 θn−1 (α0 ) πα0 0 ··· 0 Hence θ(α0 ) = α0 . Hence α0 ∈ K, and a = uIn for some u ∈ K, which shows that K is ∗ n ∗ the center of D. Finally, the equation (πD ) = πIn proves that πD is a prime element for D. r , as claimed. Thus D is a skewfield with Hasse invariant n.  ··· 0   ··· 0   ..   .  0        .

(26) 26. YANG, MING-WEN. 5. Cyclic Algebra Over Local Field Let K be a local field, meaning there exists a discrete valuation v on K. K is complete with respect to v and the residue class field k of K with respect to v is a finite field. Let R be a complete discrete valuation ring, with maximal ideal p = πR 6= 0, and let ¯ = R/p. Let K be the quotient field of R, and vK the exponental valuation on K. We R ¯ is finite, and set card R ¯ = q. assume throughout that the residue class field R Let W be the unique unramified extension of K of degree n and the ramification index e of L over K equal 1 and the associated extension of residue class field is separable. Then W = K(w), where w is a primitive (q n − 1)th root of unity over K. The galois group Gal(W/K) is cyclic of order n and has as canonical generator the Frobenius automorphism σ of W/K, defined as σ(w) = wq . As we shown before that W may be embedded in D, n−1 P such that D = W z j , where zαz −1 = σ r (α), α ∈ W , and z n = π, where z ∈ D is a j=0. prime element, (r, n) = 1, and D determines by r mod n uniquely. Further, each pair r, n with (r,n)=1 arises from some D. Hence D ' (W/K, σ r , π). Choose s ∈ Z so that rs ≡ 1( mod n). Then also (s,n)=1 and by theorem (L/K, σ, a) ' (L/K, σ s , as ) for each s ∈ Z such that (s,n)=1. D ' (W/K, σ r , π) ' (W/K, σ rs , π s ) = (W/K, σ, π s ). Furthermore, we could have restricted s to lie in the range 1 ≤ s ≤ n. Let ϕ(n) denote the number of such integer s.. Theorem 5.1. Let W/K be an unramified extension of degree n, and let σ be the Frobenius automorphism of W/K. Let {as } be any set of ϕ(n) elements of K ∗ , such that the value {vk (as )} are relatively prime to n, and are incongruent. mod n. Then the ϕ(n) cyclic. algebras {(W/K, σ, as )} give a full set of non-isomorphic skewfields with center K and index n.. Definition 5.2. We denote by exp[A] the exponent of [A] in the Brauer group Br(K), that is, the least positive integer t such that [At ] = 1 in Br(K), and by index[A] the idnex of the skewfield part of A.. Theorem 5.3. Let D be a skewfield with center K and index n. Then n = exp[D]. Hence for each [A] ∈ Br(K), exp[A]=index[A]..

(27) ON THE COMPUTATION OF THE INVARIANTS OF SOME CYCLIC ALGEBRAS OVER LOCAL FIELDS 27. Theorem 5.4. Let W/K be an unramified extension of degree n, with Frobenius automor0. 0. 0. 0. phism σ, and let s ∈ Z. Write s/n = s /n , where (s , n ) = 1. Then (W/K, σ, π s ) ∼ 0. 0. 0. 0. 0. (W /K, σ , π s )=the skewfield of index n , where W /K is an unramified extension of degree 0. 0. n , with Frobenius automorphism σ . Theorem 5.5. We now define the Hasse invariant of A, denoted by inv A, by the formula s ∈ Q/Z. inv(W/K, σW/K , π s ) = m Therefore every class in Br(K) is represented for some cyclic algebra (W/K, σW/K , π s ) with W/K unramified, and hence there is a well-defined map. inv : Br(K) → Q/Z. Caution: Let L/K be a cyclic extension with galois group < σ > cyclic of degree n, and let a ∈ K ∗ . Then the cyclic algebra B = (L/K, σ, a) determines a class [B] in Br(k). However, v (a) it is not necessarily true that inv[B] = kn . Even when L/K is unramified, the formula is valid only when σ equals the Frobenius automorphism σL/K . Further, in the ramified case the Frobenius automorphism σL/K is not defined. In order to compute inv[B] when L/K is ramified, we must first write B ∼ (W/K, σW/K , π s ) = A for some unramified extension W/K, and then inv[B]=inv[A]=s/n. Theorem 5.6. inv : Br(K) ' Q/Z. Proof. 10 The map is onto: Since each element of Q/Z is of the form s/n, with (s, n) = 1, n ≥ 1. Now let [A] ∈ Br(K) can be represented by the cyclic algebra A=(W/K, σ, π s ), where W/K is unramified, σ = σW/K , and |W : K| = m. Then inv[A] = s/n ∈ Q/Z. 20 The map is 1-1: So inv[A] = 0 if and only if n|s. i.e. s = nk, k ∈ Z, A = (W/K, σ, π s ) = (W/K, σ, π nk ) = (W/K, σ, 1), that is , if and only if A ∼ K. 30 Finally, given [A], [B] ∈ B(K), we may choose W/K unramified so that A = (W/K, σ, π s ), B = (W/K, σ, π t ), σ = σW/K . Then [A][B] = [A ⊗k B] ∼ (W/K, σ, π s+t ). Therefore inv[A][B] = inv[A] + inv[B], which complete that inv is an isomorphism of groups. Theorem 5.7. Let E be any finite extension of K. The following diagram commutes inv. K B(K) −→ Q/Z. ↓. ↓ |E : K| invE. B(E) −→ Q/Z. .

(28) 28. YANG, MING-WEN. Where the horizontal maps are isomorphisms, and where the second vertical map is defined to be multiplication by |E : K|. Proof. Let A = (W/K, σ, π s ) represent a class [A] ∈ B(K), where W/K is unramified, σ = σW/K , and |W : K| = n. We may view W and E as embedded in some algebric closure of K, and then we may form their intersection F = E ∩ W and their composite EW . If k is the least positive integer such that σ k fixes F , then F/K is unramified and |F : K| = k. By < σ k >= Gal(W/W ∩ E) ' Gal(EW/W ) and then we have E ⊗k (W/K, σ, πs ) ∼ (EW/E, σ k , π s ), so it remains for us to compute the Hasse invariant of the right hand expression. As the notation before. Since |E : K| = ef , vE = evk on K. Now that ¯W : O ¯ K |. Let us put d = (n, f ), n = dn0 , f = df 0 , where (n0 , f 0 ) = 1. n = |W : K| = |O ˜ ⊂O ¯W ∩ O ¯ E with (O ˜ :O ¯ K ) = d. It follows Since d divides both n and f, there is a field O that there is a field F˜ ⊂ W ∩ E = F . i.e. K ⊂ F˜ ⊂ F such that F˜ /K is unramified, ¯ F is contained in ˜ : O ¯ K | = d, O ˜ = O ¯ ˜ . On the other hand, the field O |F˜ : K| = |O F ¯ W and O ¯ E , whence k = |O ¯F : O ¯ K | divides both n and f, hence divides d. But both O ¯F : O ¯ K |, d = |F˜ : K| and F ⊃ F˜ , hence d|k. Since we have just show that k = |F : K| = |O 0 f k|d ⇒ k = d and F = F˜ . Consequently we obtain f (E/F ) = f (E/K)/f (F/K) = = f . d Diagrammatically, we have EW \. / W. Gal(EW/E) ' Gal(W/F ) =< σ k >. E \. / 0. F =E∩W. n = |W : F | = |EW : E|. K Now that τ be the Frobenius automorphism of EW/E, and view σ k as element of Gal(EW/E). 0. 0. 0. 0. We have τ = (σ k )f (E/F ) = σ kf . But (f , n ) = 1 and n = |EW : E|, so by theorem:(L/K, σ, a) ' 0. 0. (L/K, σ s , as ). when (s,n)=1. We obtain E ⊗K A ∼ (EW/E, σ k , π s ) ' (EW/E, σ kf , π sf ) ' 0. 0. (EW/E, τ, π sf ). Therefore E ⊗K A ∼ (EW/E, τ, π sf ), and so 0 0 0 0 sf vE (π) sf e sf e sf invE [E ⊗K A] = vE (π )/n = = 0 = (∗) n = |E : K|inv[A]. 0 n n f n ((∗) ∵ d = 0 = 0 ) n f 1 Z/Z. Theorem 5.8. Let E/K be any finite extension of degree n. Then Br(E/K) ' n.

(29) ON THE COMPUTATION OF THE INVARIANTS OF SOME CYCLIC ALGEBRAS OVER LOCAL FIELDS 29. Proof. By definition, B(E/K) consists of all classes [A] ∈ B(K) which are split by E. Hence there is a commutative diagram. E⊗. 1 → Br(E/K) → Br(K) −→k Br(E) ↓ invK. ↓ invE n. Q/Z −→ Q/Z where the top row is exact, the bottom map is given by multiplication by n, and the vertical maps are isomorphisms. It follows that Br(E/K) is isomorphic to the kernel of the 1 Z/Z. bottom map, that is, Br(E/K) ' n Corollary 5.9. For given m, n ∈ N the following diagram commutes: 1 1 n Z/Z −→ mn Z/Z ↓ ↓ Br(Kn /K) −→ Br(Kmn /K) Corollary 5.10. If |E : K| = n, then Br(E/K) = {[A] ∈ Br(K) : [A]n = 1} 1 Z/Z Proof. ” ⊂ ” By the theorem before Br(E/K) ' n ” ⊃ ”. Since [A]n = 1, hence exp[A]|n and exp[A] = index[A] = index[E], so index[A] | |E : K| = n By the corollary below A splits by E.. . Corollary 5.11. Let D be a skewfield with center K and index m, and let E be any finite extension of K. Then E splits D if and only if m | |E : K| s for some s relatively prime to Proof. Consider the element [D] ∈ B(K); then inv[D] = m m. Clearly E splits D if and only if invE [E ⊗K D] = 0. On the other hand, invE [E ⊗K D] = s . Since (s, m) = 1. We conclude that inv [E ⊗ D] = 0 if and only if m | |E : K|. |E : K| m E K.

(30) 30. YANG, MING-WEN. 6. Simple Algebra Over Global Fields Let A be a central simple K-algebra, where K is a global field, and let P range over the primes of K. In order to simplify the notation. We shall use Kp to denote the P -adic completion of K. Then put AP = Kp ⊗K A = P -adic completion of A. AP is a central simple KP -algebra, thus brauer groups Br(K) → Br(KP ) : [A] → [AP ] is a homomorphism. Definition 6.1. We call mp the local index of A at P, and we write mp = index[A], clearly Ap ∼ KP if and only if mP = 1. We say that A ramifies at P, or that P is ramified in A, if mP > 1. If P is any finite prime of K, then KP is a complete field relative to a discrete valuation, and a finite residue class field. We define the Hasse invariant inv[Ap ] of a center simple K p -algebra, thereby obtaining an isomorphism inv: B(KP ) ' Q/Z. we can show that  inv[A ] = sP P mP , where m = index[A ], (s , m ) = 1. P P P P  exp[AP ] = mP We would like to have the same formulas true for the case of finite primes. First we must define Hasse invariants when P is an infinite prime, it is sufficient to define these invariants for the three cases C, R, H. We set inv[C] = 0, inv[R] = 1, inv[H] = 12 . The next theorem is of fundamental important for the entire theory of simple algebras over global fields. The proof depends on class field theory. Theorem 6.2. (Hasse Norm Theorem) Let L be a finite cyclic extension of the global field K, and let a ∈ K. For each prime P of K, we choose a prime p of L which extends P . Then a ∈ NL/K (L) ⇔ a ∈ NLp /KP (Lp ) for each P . Remarks. (i)The theorem asserts that a is a global norm (from L to K) if and only if at each p, a is a local norm (from Lp to KP ). 0. (ii) If p and p are primes of L, both of which extends P, then there a Kp -isomorphism Lp ' Lp0 , and therefore NLp /KP (Lp ) = NLp0 /KP (Lp0 ). This shows that in determining local norms at P , it does not matter which prime p of L we use, provided only that p is an extension of the valuation P from K to L. (iii)The theorem breaks down if we drop the hypothesis that L/K be cyclic. There are counterexamples even when L/K is abelian. (see Cassels-Frohlich [1,exercise 5]..

(31) ON THE COMPUTATION OF THE INVARIANTS OF SOME CYCLIC ALGEBRAS OVER LOCAL FIELDS 31. Corollary 6.3. Let A = (L/K, σ, a) be a cyclic algebra, where Gal(L/K) =< σ > and a ∈ K ∗ . Then A ∼ K if and only if AP ∼ KP for each prime P of K. The following result may be regarded as a generalization of the Hasse Norm Theorem. Theorem 6.4. (Hasse − Brauer − Nother − Albert) Let A be a central simple K-algebra. Then A ∼ K ⇔ AP ∼ KP for each prime P of K. Proof. ” ⇒ ” If A ∼ K, then clearly AP ∼ KP ” ⇐ ” By assume that AP ∼ KP for each P, but that m = index[A] > 1, and obtaining S a contradiction. By Br(K) = Br(L/K), there exists a finite galois extension L/Kf inite,Galois. L/K such that L splits A, that is, L ⊗k A ∼ L. Let G = Gal(L/K), let p be some rational prime dividing m, and let H be a sylow p-subgroup of G. Then p - [G : H], whereas |H| is a power of p. It is well know from group theory that H has a descending chain of subgroups. H = H0 ⊃ H1 ⊃ · · · ⊃ Hn = 1, [Hi : Hi+1 ] = p . With Hi+1 4Hi for each i. Let Ei be the subfield of L fixed by Hi , so En = L, and each Ei+1 /Ei is a cyclic extension of degree p. |. |. |H|. Hn−1. L = En F = En−1. G|. E0. |. K. Further, |E0 : K| = [G : H] 6= 0( mod p). Set F = En−1 , so L/K is a cyclic extension of degree p. Let us set B = F ⊗K A, a central simple F -algebra. Then L ⊗F B ' L ⊗F (F ⊗K A) ' L⊗K A ∼ A. So L splits B. Hence B is similar to a cyclic algebra C = (L/K, σ, b), where Gal(L/K) =< σ > and b ∈ F ∗ . We intend to show that B ∼ F , so it suffices for us to prove that C ∼ F . By the theorem before C ∼ F ⇔ Cp ∼ Fp for every prime p of F . Given a prime p of F, its restriction to K is a prime P of K, and we have Cp ∼ Bp = Fp ⊗F B = Fp ⊗F (F ⊗K A) ' Fp ⊗K A ' Fp ⊗Kp Ap ∼ FP . This proves that C splits locally everywhere, whence C ∼ F , and so B ∼ F . We have thus shown that F splits the algebra A. Repeat the argument, using the cyclic extension F/En−2 in place of L/F . Then we find that En−2 also splits A. Continuing in this manner, we see that E0 splits A. But then m | |E0 : K|. This is impossible since p | m but p - |E0 : K|, and so m = 1, i.e.A ∼ K, and so theorem is established.. .