RECIPROCITY, BRAUER GROUPS AND QUADRATIC FORMS OVER NUMBER FIELDS

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1 RECIPROCITY, BRAUER GROUPS AND QUADRATIC FORMS OVER NUMBER FIELDS THOMAS POGUNTKE Abstract. Artin s recirocity law is a vast generalization of quadratic recirocity and contains a lot of information about abelian extensions of number fields. We outline its roof and use it to show the theorems of Brauer-Hasse-Noether about central simle algebras and Hasse-Minkowski in the theory of quadratic forms. Contents 1. Introduction 1 2. Idele Invariants 2 3. Recirocity Laws 6 4. The Brauer grou 8 5. Quadratic Forms 13 References Introduction In the whole thesis, K will denote an algebraic number field, L K a normal extension, and Ω = K the algebraic numbers, unless exlicitly stated otherwise. G L K denotes the Galois grou of L K, G P G L K the decomosition grou of a rime P of L over K, and C L = I L /L the idele class grou. (G Ω K, C Ω ) is a class formation, i.e. for every L K we have H 1 (G L K, C L ) = 1, and there are canonical isomorhisms inv L K : H 2 (G L K, C L ) 1 [L:K] Z/Z with certain comatibilities. Abstract class field theory thus yields Artin s recirocity law G ab L K = C K/N L K C L. We will resent the crucial stes in the roof. In articular, we will easily obtain the classical recirocity laws by Hilbert and Gauss on the way. A very imortant corollary will be the theorem of Brauer-Hasse-Noether, which constitutes a local-global-rincile for the Brauer grou Br(K), namely that the following sequence is exact: 1 Br(K) Br(K ) Q/Z 0. Lastly, we will use this result to rove the Hasse-Minkowski theorem, which states that a quadratic form over K is isotroic if and only if it is isotroic over all comletions K. We will resuose grou cohomology and local class field theory ([3], arts I and II), as well as some results from global class field theory. The following result allows us to lift local class field theory to the global situation: Date: June 16, Key words and hrases. Class Field Theory, Artin Recirocity, Brauer-Hasse-Noether Theorem, Hasse- Minkowski Theorem. 1

2 2 THOMAS POGUNTKE Proosition 1.1. ([3], III.3.2) We have the decomosition H 2 (G L K, I L ) H2 (G LP K, L P ), c (c ), where in each comonent we take any P. We also assume the first axiom of our class formation: Proosition 1.2. ([3], III.4.7) We have H 1 (G L K, C L ) = 1. Thirdly, a result towards the second axiom: Proosition 1.3. ([3], III.4.9) We have ord H 2 (G L K, C L ) [L : K]. Finally, a dee result about the slitting of rimes, which will rove very useful. Note: We only use the existence of one non-slit rime. Theorem 1.4. ([3], III.4.3) If L K is cyclic of degree e ( rime), infinitely many laces of K do not slit in L. 2. Idele Invariants Definition 2.1. The invariant homomorhism is given by inv L K : H 2 (G L K, I L ) 1 [L:K] Z/Z, c = (c ) inv L P K c where inv LP K are the local invariants. Lemma 2.2. inv L K is well-defined. (P ), Proof. Since (c ) H2 (G LP K, L P ) as in roosition 1.1, c = 1, i.e. inv LP K c = 0, for almost all. Hence the summation makes sense. Let P and P be rimes of L, then there is σ G L K such that σp = P. This yields a K -isomorhism L σ P L P, so that [L P : K ] = [L P : K ] (cf. [3],.127). Let L K be the unramified extension of same degree, then by definition ([3], II.5.5) we have: inv LP K = inv L K = inv LP K. Proosition 2.3. The homomorhism inv L K has the roerties of the invariant ma of a class formation, i.e. if N L K are any extensions such that N K is normal, we have: (a) inv N K c = inv L K c for c H 2 (G L K, I L ), L K normal. (b) inv N L (Res L c) = [L : K] inv N K c for c H 2 (G N K, I N ). (c) inv N K (Kor K c) = inv N L c for c H 2 (G N L, I N ). Proof. We use the analogous local results to calculate: (a) (b) inv N K c = inv NP K c (2) = inv LP K c = inv L K c (P P ). inv N L (Res L c) = P inv N P L P (Res L c) P (1) = P inv N P L P (Res LP c ) (2) = P [L P : K ] inv NP K c (3) = ( P [L P : K ]) inv NP K c (4) = [L : K] inv N P K c = [L : K] inv N K c (P P ).

3 RECIPROCITY, BRAUER GROUPS AND QUADRATIC FORMS OVER NUMBER FIELDS 3 Here, we have (1) by [3], III.3.3, and (2) from the local theory. For (3), see the roof of lemma 2.2, and (4) is the fundamental equation [L : K] = e(p )f(p ) = ord G P = ord G LP K = P : K ], P P P P [L where G P = GLP K, since whenever σp = P, we obtain a unique K -morhism L σ P L P (the inverse ma is given by restricting σ G LP K to L). (c) Finally, inv N K (Kor K c) = inv N P K (Kor K c) (1) = P inv N P K (Kor K c P ) (2) = P inv N P L P c P = inv N L c. Definition 2.4. For L K abelian, the norm residue symbol is given by where ( Lemma 2.5. ( (, L K): I K G L K ( G LP K ), a = (a ) (a, L P K ), L P K ) are the local norm residue symbols., L K) is well-defined. Proof. It is well-known that L P K is unramified for almost all, and by [3], II.4.4, U K = N LP K U LP N LP K L P ( ) = ker(, L P K ) (where ( ) holds by definition). Since we also have a U K for almost all, we obtain that (a, L P K ) = 1 for almost all. Hence the roduct makes sense. In addition, the definition is indeendent of ordering, because L K is abelian. Proosition 2.6. Let L K abelian, a = (a ) I K, and a = an L K I L H 0 (G L K, I L ) its cohomology class. For every character χ χ(g L K ) = H 1 (G L K, Q/Z), we have: χ(a, L K) = inv L K (a δχ), where δ : H 1 (G L K, Q/Z) H 2 (G L K, Z) is the connecting morhism, which is bijective by the cohomological triviality of Q ([3], I.3.17). Proof. Let χ := χ and a GLP K := a N LP K L P H0 (G LP K, L P ). Because we already have the local analogue ([3], II.5.11), we can deduce: χ(a, L K) = χ( (a, L P K )) = χ(a, L P K ) = inv L P K (a δχ ) ( ) = inv L P K (a δχ) = inv L K (a δχ), where ( ) is obtained from the remarks after [3], III.3.2 as follows: (a δχ) = π(res(a δχ)) I.5.4 = π(res(a) Res(δχ)) = π(res(a) δ(res(χ))) = π(an LP K I L δχ ) I.5.2 = π(a d) = π(a) d (d = δχ ) = a d I.5.2 = a δχ, in which π : I L I K L P is the rojection to the P-comonent.

4 4 THOMAS POGUNTKE Proosition 2.7. We have H 2 (G L K, L ) ker(inv L K ). Proof. If c H 2 (G L K, L ) and L N Q normal, we can take c H 2 (G L K, L ) H 2 (G N K, N ) H 2 (G N K, I N ). Then Kor Q c H 2 (G N Q, N ), and inv L K c = inv N K c = inv N Q (Kor Q c), by roosition 2.3. Therefore, we can take K = Q. By [3], III.3.6, there is a cyclic extension N Q with N Q(ζ), ζ a root of unity, such that c H 2 (G N Q, N ). To show inv L Q c = 0, we can thus take L = N. If χ generates χ(g L Q ) = G L Q, then δχ generates the grou H 2 (G L Q, Z), where the ma δ : H 1 (G L Q, Q/Z) H 2 (G L Q, Z) is the connecting isomorhism, as above. Now Tate s theorem ([3], I.7.3) imlies that the ma δχ : H 0 (G L Q, L ) H 2 (G L Q, L ) is an isomorhism, because in the statement we can relace Z with L, since by [3], I.7.2, we have H q (G L Q, L ) = H q (G L Q, Z). Setting a = an L Q L H 0 (G L Q, L ) for a Q, we can write any element c H 2 (G L Q, L ) as c = a δχ. With roosition 2.6, we obtain: inv L Q c = inv L Q (a δχ) = χ(a, L Q). Now [3], II.5.10(a) imlies that (a, L Q) = π(a, Q(ζ) Q), where π is the natural rojection, therefore we only have to show (a, Q(ζ) Q) = 1. Moreover, Q(ζ) is generated by rimitive n th roots of unity ( rime), so it suffices to check the statement for these. Now we might as well take ζ to be a rimitive n th root of unity itself. Altogether, we really have to show that for all a Q, we have: (a, Q(ζ) Q) = (a, Q (ζ) Q ) = 1, and since (, Q(ζ) Q) is a homomorhism, we can take a Z. (1) For and (the infinite lace), Q (ζ) Q is unramified (by the law of decomosition for rimes in Q(ζ)), so by [3], II.4.8 we know: (a, Q (ζ) Q )(ζ) = ϕ ν(a) (ζ). Here, ϕ is the Frobenius homomorhism, that is ϕ(ζ) = ζ, since Z /Z = F, and we get: (a, Q (ζ) Q )(ζ) = ζ ν(a). (2) If =, we write a as a = u ν(a), u Z. In this case, [3], II.7.16 says that (a, Q (ζ) Q )(ζ) = ζ r, with r N determined by r a 1 ν(a) mod n. (3) If =, then (a, Q (ζ) Q ) = (a, C R) G C R, and we know ([3],.105): From (1), (2), (3) combined, we obtain: (a, Q (ζ) Q )(ζ) = ζ sgn a. (a, Q(ζ) Q)(ζ) = (a, Q (ζ) Q )(ζ) = ζ ω, where ω = sgn(a) ν(a) r. Since the roduct formula tells us that ω sgn(a) ν(a) ν(a) a 1 = ( a ) 1 = 1 mod n, we get (a, Q(ζ) Q)(ζ) = ζ, i.e. (a, Q(ζ) Q) = 1, so finally inv Q(ζ) Q c = 0.

5 RECIPROCITY, BRAUER GROUPS AND QUADRATIC FORMS OVER NUMBER FIELDS 5 Proosition 2.8. Let L K be cyclic, then the sequence is exact. 1 H 2 (G L K, L ) H 2 (G L K, I L ) inv L K 1 [L:K] Z/Z 0 Proof. (1) If [L : K] = e ( rime), then inv L K is onto: We shall construct c H 2 (G L K, I L ) such that inv L K c = 1 [L:K] + Z. Since L K is cyclic and [L : K] = e, by theorem 1.4, there is a rime which does not slit in L, say. Then for any P, we have: G L K = G P = GLP K, i.e. [L P : K ] = [L : K]. Locally, the invariants are isomorhisms, whence we obtain c H 2 (G LP K, L P ) such that inv LP K c = 1 [L P :K ] + Z = 1 [L:K] + Z. Let c := (c q ) H 2 (G L K, I L ) with c q = 1 for q, then we have: inv L K c = q inv L Q K q c q = inv LP K c = 1 [L:K] + Z. (2) inv L K is onto for every [L : K] = n = e1 1 er r ( i j rime). To show this, let L L i K be a cyclic subextension of degree [L i : K] = ei i, i.e. a field fixed by a (cyclic) subgrou of index ei i. Let further 1 n = n1 e nr be the artial fraction decomosition, then from (1) we obtain 1 er r elements c i H 2 (G Li K, I Li ) such that inv Li K c i = ni [L + Z. i:k] If then we define c := c 1 c r H 2 (G L K, I L ), it follows (with roosition 2.3) that: inv L K c = r inv L K c i = r inv Li K c i = r i=1 i=1 i=1 n i e i + Z = 1 [L:K] + Z. i (3) We have H 2 (G L K, L ) = ker(inv L K ): is roosition 2.7. We show ord(ker inv L K ) ord H 2 (G L K, L ) in order to obtain the equality, which by (2) is the same as ord(h 2 (G L K, I L )/H 2 (G L K, L )) ord(h 2 (G L K, I L )/ ker inv L K ) = ord( 1 [L:K] Z/Z), i.e. the degree [L : K]. To show this, we obtain from the exact sequence 1 L I L C L 1 together with theorem 1.2 the exact cohomology sequence 1 H 2 (G L K, L ) H 2 (G L K, I L ) j H 2 (G L K, C L ), from which in turn we get ord(h 2 (G L K, I L )/H 2 (G L K, L )) = ord(im j) ord H 2 (G L K, C L ). But from theorem 1.3, we know that ord H 2 (G L K, C L ) [L : K]. Proosition 2.9. There is a unique homomorhism inv K : H 2 (G Ω K, I Ω ) Q/Z such that inv K H2 (G L K,I L )= inv L K, and inv K is onto. Proof. For N L K normal, we know: inv N K H 2 (G L K,I L )= inv L K, which means that inv K := lim inv L K : lim H 2 (G L K, I L ) = H 2 1 (G Ω K, I Ω ) Q/Z = lim [L:K] Z/Z, where L runs through all normal extensions L K, exists and is uniquely determined, and we have inv K = inv L K on H 2 (G L K, I L ) by definition. Furthermore, we know that for each n N, there exists a cyclic extension L K such that 1 n [L : K] (by [3], III.3.7), so every a Q/Z is contained in some [L:K] Z/Z with L K cyclic. Since then inv L K is onto, we obtain c H 2 (G L K, I L ) such that a = inv L K c = inv K c.

6 6 THOMAS POGUNTKE 3. Recirocity Laws Theorem 3.1. (a) (G Ω K, C Ω ) is a class formation. This yields a canonical isomorhism H q (G L K, Z) H q+2 (G L K, C L ). In articular, we have H 4 (G L K, C L ) = χ(g L K ), and H 3 (G L K, C L ) = 1. (b) (Artin s recirocity law) There is a canonical isomorhism θ L K : G ab L K C K /N L K C L. (c) θ 1 L K is exact. induces the norm residue symbol (, L K), such that the sequence 1 N L K C L C K (,L K) G ab L K 1 Proof. (Sketch) For (a), first of all H 1 (G L K, C L ) = 1 by theorem 1.2. Now it remains to show that the invariants inv L K : H 2 (G L K, I L ) 1 [L:K] Z/Z become isomorhisms when lifting them to H 2 (G L K, C L ). If L K is cyclic, we have H 3 (G L K, L ) = H 1 (G L K, L ) = 1 (see [3], I.6.1, and Hilbert s theorem 90), whence we get the exact cohomology sequence 1 H 2 (G L K, L ) H 2 (G L K, I L ) j H 2 (G L K, C L ) 1. Then for c H 2 (G L K, C L ) with jc = c, let inv L K c := inv L K c. This is well-defined, because by exactness, each two c, c with jc = jc only differ by an element of the kernel H 2 (G L K, L ) = ker inv L K (roosition 2.7). Now, by [3], III.6.4, the cyclic extensions L K are enough to generate H 2 (G Ω K, C Ω ) = lim H 2 (G L K, C L ). So the morhisms j : H 2 (G L K, I L ) H 2 (G L K, C L ) combine to an eimorhism j : H 2 (G Ω K, I Ω ) H 2 (G Ω K, C Ω ), since each j commutes with inflation and is onto for cyclic L K. This way, we obtain the absolute invariant ma inv K : H 2 (G Ω K, C Ω ) Q/Z, where again c = jc H 2 (G Ω K, C Ω ) mas to inv K c := inv K c. As above, this is well-defined. For arbitrary normal extensions L K, restricting inv K to H 2 (G L K, C L ) yields a ma inv L K : H 2 (G L K, C L ) 1 [L:K] Z/Z, as desired: The roerties from roosition 2.3 carry over in the cyclic case. For other L K, take L K cyclic of the same degree, then by [3], III.6.3 we have H 2 (G L K, C L ) = H 2 (G L K, C L ). Therefore, if c H 2 (G L K, C L ) satisfies c = jc, once more we can define the invariant ma to be inv L K c := inv L K c = inv L K c. From this, we also obtain surjectivity. Let α 1 [L:K] Z/Z, then by roosition 2.8, there is c H 2 (G L K, I L ) such that inv L K c = inv L K c = α. Finally, since ord H 2 1 (G L K, C L ) [L : K] = ord( [L:K] Z/Z) by theorem 1.3, the ma inv L K : H 2 (G L K, C L ) 1 [L:K] Z/Z is indeed an isomorhism. Then abstract class field theory (see [3], II.1.7) tells us that the fundamental class u L K := inv 1 L K ( 1 [L:K] + Z) induces an isomorhism u L K : H q (G L K, Z) H q+2 (G L K, C L ).

7 RECIPROCITY, BRAUER GROUPS AND QUADRATIC FORMS OVER NUMBER FIELDS 7 In articular, if q = 2: H 4 (G L K, C L ) = H 2 (G L K, Z) = H 1 (G L K, Q/Z) = χ(g L K ) by the cohomological triviality of Q, and for q = 1: H 3 (G L K, C L ) = H 1 (G L K, Z) = 1. (b) If q = 2, we obtain our main result (cf. [3], I.3.19): θ L K : G ab L K = H 2 (G L K, Z) H 0 (G L K, C L ) = C K /N L K C L. (c) is just the definition of (, L K). Proosition 3.2. Let L K be abelian, A C K, and a I K any reresentative ak = A. Then (A, L K) = (a, L P K ) G L K. Proof. Set A = AN L K C L H 0 (G L K, C L ), and similarly a = an L K I L H 0 (G L K, I L ). Then, by definition, the ma j : H q (G L K, I L ) H q (G L K, C L ), a q=0 A, thus a δχ q=2 A δχ, for all characters χ χ(g L K ). Hence, by [3], II.1.10 and roosition 2.6, χ(a, L K) = inv L K (A δχ) = inv L K (a δχ) = χ(a, L K) (recall the construction of inv L K ), again for all χ. This means that indeed (A, L K) = (a, L K) = (a, L P K ). Definition 3.3. Let a, b K, a lace of K. Then define { 1 if ax (a, b) := 2 + by 2 = z 2 has a non-trivial solution in K, 1 otherwise, the -adic Hilbert symbol. We obviously have (a, b) = (b, a). Theorem 3.4. (Hilbert s recirocity law) Let a, b K. laces of K, we have: (a, b) = 1. For running through all Proof. Let L := K( b). First, we observe: (a, b) = 1 if and only if a N LP K L P, P : If L P = K, then a b 1 2 = ( b) 2, and a K = N LP K L P. Hence, we can take b / K, and let ax 2 + by 2 = z 2 for some non-trivial x, y, z K. Then x 0, otherwise by 2 = z 2, i.e. b = ( z y )2 (K ) 2, where y 0, since otherwise we would have z = y = x = 0. Thus, a = ( z x )2 b( y x )2 = N LP K ( z x + b y x ) is a norm element. Conversely, for a = z 2 by 2 with y, z K, we have: a by 2 = z 2. Since N LP K L P = ker(, L P K ), we can write (a, b) = (a, L P K ), where we identify {1, 1}. Obviously, 1 = A := ak C K, so roosition 3.2 yields: G LP K 1 = (A, L K) = (a, L P K ) = (a, b).

8 8 THOMAS POGUNTKE Definition 3.5. Let be an odd rime in Q, a Z. We say that a is a quadratic residue mod, if a 0 mod and a x 2 mod for some x Z. We define a 1 if a is a quadratic residue mod, := 1 if a is a quadratic non-residue mod, 0 if a 0 mod, the Legendre symbol. Theorem 3.6. (Quadratic recirocity law) Let q be odd rimes in Q. We have = ( 1) 1 q 1 q 2 2. q Proof. Let L = Q( q), then is unramified in L, so that (, L P Q ) = ϕ, by [3], II.4.8. The Frobenius ϕ G LP Q = G k(p) k() is given by ϕ : F ( q) F ( q), q q = q 1 q 2 q, q that is, (, q) = (, L P Q ) = q. Analogously, we obtain: (, q) q = (q, ) q = q. Next, we show: (, q) 2 = 1 1 mod 4 or q 1 mod 4. That is, (, q) 2 = ( 1) 1 2 : Let x 0, y 0, z 0 Q 2 non-trivial with x qy0 2 = z0. 2 Then x = 2 h x 0, y = 2 h y 0, z = 2 h z 0 Z 2, where h := inf(ν 2 (x 0 ), ν 2 (y 0 ), ν 2 (z 0 )), and one of the x, y, z is even a unit. Suose 1 mod 4 and q 1 mod 4, then z 2 + x 2 + y 2 0 mod 4. Since 0 and 1 are the only squares modulo 4, it follows that x 2, y 2, z 2 0 mod 4, i.e., x, y, z 0 mod 2. But this means x, y, z Z 2 Z 2. : If 1 mod 4, then 2 is unramified in N = Q( ). Therefore, (, q) 2 = (q, ) 2 = (q, N Q Q 2 ) = ϕ ν2(q) 2 = 1, since ν 2 (q) = 0. If q 1 mod 4, the same way we get (, q) 2 = (, L P Q 2 ) = 1. The remaining factors are trivial: q, R = Q, so we have (, q) = 1, and for all other rimes, once more we use (, q) = (, L P Q ) = ϕ ν() = 1. Taking all this together, by Hilbert s recirocity law, indeed we obtain: 1 = (, q) = ( 1) 1 q 1 q 2 2. q q The Brauer grou Definition 4.1. Let K be a field, A an algebra over K (that is, a ring with comatible K-vector sace structure). (a) Define A o to be the K-algebra with the same vector sace structure A = A o, and multilication as follows: A o A o A o, (a, b) ba. (b) A is said to be simle, if it has no non-trivial two-sided ideals. (c) The centralizer of some B A is defined to be Z A (B) := {a A ba = ab b B}. A is called central, if its centre is trivial, which means Z(A) := Z A (A) = K. (d) A central simle K-algebra of dim K A < shall be referred to as an Azumaya algebra over K.

9 RECIPROCITY, BRAUER GROUPS AND QUADRATIC FORMS OVER NUMBER FIELDS 9 Lemma 4.2. Let A be an Azumaya algebra over K. We have A K A o = Mn (K), for n = dim K A. Proof. The universal roerty of the tensor roduct yields the linear ma ϕ: A K A o End K (A) = M n (K) a b (x axb). A simle calculation shows that ϕ is an algebra homomorhism as well. Since A, A o are simle and A is central, A K A o is simle by [1], 2.6. Because ker ϕ A K A o is an ideal, ϕ must be injective. On the other hand, we have dim K M n (K) = n 2 = dim K (A) dim K (A o ) = dim K (A K A o ), which means ϕ must be bijective. Definition 4.3. Let A be a finite dimensional K-algebra. We say, A slits over a field L K, if A K L = M n (L) for some n N. Proosition 4.4. A is an Azumaya algebra if and only if A slits over a finite Galois extension L K. This L is called the slitting field of A. Proof. : We only use the results of this aragrah in characteristic zero (though they hold in general), hence for simlicity, we shall assume that K is erfect. First note that if L K is a maximal subfield of A, we have Z A (L ) = L. Assume there exists a Z A (L ) L, then L [a] Z A (L ) A, and a is algebraic, since dim L L [a] <, so that L (a) A, L is maximal. Hence, using [1], 2.5, A K L = Z A (K) K Z A o(l ) = Z A K A o(k K L ) = Z EndK (A)(L ) = End L (A), the matrix algebra M n (L ), with n = dim L A. Now we simly have to take the normal closure L of L to get our desired Galois extension A K L = A K L L L = M n (L ) L L = M n (L). : Since Z(A) K L = Z(A K L) = Z(M n (L)) = L, we get dim K Z(A) = dim L (Z(A) K L) = 1, therefore A is central. If there was a non-trivial two-sided ideal I A, the same would hold for I K L A K L. Lemma 4.5. (a) For Azumaya algebras A, B over K, the relation defines an equivalence relation. A B : m, n N: A K M n (K) = B K M m (K) (b) On the set of equivalence classes Br(K), the tensor roduct yields the structure of an abelian grou. [A] [B] := [A K B] Definition 4.6. Br(K) is called the Brauer grou of K. Proosition 4.7. (a) (Wedderburn s structure theorem) Let A be an Azumaya algebra over K. Then there is a unique n N, and a division algebra D K, unique u to isomorhism, such that A = M n (D). (b) For A = M n (D) and B = M m (D ) as in (a), we have: A B D = D. (c) We have A = B if and only if dim K A = dim K B and A B.

10 10 THOMAS POGUNTKE Proof. (a) For a slightly different version of this roof, and with more details, see [1], If A is a division algebra itself, we are done. Otherwise, there must exist a non-trivial right ideal I A (e.g. αa generated by a non-unit). Choose I such that dim K I is minimal, and let D := End A (I). Let L be the slitting field of A. As an ideal I K L A K L = M n (L), we know that we must have I K L = M m n (L) for some m < n. Thus, D K L = End A K L(I K L) = End Mn(L)(M m n (L)) = M m (L), and D is an Azumaya algebra over K. For 0 ϕ D, since 0 ϕ(i) I and I is minimal, we get ϕ(i) = I. Because dim K I dim K A < (and ϕ is in articular K-linear), ϕ is also injective, thus invertible. So D is a division algebra. Now, we have an injective K-morhism A End D (I), a (b ba), because its kernel is a two-sided ideal and A is simle. Base change to L yields M n (L) = A K L End D K L(I K L) = End Mm(L)(M m n (L)) = M n (L), thus dim K A = dim L (A K L) = dim L (End D (I) K L) = dim K End D (I), and our morhism is surjective as well. Hence, we obtain A = End D (I) = M n (D), since I = D n is free. (It is certainly finitely generated, say, D k I, k minimal. Assume its kernel R 0, then the same is true under one of the rojections π : D k D, so that π(r) is a non-zero left ideal of a division algebra, so π(r) = D, and D k 1 I is still surjective. ) Finally, assume A = M n (D) = M k (E) as well. There is a unique simle A-module, namely an ideal generated by a column, D n = E k (cf. [1], 1.8). Then we finally get D o = End A (D n ) = End A (E k ) = E o, hence D = E, and n = k. (b) : If A K M r (K) = B K M s (K), we get M n (D) K M r (K) = M m (D ) K M s (K), which yields M nr (D) = M ms (D ). By (a), this imlies D = D. : Let D = D, then of course D K M nm (K) = D K M mn (K), and hence M n (D) K M m (K) = M m (D ) K M n (K), so that A = M n (D) M m (D ) = B. (c) Let A = M n (D) and B = M m (D ), as above. Then M n (D) = M m (D ) immediately imlies m = n and D = D by (a). Thus A B by (b). Conversely, (b) says that D = D, so since dim K A = dim K B, we obtain m = n, and thus A = B. Proof. (of lemma 4.5) Note that (a) directly follows from roosition 4.7 (b). (b) To rove that the roduct is well-defined, assume that A K M n (K) = A K M m (K) and B K M r (K) = B K M s (K). Then A K B K M nr (K) = A K M n (K) K B K M r (K) = A K M m (K) K B K M s (K) = A K B K M ms (K). The roduct is associative and commutative because the tensor roduct is, [K] is the identity since A K K = A, and [A o ] is the inverse of [A] by lemma 4.2. Proosition 4.8. Let L K be a field extension, A an Azumaya algebra over K. Then A K L is an Azumaya algebra over L, and there is a grou homomorhism such that r N K = r N L r L K for N L K. r L K : Br(K) Br(L), [A] [A K L],

11 RECIPROCITY, BRAUER GROUPS AND QUADRATIC FORMS OVER NUMBER FIELDS 11 Proof. We have Z(A K L) = Z(A) K Z(L) = K K L = L (by [1], 2.5), hence A K L is central. Since A and L are simle and A is central, A K L is simle ([1], 2.6). Finally, dim L (A K L) = dim K A <, so indeed A K L is an Azumaya algebra over L. Now, r L K is a homomorhism, simly because A K B K L = (A K L) L (B K L). It is well-defined on Br(K), since from A K M n (K) = A K M m (K) it follows that: (A K L) L M n (L) = (A K L) L (M n (K) K L) = A K L K M n (K) = A K L K M m (K) = (A K L) L (M m (K) K L) = (A K L) L M m (L). Lastly, A K N = (A K L) L N imlies r N K = r N L r L K. Definition 4.9. Br(L K) := ker(r L K ) is called the relative Brauer grou of L K. We decomose it with resect to the dimension over L, i.e. Br(L K) = n N Br(n, L K) with Br(n, L K) := {[A] Br(L K) A K L = M n (L)}. Note that in Br(n, L K), Brauer equivalent is the same as K-isomorhic (roosition 4.7 (c)). Proosition (a) The Brauer grou is the union of the relative Brauer grous Br(K) = lim Br(L K), where L runs through all finite Galois extensions of K. (b) We have canonical isomorhisms Br(L K) = H 2 (G L K, L ), inducing Br(K) = lim H 2 (G L K, L ). Proof. (a) is roosition 4.4. (b) We mainly care about defining the isomorhism and will omit most calculations for brevity. For more details, see [6], X.5. Set P n (L) := Aut L (M n (L)). First, we construct a bijective ma ρ n : Br(n, L K) H 1 (G L K, P n (L)). Note that this is non-abelian cohomology. We have G L K acting via b α b σα on the scalar extensions M n (L) = M n (K) K L and A K L, resectively, and then its action on P n (L) is given by g σ g σ 1. Now, fixing [A] Br(n, L K), whenever we have f : M n (L) A K L, we obtain a 1-cocycle σ f 1 σ f σ 1 P n (L), defining the same cohomology class. So we have a well-defined ma ρ n. If ρ n ([A]) = ρ n ([B]), we can choose f : M n (L) such a way that they give the same cocycle A K L and g : M n (L) f 1 σ f σ 1 = g 1 σ g σ 1, thus σ 1 g f 1 σ = g f 1. For any basis b 1,..., b n 2 B, we get bi β i = (g f 1 )(a 1) = σ 1.(g f 1 )(a 1) = b i σ 1 β i, B K L in for some β i L. Hence σβ i = β i σ G L K, and thus β i K. This way, we obtain a K- morhism h : A B, a β i b i, with inverse induced analogously by f g 1. Therefore, we get [A] = [B]. For surjectivity, see [6], X.2, roosition 5. Every automorhism of M n (L) is inner (weak Skolem-Noether theorem), which means that the ma GL n (L) P n (L) induced by conjugation yields an exact sequence 1 L GL n (L) P n (L) 1.

12 12 THOMAS POGUNTKE From [6], VII, we thus get the connecting ma n : H 1 (G L K, P n (L)) H 2 (G L K, L ). The comositions δ n := n ρ n are comatible in the sense that δ mn ([A][B]) = δ m ([A]) + δ n ([B]) for all [A] Br(m, L K), [B] Br(n, L K). Furthermore, δ m ([A]) = 0 A = M m (K). Thus, they combine to an injective homomorhism δ : Br(L K) H 2 (G L K, L ). In [6], X.5, lemma 1, it is shown by a simle calculation that δ [L:K] is surjective. Finally, for Galois extensions N L K, n N, it is clear that the diagram Br(n, L K) H 2 (G L K, L ) Br(n, N K) H 2 (G N K, N ) commutes, thus giving our desired isomorhism δ : Br(K) lim H 2 (G L K, L ). Theorem (Brauer-Hasse-Noether) Let K be an algebraic number field. We have a canonical exact sequence 1 Br(K) Br(K ) inv K Q/Z 0. Proof. For L K cyclic, so roosition 2.8 tells us, we have exact sequences 1 H 2 (G L K, L ) H 2 (G L K, I L ) inv L K 1 [L:K] Z/Z 0. Alying the exact functor lim, by [3], III.3.6 (and roosition 4.10), we get 1 Br(K) H 2 (G Ω K, I Ω ) inv K Q/Z 0. Finally, we adjust the middle term via the isomorhism lim H2 (G L K, I L ) = lim H 2 (G LP K, L P ) = lim H2 (G LP K, L P ) = Br(K ). Theorem (Brauer-Hasse-Noether, classical version) Let K be a number field, and A an Azumaya algebra over K. Then A slits over K if and only if the localization A := A K K slits over K at all laces. Proof. : If A = M n (K), then A = Mn (K) K K = M n (K ). : We have to rove that the morhism (r K K) : Br(K) Br(K ), [A] ([A ]), is injective. By theorem 4.11, it suffices to show that the following diagram commutes: Br(L K) H 2 (G L K, L ) r K K Br(L P K ) H 2 (G P, L P ).

13 RECIPROCITY, BRAUER GROUPS AND QUADRATIC FORMS OVER NUMBER FIELDS 13 Namely, if we aly lim to it (and to the lower arrow), we obtain our result: (r K K ) Br(K) Br(K ) lim H2 (G L K, L ) lim H 2 (G L K, I L ). In order to show commutativity, recall that the horizontal isomorhisms are given in each dimension n N by mas Br(n, L K) ρ n H 1 (G L K, P n (L)) H 2 (G L K, L ) r K K Res H 1 (G P, P n (L)) Res H 2 (G P, L ) Br(n, L P K ) ι ρ n H 1 (G P, P n (L P )) H 2 (G P, L P ). κ First, we make sure that all r K K([A]) = [A ] Br(n, L P K ). Indeed, if we have an isomorhism f : M n (L) A K L, it induces an isomorhism f P : M n (L P ) = M n (L) L L P f L P A K L L L P = A K L P = A K L P. Considering M n (L P ) = M n (L) L L P, the ma ι is induced by P n (L) P n (L P ), g g L P, and the morhism κ simly comes from the inclusion L L P. It is then clear that both diagrams on the right are commutative. The 1-cocycle (ρ n r K K)([A]) is given at σ G P = G LP K by f 1 P σ f P σ 1, whereas (ι Res ρ n )([A]) sends σ G P to the ma (f 1 σ f σ 1 ) L L P. It suffices to show that these coincide on elements X M n (L P ) of the standard basis. To simlify the notation, assume wlog that f(x) = a β A K L. Of course, both σ and σ L L P fix X M n (K). After that, it is maed under both our endomorhisms in P n (L P ) as follows: M n (L) L L P f L P A K L L L P σ A K L L L P f 1 L P M n (L) L L P, X 1 a β 1 a σ(β) 1 X β 1 σβ. 5. Quadratic Forms Definition 5.1. Let K be a field, a, b K. The quaternion algebra A = defined to have basis {1, i, j, k}, such that i 2 = a, j 2 = b, and k = ij = ji. We denote by A 0 := san(i, j, k) the ure quaternions of A. ) Lemma 5.2. For a, b K, A = is an Azumaya algebra over K. ( a,b K a,b K over K is

14 14 THOMAS POGUNTKE Lemma 5.3. Let A = a,b K, a, b K. The following are equivalent: (i) A slits over K, (ii) A is not a division algebra, (iii) w, x, y, z K non-trivial: N(w + xi + yj + zk) := w 2 ax 2 by 2 + abz 2 = 0, (iv) (a, b) = 1, that is, ax 2 + by 2 = z 2 has a non-trivial solution over K. Proof. (i) (ii) is obvious. (ii) (i) : Let A = M n (D), as er Wedderburn s structure theorem. Then 4 = dim K A = dim K M n (D) = dim K (M n (K) K D) = n 2 dim K D, so n {1, 2}. But if n = 1, then A = D is a division algebra. Hence n = 2, i.e. A = M 2 (K). (ii) (iii) : To see this, we will show that α A is invertible if and only if N(α) 0. Namely, if α 1 A, then 1 = N(α 1 α) = N(α 1 )N(α), i.e. N(α) K. Conversely, if an element α = w + xi + yj + zk has norm N(α) = w 2 ax 2 by 2 + abz 2 = αα = c 0, where α := (w xi yj zk), then α is invertible: α(c 1 α) = 1. (i) (iv) : Let ϕ : M 2 (K) =: B A. Noting that B = 1,1 K, we see that certainly there exists 0 β B 0 with trivial norm N B (β) = 0. Then we also have N(ϕ(β)) = ϕ(β)ϕ(β) = ϕ(β)ϕ(β) = ϕ(ββ) = ϕ(n B (β)) = 0. Moreover, ϕ(β) A 0, otherwise c K with ϕ(β) c A 0, hence β c B 0. Thus, ax 2 by 2 + abz 2 = 0 for some non-trivial x, y, z K, and so a( y a )2 + b( x b )2 = z 2. (iv) (iii) : We get N(z + xi + yj) = z 2 ax 2 by 2 = 0. Corollary 5.4. Let K be an algebraic number field, A = a,b K, a, b K, then A = a,b K is a division algebra for a finite odd number of laces, and slits otherwise. Proof. The revious lemma imlies that A is a division algebra (a, b) = 1, and slits otherwise. Hence the statement is just Hilbert s recirocity law (a, b) = 1. Definition 5.5. (a) A quadratic form over K is a homogenous olynomial q K[x 1,..., x n ] of degree 2, that is, q(x 1,..., x n ) = a ij x i x j 1 i j n with a ij K. We can diagonalize q, so that q = n i=1 a ix 2 i, and we write q = a 1,..., a n. The dimension of q is simly defined to be dim q := n. (b) We say that q reresents some a K, if there are non-trivial y 1,..., y n K with q(y 1,..., y n ) = a. The form q is said to be isotroic, if it reresents 0. Proosition 5.6. (Global square theorem) Let K be an algebraic number field, x K. Then we have x (K ) 2 if and only if x (K ) 2 for all laces of K. Proof. is trivial, since K K. : Conversely, assume that x is not a square in K. Let L := K( x). For each rime, we have the fundamental equation 2 = [L : K] = P [L P : K ], but L P = K ( x) = K for all, hence all slit comletely. This is a harsh contradiction to theorem 1.4.

15 RECIPROCITY, BRAUER GROUPS AND QUADRATIC FORMS OVER NUMBER FIELDS 15 Theorem 5.7. (Hasse-Minkowski) Let K be an algebraic number field. A quadratic form q is isotroic over K if and only if q is isotroic over K at all laces. Proof. is trivial, since K K. : If dim q = 1, there is nothing to say. Let dim q = 2. Since ax 2 + by 2 (a, b 0) is isotroic if and only if x 2 + b a y2 is, we can assume q to be q = 1, a. Then by the global square theorem, we have: q isotroic over K a (K ) 2 a (K ) 2 for all q isotroic over K for all. Now let dim q = 3, and wlog q = 1, a, ( b, ) as above. Then if q is isotroic over all K, lemma 5.3 imlies that the algebra A = a,b K slits over K for all. ) By Brauer-Hasse-Noether, A = slits over K, so q is isotroic over K by lemma 5.3. ( a,b K Now let dim q = 4, and assume q = 1, a, b, c, as before. Let L := K( abc). If q is isotroic over K for all, of course the same is true over L P for all P. Since abc (L ) 2, we can assume q = 1, a, b, ab over L (and L P, resectively): for if abc = d 2 and w 2 ax 2 by 2 + abz 2 = 0, we have w 2 ax 2 by 2 + c(c 1 dz) 2 = 0. By lemma 5.3, a,b L P slits over L P for all P, hence Brauer-Hasse-Noether tells us that a,b L slits over L. Again by lemma 5.3, q is isotroic over L, and thus over K as well ([2], VII.3.1). Lastly, let dim q 5, and use induction over dim q. We decomose q = q 0 q 1 orthogonally (cf. [4], I.3.13), such that q 0 = a, b and dim q 1 3. Since for almost all finite laces (with char k() 2), the coefficients of q 1 are units, q 1 is isotroic there ([4], VI.2.6). Hence, there are only finitely many laces 1,..., r, at which q 1 is not isotroic. But because q is isotroic there, we have c i K i, which are reresented by both q 1 and q 0. This means that we have equations ax 2 i + byi 2 = c i for some x i, y i K i. By the strong aroximation theorem, there are x, y K with x i x i, y i y i arbitrarily small for all i. Set c := ax 2 +by 2, then c i c i and therefore c 1 c i 1 i and 1 c 1 i c i get arbitrarily small. This means that c 1 c i U n := 1 + n i U K i or c 1 i c U n, resectively, for n N arbitrarily large. Now [3], II.3.5 tells us that we can find some n N such that U n (K i ) 2, from whence we get c i c K i /(K i ) 2 (this is sometimes referred to as the Local Square Theorem). Altogether, we see that c q 1 is everywhere locally isotroic, thus by induction globally isotroic, and by the definition of c, so is then q. References [1] I. Kersten, Brauergruen. Universitätsverlag Göttingen, [2] T. Y. Lam, Introduction to Quadratic Forms over Fields. Graduate Studies in Mathematics 67, AMS, [3] J. Neukirch, Klassenkörertheorie. Elektronische Ausgabe, herausgegeben von Alexander Schmidt, [4] W. Scharlau, Quadratic and Hermitian Forms. Grundlehren der mathematischen Wissenschaft 270, Sringer-Verlag, [5] J.-P. Serre, A Course in Arithmetic. Graduate Texts in Mathematics, Sringer-Verlag, [6] J.-P. Serre, Local Fields. Graduate Texts in Mathematics, Sringer-Verlag, 1979.