Notes to Cross Product Algebras [version 4 May 2018 ed.]

Table of contents :
1. A cross product algebra is central simple......Page 1
2. A cross product algebra is a Brauer representative......Page 4
3. Multiplication of cross product algebras......Page 9
4. The period-index problem......Page 14
References......Page 18

Citation preview

NOTES TO CROSS PRODUCT ALGEBRAS ZHENGYAO WU (吴正尧)

Abstract. Lecture notes for my talk on 4 May 2018 at Shantou University. Main reference: [Her68, Ch. 4, Sec. 4].

Contents 1.

A cross product algebra is central simple

1

2.

A cross product algebra is a Brauer representative

4

3.

Multiplication of cross product algebras

9

4.

The period-index problem

14

References

18

1. A cross product algebra is central simple Notation 1.1. Let F be a field. Let K/F be a Galois extension. Let K ∗ = K − {0} be the multiplicative group of K. The multiplication of the Galois group G = Gal(K/F ) is the composition of K/F -automorphisms. For x ∈ K ∗ and σ ∈ G, we write xσ for the image of x under σ. Definition 1.2. A map f : G × G → K ∗ is called a factor set if f (σ1 , σ2 σ3 )f (σ2 , σ3 ) = f (σ1 σ2 , σ3 )f (σ1 , σ2 )σ3 , ∀σ1 , σ2 , σ3 ∈ G. Lemma 1.3. Let f : G × G → K ∗ be a factor set. Then (1) f (σ, 1) = f (1, 1) for all σ ∈ G. (2) f (1, σ) = f (1, 1)σ for all σ ∈ G. Proof. (1) Let σ1 = σ, σ2 = σ3 = 1 in Definition 1.2. Then f (σ, 1)f (1, 1) = f (σ, 1)f (σ, 1) and hence f (σ, 1) = f (1, 1). (2) Let σ1 = σ2 = 1, σ3 = σ in Definition 1.2. σ

σ

f (1, σ)f (1, 1) and hence f (1, σ) = f (1, 1) . 1

Then f (1, σ)f (σ, 1) = 

2

ZHENGYAO WU (吴正尧)

Definition 1.4. Let f : G × G → K ∗ be a factor set. An algebra A over F is called a cross product if it has a basis (eσ )σ∈G over K such that (1) xeσ = eσ xσ for all x ∈ K and σ ∈ G. (2) eσ1 eσ2 = eσ1 σ2 f (σ1 , σ2 ) for all σ1 , σ2 ∈ G. We write A = (K, G, f ). Lemma 1.5. Let A = (K, G, f ) be a cross product. Then e1 f (1, 1)−1 is the unit element of A. Proof. First, eσ · (e1 f (1, 1)−1 ) = eσ for all σ ∈ G. eσ e1 f (1, 1)−1 = eσ f (σ, 1)f (1, 1)−1 ,

by Definition 1.4(2).

= eσ ,

by Lemma 1.3(1).

Also, (e1 f (1, 1)−1 ) · eσ = eσ for all σ ∈ G. e1 f (1, 1)−1 eσ = e1 eσ (f (1, 1)−1 )σ ,

by Definition 1.4(1).

−1 σ

= eσ f (1, σ)(f (1, 1) σ

= eσ f (1, 1) (f (1, 1) = eσ (f (1, 1)f (1, 1)

) ,

−1 σ

) ,

by Definition 1.4(2). by Lemma 1.3(2).

−1 σ

) = eσ . 

Lemma 1.6. Let A = (K, G, f ) be a cross product. Then eσ is invertible for all σ ∈ G. −1 −1 Proof. We show that e−1 ) f (1, 1)−1 for all σ ∈ G. σ = eσ −1 f (σ, σ

First, eσ eσ−1 f (σ, σ −1 )−1 f (1, 1)−1 = e1 f (σ, σ −1 )f (σ, σ −1 )−1 f (1, 1)−1 ,

by Definition 1.4(2).

−1

= e1 f (1, 1)

Let σ1 = σ3 = σ and σ2 = σ −1 in Definition 1.2. Then f (σ, 1)f (σ −1 , σ) = f (1, σ)f (σ, σ −1 )σ . Apply Lemma 1.3, we obtain and call the following ? f (1, 1)f (σ −1 , σ) = f (1, 1)σ f (σ, σ −1 )σ .

NOTES TO CROSS PRODUCT ALGEBRAS

3

Then we have eσ−1 f (σ, σ −1 )−1 f (1, 1)−1 eσ = eσ−1 eσ (f (σ, σ −1 )−1 )σ (f (1, 1)−1 )σ , = e1 f (σ

−1

= e1 f (σ

−1

by Definition 1.4(1).

, σ)(f (σ, σ

−1 −1 σ

−1 σ

, σ)(f (σ, σ

−1 σ −1

σ −1

)

) (f (1, 1)

) )

(f (1, 1) )

−1

f (1, 1) (f (1, 1) )

−1

.

= e1 f (1, 1)

= e1 f (1, 1)

σ −1

σ

) ,

,

by Definition 1.4(2)

, by ? 

Lemma 1.7. Let A = (K, G, f ) be a cross product. Then A is associative. Proof. We show that (eσ1 eσ2 )eσ3 = eσ1 (eσ2 eσ3 ) for all σ1 , σ2 , σ3 ∈ G. (eσ1 eσ2 )eσ3 = eσ1 σ2 f (σ1 , σ2 )eσ3 ,

by Definition 1.4(2).

σ3

= eσ1 σ2 eσ3 f (σ1 , σ2 ) ,

by Definition 1.4(1). σ3

= eσ1 σ2 σ3 f (σ1 σ2 , σ3 )f (σ1 , σ2 ) ,

by Definition 1.4(2).

= eσ1 σ2 σ3 f (σ1 , σ2 σ3 )f (σ2 , σ3 ),

by Definition 1.2.

= eσ1 eσ2 σ3 f (σ2 , σ3 ),

by Definition 1.4(2).

= eσ1 (eσ2 eσ3 ),

by Definition 1.4(2). 

Lemma 1.8. Let A = (K, G, f ) be a cross product. Then A is central. Proof. We show that the center Z(A) of A is F e1 f (1, 1)−1 . We have F e1 f (1, 1)−1 ⊂ Z(A) since e1 f (1, 1)−1 a = ae1 f (1, 1)−1 for all a ∈ A. P Conversely, suppose a = eσ aσ ∈ Z(A) for aσ ∈ K. Then for all x ∈ K, we σ∈G

have X

xa =

xeσ aσ =

σ∈G

X

eσ xσ aσ , by Definition 1.4(1).

σ∈G

ax =

X

eσ aσ x.

σ∈G

Then (xσ − x)aσ = 0 for all σ ∈ G. If aσ 6= 0, then xσ = x for all σ ∈ G, i.e. σ = 1. Hence a = e1 a1 . Next for all σ ∈ G, aeσ = e1 a1 eσ = e1 eσ aσ1 , = =

eσ f (1, σ)aσ1 , eσ f (1, 1)σ aσ1 ,

by Definition 1.4(1). by Definition 1.4(2). by Lemma 1.3(2).

4

ZHENGYAO WU (吴正尧)

Also eσ a = eσ e1 a1 = eσ f (σ, 1)a1 ,

by Definition 1.4(2).

= eσ f (1, 1)a1 ,

by Lemma 1.3(1).

Since aeσ = eσ a for all σ ∈ G, we have (f (1, 1)a1 )σ = f (1, 1)a1 for all σ ∈ G. Thus f (1, 1)a1 ∈ F and a = e1 f (1, 1)−1 f (1, 1)a1 ∈ F e1 f (1, 1)−1 . Therefore Z(A) ⊂ F e1 f (1, 1)−1 .



Lemma 1.9. Let A = (K, G, f ) be a cross product. Then A is simple. Proof. Let I be a two-sided ideal of A. Suppose I 6= 0. Take 0 6= a ∈ A, P a= eσ aσ with the shortest length |{σ ∈ G | aσ 6= 0}| > 0. If a1 = 0, then σ∈G

aτ 6= 0 for some 1 6= τ ∈ G. By Definition 1.4(2), eτ −1 eτ aτ = e1 f (τ −1 , τ )aτ . Then coefficient of e1 in eτ −1 a is f (τ −1 , τ )aτ 6= 0. We replace a by eτ −1 a. Assume that a1 6= 0. Then for all x ∈ K, P P xa − ax = x eσ aσ − eσ aσ x σ∈G σ∈G P = eσ (xσ − x)aσ ,

by Definition 1.4(1).

σ∈G

The coefficient of e1 in xa − ax is (x1 − x)a1 = 0. The coefficient of eσ is (xσ − x)aσ = 0 if aσ = 0. Thus xa − ax is shorter than a, a contradiction. Therefore I = 0.



Theorem 1.10. If A = (K, G, f ) be a cross product, then A is a unital associative central simple algebra over F such that dimF A = |G|2 , if G is finite. Proof. It follows from Lemma 1.5, Lemma 1.6, Lemma 1.7, Lemma 1.8 and Lemma 1.9. Also, dimK A = |G| = [K : F ], hence dimF A = (dimK A)[K : F ] = |G|2 .



2. A cross product algebra is a Brauer representative Definition 2.1. Two central simple F -algebras A, B are Brauer equivalent if Mm (A) ' Mn (B) for some integers m, n > 0. It is an equivalence relation on the set of isomorphism classes of central simple algbras. The Brauer group of F is an abelian group such that (1) The underlying set Br(F ) is the quotient of the set of isomorphism classes of central simple algbras modulo Brauer equivalence.

NOTES TO CROSS PRODUCT ALGEBRAS

5

(2) The addition [A] + [B] = [A ⊗F B] is well-defined by [DK94, Cor. 4.3.6]. It is associative and commutative since ⊗ is so . (3) The element 0 = [F ] = [Mn (F )] for all n ≥ 1. (4) The inverse −[A] = [Aop ] where Aop is the opposite algebra of A.

Review 2.2. Wedderburn’s theorem (See [DK94, Th. 2.4.3]). Let A be a central simple algebra over a field F . Then A ' Mk (D) for a central division algebra D over F and an integer k ≥ 1. Then two central simple algebra are Brauer equivalent over F iff they have isomorphic underlying central division algebras.

Review 2.3. Skolem-Noether theorem (see [DK94, Th. 4.4.1]) Let A be a central simple algebra. Let B be a simple algebra. If f, g : B → A are two homomorphisms, then there exists a ∈ A∗ such that g(b) = af (b)a−1 for all b ∈ B.

Lemma 2.4. Noether. This is [DK94, Lem. 5.6.1]. A finite dimensional central division algebra D over F has a separable maximal subfield.

Proof. For char F = 0, every extension of F is separable, the statement is trivial.

For char F = p > 0, we show that there exists a separable element in D − F . Take a ∈ D − F with minimal polynomial f (X) ∈ F [X]. (1) If a is separable, then we are done. (2) If a is not separable, then f (X) = g(X p ) for some g(X) ∈ F [X] and deg f deg g = < deg(f ). Since dimF D < ∞ there exists an integer m > 0 such p m+1 m that b = ap ∈ D − F is non-separable and bp = ap is separable. (2.1) If bp 6∈ F , then we are done. (2.2) If bp ∈ F , then there exists d0 ∈ D such that d0 b 6= bd0 and d0 bp = bp d0 since F = Z(D). Let δ : D → D be the map such that δ(d) = db − bd. Then n
P δ n (d) = (−1)i ni bi dbn−i . We have δ(d0 ) 6= 0 and δ p (d0 ) = d0 bp − bp d0 = 0. i=0

Let m ≤ p be the least integer such that δ m (d0 ) = 0. Let δ m−1 (d0 ) = t and

6

ZHENGYAO WU (吴正尧)

δ m−1 (d0 ) = w. Then b = tt−1 b = (wb − bw)t−1 b, −1

= wbt = wt

b

−1

b,

b − bwt

−1 2

since t = δ(w) = bw − wb.

−1

b − bwt

since bt = tb =⇒ bt−1 = t−1 b. let c = wt−1 b.

= cb − bc,

Hence c = 1 + bcb−1 . We have c 6∈ F otherwise b = 0. (2.2.1) If c is separable, then we are done. (2.2.2) If c is not separable, then similar to (2), there exists n ≥ 1 such that n

n

n

n

cp is separable. Since cp = 1 + bcp b−1 , we have cp 6∈ F .



Theorem 2.5. Every finite dimensional central simple algebra A over a field F is Brauer equivalent a cross product algebra (K, G, f ) over F . In particular [A] = [(K, G, f )] in Br(F ). Proof. Suppose A ' Mk (D) for some central division algebra D by Review 2.2. Let L be a maximal separable subfield of D. Let K be the Galois closure of L, i.e. the splitting field of a where L = F (a) by [DK94, Cor. 5.5.6]. Suppose [K : L] = m and B = Mm (D). Then A is Brauer equivalent to B and hence [A] = [B] in Br(F ). Let [L : F ] = n. Then dimF D = n2 , [K : F ] = mn and dimF B = m2 n2 . Then K is a maximal subfield of B. Let G = Gal(K/F ). Then |G| = [K : F ] = mn For every K/F automorphism σ ∈ G, by Review 2.3, there exists eσ ∈ B ∗ such that xσ = e−1 σ xeσ for all x ∈ K. Then Definition 1.4(1) is verified and P { eσ aσ | aσ ∈ K} is a F -subspace of B of dimension m2 n2 . Since dimF B = σ∈G

m2 n2 , this subspace is all of B. Next, elements (eσ )σ∈G are linearly independent over K. Suppose

P

eσ aσ = 0.

σ∈G

We need to show that aσ = 0 for all σ ∈ G. We may assume that K 6= F . Then |G| > 1. Suppose to the contrary that P eσ aσ = 0 with the shortest length |{σ ∈ G | aσ 6= 0}| > 0. Similar to the σ∈G

proof of Lemma 1.9, we may assume that a1 6= 0. Since e1 a1 6= 0, there exists 1 6= τ ∈ G such that aτ 6= 0. Since τ 6= 1, there exists x ∈ K not fixed by τ . Similar to the proof of Lemma 1.9, xa−ax is shorter than a and has a nonzero summand eτ (xτ −x)aτ , a contradiction. Thus (eσ )σ∈G is a basis of B over K.

NOTES TO CROSS PRODUCT ALGEBRAS

7

Next, we show the existence of a factor set. For all x ∈ K and σ1 , σ2 ∈ G, σ1 σ1 σ2 = e−1 e−1 σ1 σ2 eσ1 eσ2 x σ1 σ2 eσ1 x eσ2 ,

σ1 since xσ1 σ2 = e−1 σ 2 x eσ 2 .

= e−1 σ1 σ2 xeσ1 eσ2 ,

since xσ1 = e−1 σ1 xeσ1 .

= xσ1 σ2 e−1 σ1 σ2 eσ1 eσ2 ,

since xσ1 σ2 = e−1 σ1 σ2 xeσ1 σ2 .

∗ ∗ Hence e−1 σ1 σ2 eσ1 eσ2 ∈ CB (K) = K, so it is in B ∩ K = K . Suppose f (σ1 , σ2 ) =

e−1 σ1 σ2 eσ1 eσ2 . Then eσ1 σ2 f (σ1 , σ2 ) = eσ1 eσ2 and Definition 1.4(2) is verified. We need to show that f is a factor set. The associativity gives eσ1 (eσ2 eσ3 ) = (eσ1 eσ2 )eσ3 where eσ1 (eσ2 eσ3 ) = eσ1 eσ2 σ3 f (σ2 , σ3 ) = eσ1 σ2 σ3 f (σ1 , σ2 σ3 )f (σ2 , σ3 ), (eσ1 eσ2 )eσ3 = eσ1 σ2 f (σ1 , σ2 )eσ3 = eσ1 σ2 eσ3 f (σ1 , σ2 )σ3 = eσ1 σ2 σ3 f (σ1 , σ2 )f (σ1 , σ2 )σ3 . Therefore f (σ1 , σ2 σ3 )f (σ2 , σ3 ) = f (σ1 , σ2 )f (σ1 , σ2 )σ3 and B ' (K, G, f ) is a cross product.



Definition 2.6. Let f, f 0 : G × G → K ∗ be two factor sets. We say that f and f 0 are equivalent if there exists λ : G → K ∗ such that f 0 (σ1 , σ2 ) =

λ(σ1 )σ2 · λ(σ2 ) f (σ1 , σ2 ) λ(σ1 σ2 )

for all σ1 , σ2 ∈ G. Proof. We show that if f is a factor set, then so is f 0 . For all σ1 , σ2 , σ3 ∈ G,

= = = = =

f 0 (σ1 , σ2 σ3 )f 0 (σ2 , σ3 ) λ(σ1 )σ2 σ3 · λ(σ2 σ3 ) λ(σ2 )σ3 · λ(σ3 ) f (σ1 , σ2 σ3 ) f (σ2 , σ3 ), λ(σ1 σ2 σ3 ) λ(σ2 σ3 ) σ2 σ3 σ3 λ(σ1 ) · λ(σ2 σ3 ) · λ(σ2 ) · λ(σ3 ) f (σ1 , σ2 σ3 )f (σ2 , σ3 ) λ(σ1 σ2 σ3 ) · λ(σ2 σ3 ) σ2 σ3 σ3 λ(σ1 ) · λ(σ2 ) · λ(σ3 ) f (σ1 σ2 , σ3 )f (σ1 , σ2 )σ3 , λ(σ1 σ2 σ3 ) λ(σ1 σ2 )σ3 · λ(σ3 ) λ(σ1 )σ2 σ3 · λ(σ2 )σ3 f (σ1 σ2 , σ3 ) f (σ1 , σ2 )σ3 λ(σ1 σ2 σ3 ) λ(σ1 σ2 )σ3 f 0 (σ1 σ2 , σ3 )f 0 (σ1 , σ2 )σ3 ,

Hence f 0 is a factor set by Definition 1.2.

by Definition 2.6.

by Definition 1.2.

by Definition 2.6. 

Lemma 2.7. Two cross product algebras (K, G, f ) ' (K, G, f 0 ) iff f and f 0 are equivalent. In particular, “equivalent” as in Definition 2.6 is an equivalence relation.

8

ZHENGYAO WU (吴正尧)

Proof. Let (eσ )σ∈G be a K-basis of A = (K, G, f ) and let (e0σ )σ∈G be a Kbasis of A0 = (K, G, f 0 ). Suppose there is an F -isomorphism Ψ : A0 → A. Since Ψ(Ke01 f 0 (1, 1)−1 ) is a 1-dimensional vector space over K, there exists an automorphism Ψ0 of A such that Ψ0 ◦ Ψ(e01 f 0 (1, 1)−1 K) = e1 f (1, 1)−1 K. Here Φ0 is inner by Review 2.3. Let Ψ00 be an automorphism of A extending a K/F -automorphism such that Ψ00 ◦ Ψ0 ◦ Ψ(e01 f 0 (1, 1)−1 x) = e1 f (1, 1)−1 x for all x ∈ K. Let Φ = Ψ00 ◦ Ψ0 ◦ Ψ. By Lemma 1.5, we simply write Φ(x) = x for all x ∈ K and we call this property ?. 0 Next, we show that e−1 σ Φ(eσ ) ∈= K. For all x ∈ K, 0 σ e−1 σ Φ(eσ )x 0 σ = e−1 σ Φ(eσ x ),

= = =

0 e−1 σ Φ(xeσ ), 0 e−1 σ xΦ(eσ ), xσ eσ−1 Φ(e0σ ),

by ? by Definition 1.4(1). by ? . by xeσ = eσ xσ .

0 0 Hence e−1 σ Φ(eσ ) ∈ CA (K) = K and Φ(eσ ) = eσ λ(σ) for some λ(σ) ∈ K. Then

Φ(e0σ1 )Φ(e0σ2 ) = eσ1 λ(σ1 )eσ2 λ(σ2 ) ⇐⇒

Φ(e0σ1 e0σ2 ) = eσ1 eσ2 λ(σ1 )σ2 · λ(σ2 ),

by Definition 1.4(1)

⇐⇒

Φ(e0σ1 σ2 f 0 (σ1 , σ2 )) = eσ1 σ2 f (σ1 , σ2 )λ(σ1 )σ2 · λ(σ2 ), eσ1 σ2 λ(σ1 σ2 )f 0 (σ1 , σ2 ) = eσ1 σ2 f (σ1 , σ2 )λ(σ1 )σ2 · λ(σ2 ),

by Definition 1.4(2)

⇐⇒ ⇐⇒ ⇐⇒

0

σ2

λ(σ1 σ2 )f (σ1 , σ2 ) = f (σ1 , σ2 )λ(σ1 ) · λ(σ2 ), λ(σ1 )σ2 · λ(σ2 ) f 0 (σ1 , σ2 ) = f (σ1 , σ2 ). λ(σ1 σ2 )

by ? . by Lemma 1.6.

Conversely, suppose f and f 0 are equivalent by λ. Then Φ : A0 → A such that P 0 P Φ( eσ aσ ) = eσ λ(σ)aσ is an isomorphism of cross product algebras.  σ∈G

σ∈G

Definition 2.8. Let g : G × G → K ∗ be a factor set such that g(1, σ) = g(σ, 1) = 1. Then we call g a normalized factor set. Lemma 2.9. Every factor set is equivalent to a normalized one. Proof. Let f be a factor set. Let λ : G → K ∗ be a map such that λ(σ) = 1 λ(σ1 )σ2 · λ(σ2 ) . Let g(σ , σ ) = f (σ1 , σ2 ). Then f is equivalent to g. 1 2 f (1, 1)σ λ(σ1 σ2 )

NOTES TO CROSS PRODUCT ALGEBRAS

9

We need to show that g is normalized. By definition, g(σ1 , σ2 ) =

1 σ2 ( f (1,1) · σ1 )

1 f (1,1)σ2

1 f (1,1)σ1 σ2

f (σ1 , σ2 ) =

f (σ1 , σ2 ) . f (1, 1)σ2

By Lemma 1.3(1), we have f (σ, 1) f (1, 1) = = 1. 1 f (1, 1) f (1, 1)

g(σ, 1) = By Lemma 1.3(2), we have g(1, σ) =

f (1, 1)σ f (1, σ) = = 1. f (1, 1)σ f (1, 1)σ

Therefore g is normalized.



Remark 2.10. The set of factor sets form an abelian group, called Z 2 with • multiplication (f g)(σ1 , σ2 ) = f (σ1 , σ2 )g(σ1 , σ2 ), • unit element 1(σ1 , σ2 ) = 1, • inverse f −1 (σ1 , σ2 ) = (f (σ1 , σ2 ))−1 , for all σ1 , σ2 ∈ G. It has a normal subgroup formed of factor sets equivalent to 1, called B 2 = {g ∈ Z 2 (G, K ∗ ) | g(σ1 , σ2 ) =

λ(σ1 )σ2 λ(σ2 ) for some λ : G → K ∗ .} λ(σ1 σ2 )

By Definition 1.2 and Definition 2.6, Z 2 /B 2 is the set of equivalence classes of factor sets. By Definition 1.4 and Lemma 2.7 it corresponds bijectively to the set of isomorphism classes of cross product algebras and hence a subset of Br(F ). In the next section, we will show that Z 2 /B 2 ' Br(F ) as abelian groups. 3. Multiplication of cross product algebras Lemma 3.1. Let [K : F ] = n. Then (K, G, 1) ' Mn (F ). Proof. Let (eσ )σ∈G be a K-basis of A = (K, G, 1). Define ϕ : A → (EndF K)op P P σ such that ϕ( eσ aσ )(x) = x aσ for all aσ , x ∈ K. First, ϕ is a homomorσ∈G

σ∈G

phism. ϕ(eσ1 eσ2 )(x) = ϕ(eσ1 σ2 )(x), = xσ1 σ2 = (xσ1 )σ2 = (ϕ(eσ1 )(x))σ2 = ϕ(eσ2 ) ◦ ϕ(eσ1 )(x)

by Definition 1.4(2).

10

ZHENGYAO WU (吴正尧)

Injectivity follows from Lemma 1.9. By Theorem 2.5, dimF A = n2 = dimF (EndF K), then ϕ is an isomorphism. Finally, A ' EndF K op ' Mn (F ).



Lemma 3.2. Let B an F -algebra. Suppose Mm (F ) ⊂ B. Then B ' Mm (CB (Mm (F ))) where CB (Mm (F )) denotes the centralizer of Mm (F ) in B. Proof. Let eij ∈ Mm (F ) whose (i, j)-th entry is 1 and all other entries are 0. We show that ϕ : B → Mm (CB (Mm (F ))), ϕ(b) = (bij )1≤i,j≤m such that m P bij = eki bejk is an isomorphism. k=1

First, we show that ϕ is well-defined, i.e. bij ∈ CB (Mm (F )) for all 1 ≤ i, j ≤ m. We verify that bij est = est bij for all 1 ≤ s, t ≤ m. bij est =

m X

eki bejk est =

m X

k=1

est bij = est

eki bejt δks = esi bejt ,

k=1

m X

eki bejk =

k=1

m X

esi δtk bejk = esi bejt .

k=1

Next, the inverse of ϕ is ϕ−1 ((bij )1≤i,j≤m ) =

m P

bij eij .

i,j=1 m P

ϕ−1 (ϕ(b)) =



m P

 eki bejk eij

i,j=1 k=1 m P m P

=

eki bejj δki

i,j=1 k=1 m P

=

i,j=1 m P

=

eii bejj  m  P eii b ejj = b

i=1

ϕ(ϕ−1 ((bij )1≤i,j≤m )) =

m P

= =

m P

eks bij

m P

i,j=1

k=1

m P

m P

bij eij

k=1

etk !

1≤s,t≤m

eks eij etk !1≤s,t≤m

eks eik δjt i,j=1 k=1 m  m P P ekk δsi = bit i=1

bij

!

!

i,j=1

k=1 m P

i=1

= 1≤s,t≤m 

1≤s,t≤m

=

bst

m P

m P eks eik bit i=1 k=1  m P ekk

k=1

 1≤s,t≤m

1≤s,t≤m

= (bst )1≤s,t≤m = (bij )1≤i,j≤m 

NOTES TO CROSS PRODUCT ALGEBRAS

11

Lemma 3.3. Let A and B be algebras over F . Let A0 ⊂ A and B 0 ⊂ B be subalgebras over F . Then CA⊗B (A0 ⊗ B 0 ) = CA (A0 ) ⊗ CB (B 0 ). For its proof see [Sch85, Ch. 8, Th. 3.2(i)]. Theorem 3.4. Let B an F -algebra. Suppose A is a finite dimensional central simple algebra over F such that A ⊂ B. Then B ' A ⊗F CB (A) where CB (A) denotes the centralizer of A in B. Proof. Suppose A is a central simple algebra of dimension m over F , we have A ⊗ Aop ' Mm (F ) (see [DK94, Th. 4.3.1]). Since A ⊂ B, A ⊗ Aop ⊂ B ⊗ Aop = E. It follows from Lemma 3.2 that E ' CE (Mm (F )) ⊗ Mm (F ) ' CE (Mm (F )) ⊗ A ⊗ Aop . By Lemma 3.3, CE (Aop ) = CE (1 ⊗ Aop ) ' CCE (Mm (F ))⊗A (1) ⊗ CAop (Aop ) ' CE (Mm (F )) ⊗ A. Similarly, it follows from E = B ⊗ Aop that CE (Aop ) ' B. By Lemma 3.3 again, CB⊗Aop (A ⊗ Aop ) ' CB (A) ⊗F CAop (Aop ) ' CB (A) ⊗F F ' CB (A). Therefore B ' CE (Mm (F )) ⊗ A = CB⊗Aop (A ⊗ Aop ) ⊗ A ' CB (A) ⊗ A.



Lemma 3.5. Let K/F be a finite Galois extension with G = Gal(K/F ). Then L L (1) K ⊗F K ' iσ (K ⊗ 1) ' iσ (1 ⊗ K). σ∈G

σ∈G

(2) i2σ = iσ . (3) iσ is minimal, i.e. not the sum of two other nonzero idempotents. (4) iσ1 iσ2 = 0 for all σ1 6= σ2 in G. (5) iσ (xσ ⊗ 1) = iσ (1 ⊗ x) for all x ∈ K. (6) G acts on (iσ )σ∈G by iτσ = iτ σ for all σ, τ ∈ G. Proof. Since K/F is finite separable, by [DK94, Cor. 5.5.6], there exists a ∈ K − F such that K = F (a). Let f (X) ∈ F [X] be the minimal polynomial of a. Then K ' F [X]/(f (X)). Since K/F is normal, f (X) is split over K Q and hence f (X) = (X − aσ ). By Chinese Remainder Theorem, we have an σ∈G

isomorphism ϕ : K ⊗F K ' K[X]/(f (X)) '

M σ∈G

K[X]/(X − aσ ) '

M σ∈G

K.

12

ZHENGYAO WU (吴正尧)

such that ϕ(x ⊗ y) = (xy σ )σ∈G . where iσ ∈ K ⊗ K and ϕ(iσ ) is the element L of K whose σ-th element is 1 and other elements are 0, it is a minimal σ∈G

idempotent since ϕ(iσ ) is a minimal idempotent and iσ iτ = 0 for σ 6= τ since ϕ(iσ )ϕ(iτ ) = 0. Finally, iσ (x⊗1) = iσ (1⊗xσ ) for all x ∈ K, because ϕ(iσ (xσ ⊗1)) = ϕ(iσ (1⊗x)) L is the element of K whose σ-th element is xσ and other elements are 0.  σ∈G

Lemma 3.6. Let A be a central simple algebra over F . Let i be an idempotent of A. Then [A] = [iAi] in Br(F ). Proof. We have A ' Mm (D) for some central division algebra D over F (see Review 2.2). Since i2 = i, the ! image of i in Mm (D) has eigenvalues in {0, 1}, Ir 0 . Then so it is isomorphic to 0 0 iAi '

!

Ir

0

0

0

Mm (D)

!

Ir

0

0

0

! Mr (D) 0

'

0

0

' Mr (D).

Therefore [iAi] = [Mr (D)] = [Mm (D)] = [A].



Theorem 3.7. Let K/F be a finite Galois extension with G = Gal(K/F ). Let f, f 0 : G × G → K ∗ be factor sets. Then [(K, G, f )][(K, G, f 0 )] = [(K, G, f f 0 )] in Br(F ). In other words, (K, G, f ) ⊗ (K, G, f 0 ) and (K, G, f f 0 ) are Brauer equivalent. Proof. Let (eσ )σ∈G be a K-basis of (K, G, f ) and let (e0σ )σ∈G be a K-basis of (K, G, f 0 ). Let E = (K, G, f ) ⊗ (K, G, f 0 ) and wσ = eσ ⊗ e0σ . (1) We choose an idempotent i ∈ E such that iwσ = wσ i = iwσ i. Since E ⊃ K ⊗ K, by Lemma 3.5(2)(5), there exist iσ such that iσ (x ⊗ 1) = iσ (1 ⊗ xσ ) for all x ∈ K. Let i = i1 . Then i(x ⊗ 1) = i(1 ⊗ x). Since i2 = i, (1 ⊗ e0 σ )i(1 ⊗ e0 −1 σ ) is also an idempotent of E. For all x ∈ K, σ (1 ⊗ e0 σ )i(1 ⊗ e0 −1 σ )(x ⊗ 1)

= (1 ⊗ e0 σ )i(xσ ⊗ 1)(1 ⊗ e0 −1 σ ) = (1 ⊗ e0 σ )i(1 ⊗ xσ )(1 ⊗ e0 −1 σ ),

by i(xσ ⊗ 1) = i(1 ⊗ xσ ).

= (1 ⊗ e0 σ )i(1 ⊗ e0 −1 σ )(1 ⊗ x),

by xe0σ = e0σ xσ .

By Lemma 3.5(1)(5), (1 ⊗ e0 σ )i(1 ⊗ e0 −1 σ ) = iσ (we call it ?) because −1

τ

−1

τ

−1

ϕ((1⊗e0 σ )i(1⊗e0 σ )) = (e0 σ )τ ∈G ϕ(i)((e0 σ )τ )τ ∈G = ϕ(iσ )(e0 σ )τ ∈G ((e0 σ )τ )τ ∈G = ϕ(iσ ).

NOTES TO CROSS PRODUCT ALGEBRAS

13

Similarly, (e−1 σ ⊗ 1)i(eσ ⊗ 1) = iσ . Then iwσ = i(eσ ⊗e0σ ) = i(eσ ⊗1)(1⊗e0σ ) = (eσ ⊗1)iσ (1⊗e0σ ) = (eσ ⊗1)(1⊗e0σ )i = (eσ ⊗e0σ )i = wσ i. Thus iwσ i = wσ i2 = wσ i. (2) Next, we construct a cross product algebra (i(K ⊗ 1), G, i(f ⊗ 1)(f 0 ⊗ 1)), which is a subalgebra of iEi. Take uσ = iwσ = iwσ i ∈ iEi. The underlying vector space of (i(K ⊗1), G, i(f ⊗ P 1)(g ⊗ 1)) is uσ i(K ⊗ 1). σ∈G

First, we verify Definition 1.4(1). For all x ∈ K i(x ⊗ 1)uσ = i(x ⊗ 1)i(eσ ⊗ e0σ ), = i(x ⊗ 1)(eσ ⊗

e0σ ),

since uσ = iwσ . since (x ⊗ 1)i = i(x ⊗ 1), i2 = i.

= i(xeσ ⊗ e0σ ) = i(eσ xσ ⊗ e0σ ), = i(eσ ⊗

e0σ )(xσ

σ

= uσ i(x ⊗ 1),

by Definition 1.4(1). ⊗ 1) since uσ = iwσ = iwσ i.

Next, we verify Definition 1.4(2). uσ1 uσ2 = iwσ1 iwσ2 since wσ1 i = iwσ1 , i2 = i.

= iwσ1 wσ2 , = i(eσ1 eσ2 ⊗ e0σ1 e0σ2 ) = i(eσ1 σ2 f (σ1 , σ2 ) ⊗ e0σ2 σ2 f 0 (σ1 , σ2 )), = =

by Definition 1.4(2).

i(eσ1 σ2 ⊗ e0σ2 σ2 )(f (σ1 , σ2 )f 0 (σ1 , σ2 )) iwσ1 σ2 (f (σ1 , σ2 ) ⊗ f 0 (σ1 , σ2 ))

= wσ1 σ2 i(f (σ1 , σ2 ) ⊗ f 0 (σ1 , σ2 )),

since iwσ1 σ2 = wσ1 σ2 i.

= wσ1 σ2 i(f (σ1 , σ2 )f 0 (σ1 , σ2 ) ⊗ 1),

since i(1 ⊗ f 0 (σ1 , σ2 )) = i(f 0 (σ1 , σ2 ) ⊗ 1).

(3) Now we show that iEi is a subalgebra of (i(K ⊗ 1), G, i(f ⊗ 1)(g ⊗ 1)). It P suffices to show generators i(eσ1 x ⊗ e0σ2 x0 )i ∈ uσ e(K ⊗ 1), where x, x0 ∈ K. σ∈G

i(eσ1 x ⊗ e0σ2 x0 )i = i(eσ1 ⊗ e0σ1 )(1 ⊗ (e0σ1 )−1 e0σ2 )(x ⊗ x0 )i = i(eσ1 ⊗ e0σ1 )(1 ⊗ (e0σ1 )−1 e0σ2 )i(x ⊗ x0 ),

since (x ⊗ x0 )i = i(x ⊗ x0 ).

= uσ1 i(1 ⊗ (e0σ1 )−1 e0σ2 )i(x ⊗ x0 ),

since uσ1 = iwσ1 i = uσ1 i.

= uσ1 i(1 ⊗

(e0σ1 )−1 e0σ2 )i(xx0

⊗ 1),

since i(1 ⊗ x0 ) = i(x0 ⊗ 1) in (1).

For σ1 6= σ2 , by ? in (1) and Lemma 3.5(4), we have i(1 ⊗ (e0σ1 )−1 e0σ2 )i = i(1 ⊗ (e0σ1 )−1 )(1 ⊗ e0σ2 )i = (1 ⊗ (e0σ1 )−1 )iσ1 iσ2 (1 ⊗ e0σ2 ) = 0.

14

ZHENGYAO WU (吴正尧)

For σ1 = σ2 , i(eσ1 x ⊗ e0σ2 x0 )i = uσ1 i(xx0 ⊗ 1) ∈

X

uσ e(K ⊗ 1).

σ∈G

Finally, it follows from (2)(3) that iEi ' (i(K ⊗ 1), G, i(f ⊗ 1)(f 0 ⊗ 1)) ' (K, G, f f 0 ). Therefore by Lemma 3.6, [(K, G, f )][(K, G, f 0 )] = [E] = [iEi] = [(K, G, f f 0 )] in Br(F ).



Corollary 3.8. Let K/F be a finite Galois extension with G = Gal(K/F ). Then {(K, G, f ) | f is a factor set} is a subgroup of Br(F ). Remark 3.9. By [DK94, Th. 4.5.3], {(K, G, f ) | f is a factor set} is the kernel of Br(F ) → Br(K) [A] 7→ [A ⊗F K]. which we write Br(K/F ). 4. The period-index problem Definition 4.1. Let G be a group. Let M be a right G-module. We assume g ∈ G sends m ∈ M to mg ∈ M . Let C n be the set of all functions Gn → M . Let δ n : C n → C n+1 such that n X (δ n f )(x1 , . . . , xn+1 ) = f (x2 , . . . , xn+1 )+ (−1)i f (x1 , . . . , xi xi+1 , . . . , xn+1 )+(−1)n+1 f (x1 , . . . , xn )xn+1 , i=1

for all x1 , . . . , xn+1 ∈ G and f ∈ C n . Let Z 2 (G, M ) = ker(δ 2 ), B 2 (G, M ) = Im(δ 1 ) and H 2 (G, M ) = Z 2 (G, M )/B 2 (G, M ) is the second cohomology group of G in M . Lemma 4.2. δ 2 ◦ δ 1 = 0. (Furthermore, δ n+1 ◦ δ n = 0. We ignore its proof.) Proof. For all x1 , x2 ∈ G and f ∈ C 1 , we have (δ 1 f )(x1 , x2 ) = f (x2 ) − f (x1 x2 ) + f (x1 )x2 . For all x1 , x2 , x3 ∈ G and g ∈ C 2 , we have (δ 2 g)(x1 , x2 , x3 ) = g(x2 , x3 ) − g(x1 x2 , x3 ) + g(x1 , x2 x3 ) − g(x1 , x2 )x3 .

NOTES TO CROSS PRODUCT ALGEBRAS

15

So (δ 2 (δ 1 f ))(x1 , x2 , x3 ) = (δ 1 f )(x2 , x3 ) − (δ 1 f )(x1 x2 , x3 ) + (δ 1 f )(x1 , x2 x3 ) − (δ 1 f )(x1 , x2 )x3 = f (x3 ) − f (x2 x3 ) + f (x2 )x3 − (f (x3 ) − f (x1 x2 x3 ) + f (x1 x2 )x3 ) +(f (x2 x3 ) − f (x1 x2 x3 ) + f (x1 )x2 x3 ) − (f (x2 ) − f (x1 x2 ) + f (x1 )x2 )x3 = f (x3 ) − f (x2 x3 ) + f (x2 )x3 − f (x3 ) + f (x1 x2 x3 ) − f (x1 x2 )x3 +f (x2 x3 ) − f (x1 x2 x3 ) + f (x1 )x2 x3 − f (x2 )x3 + f (x1 x2 )x3 − f (x1 )x2 x3 = 0.  Theorem 4.3. Let K/F be a finite Galois extension with G = Gal(K/F ). (1) f : G × G → K ∗ is a factor set iff f ∈ Z 2 (G, K ∗ ). (2) f and g are equivalent factor sets iff f g −1 ∈ B 2 (G, K ∗ ). (3) We have an isomorphism of groups Br(K/F ) ' H 2 (G, K ∗ ). Proof. The composition of K ∗ is written multiplicatively (not additively as in Definition 4.1). (1) follows from Definition 1.2; (2) follows from Definition 2.6; (3) it follows from Remark 2.10 and Remark 3.9 that Br(K/F ) ' {[(K, G, f )] ∈ Br(F )} ' Z 2 (G, K ∗ )/B 2 (G, K ∗ ) ' H 2 (G, K ∗ ).  Lemma 4.4. Let G be a finite group. Let M be a right G-module. Then |G| · H 2 (G, M ) = 0. (Furthermore, |G| · H n (G, M ) = 0. We ignore its proof.) Proof. Suppose f ∈ Z 2 (G, M ). Then f (x2 , x3 ) = f (x1 x2 , x3 ) − f (x1 , x2 x3 ) + P f (x1 , x2 )x3 . Let h(x2 ) = f (x1 , x2 ). Then x1 ∈G

|G|f (x2 , x3 ) P = f (x2 , x3 ) x1 ∈G P P P = f (x1 x2 , x3 ) − f (x1 , x2 x3 ) + f (x1 , x2 )x3 x1 ∈G x1 ∈G x1 ∈G P P P = f (x1 x2 , x3 ) − f (x1 , x2 x3 ) + f (x1 , x2 )x3 x1 x2 ∈G

x1 ∈G

x1 ∈G

= h(x3 ) − h(x2 x3 ) + h(x2 )x3 ∈ B 2 (G, M ). Therefore |G| · H 2 (G, M ) = 0. Theorem 4.5. Let F be a field. Then Br(F ) is torsion.



16

ZHENGYAO WU (吴正尧)

Proof. Suppose [A] ∈ Br(F ). By Theorem 2.5, [A] = [(K, G, f )] for some cross product algebra (K, G, f ). By Lemma 4.4, |G| · [(K, G, f )] = 0 in H 2 (G, K ∗ ). Therefore [A] is |G|-torsion.



Definition 4.6. Let A be a central simple algebra over a field F . Suppose A ' Mm (D) for a central division algbra D over F (see Review 2.2). The √ index of A (written ind(A)) is the integer dimF D (see [DK94, Th. 4.5.1]). The period of A (written per(A)) is the order of [A] in Br(F ). Theorem 4.7. Let A be a central simple algebra over F . Then per(A)| ind(A). Proof. Assume ind(A) = n. Let D be the underlying central division algebra of A. Let (K, G, f ) be the cross product algebra Brauer equivalent to A as in Theorem 2.5. Then (K, G, f ) ' Mq (D) ' I1 ⊕ · · · ⊕ Iq where Ii are minimal right ideals of Mq (D). Suppose elements of I1 are first-row vectors. Then dimK I1 = dimK (Mq (D))/q = n. Suppose (eσ )σ∈G is the K-basis of the cross product algebra (K, G, f ). Suppose (ui )1≤i≤n is a K-basis of the vector space I1 . Since I1 is a right ideal of Mq (D), right multiplication by eσ is an endomorphism of I1 . Assume that ui eσ = n P uj tijσ for Tσ = (tijσ )1≤i,j≤n ∈ Mn (K). Then for σ1 , σ2 ∈ G, j=1

(ui eσ1 )eσ2 =

n X

! uj tijσ1

j=1

eσ 2 =

n X

2 uj eσ2 tσijσ 1

j=1

j=1

ui (eσ1 eσ2 ) = ui eσ1 σ2 f (σ1 , σ2 ) =

=

n n X X

n X

! uk tjkσ2

k=1

2 tσijσ 1

=

n X k=1

uk

n X j=1

uk tik(σ1 σ2 ) f (σ1 , σ2 ).

k=1

Then Tσσ12 Tσ2 = Tσ1 σ2 f (σ1 , σ2 ). Let λσ = det(Tσ ) ∈ K ∗ (since eσ is invertible by Lemma 1.6) for all σ ∈ G. We have λσσ21 λσ2 = λσ1 σ2 f (σ1 , σ2 )n . Then f n is equivalent to 1 in the sense of Definition 2.6. Therefore per(A)|n.



Lemma 4.8. Let D be a central division algebra over a field F . Let K be a splitting field of D, i.e. D ⊗F K ' Mm (K) where m = ind(D). Then m|[K : F ]. Proof. Let n = [K : F ]. Using the regular representation [DK94, Th. 1.3.1], we may assume that K ⊂ Mn (F ). Then Mm (F ) ⊂ Mm (K) ' D ⊗ K ⊂ D ⊗ Mn (F ) ' Mn (D) By Lemma 3.2, Mn (D) ' Mm (CMn (D) (Mm (F ))). By Review 2.2, CMn (D) (Mm (F )) ' Mk (D) and hence n = mk.



! 2 tjkσ2 tσijσ 1

.

NOTES TO CROSS PRODUCT ALGEBRAS

17

Lemma 4.9. Let A be a central simple algebra over a field F . If p is a prime factor of ind(A), then it is also a prime factor of per(A). Proof. By Theorem 2.5, A is Brauer equivalent to some cross product algebra p (K, G, f ). Then p| ind(A)| dimF (K, G, f ) = |G|. Let K0 be the fixed field of the Sylow p-subgroup of G. Then p - [K0 : F ]. By the contrapositive of Lemma 4.8, K0 does not split A. Hence per(A ⊗ K0 ) > 1. Since K splits A⊗K0 , per(A⊗K0 )|[K : K0 ]. Since [K : K0 ] is a positive power of p, so is per(A ⊗ K0 ). Thus p| per(A ⊗ K0 ). Since the restriction map Br(F ) → Br(K0 ), [A] 7→ [A ⊗ K0 ] is a homomorphism, we have per(A ⊗ K0 )| per(A). Therefore p| per(A).



Corollary 4.10. If per(A) = pt for some prime p and some integer t > 0, then ind(A) = ps for some integer s such that t ≤ s. Proof. It follows from Theorem 4.7 and Lemma 4.9.



Lemma 4.11. If D1 and D2 are central division algebras over a field F such that ind(D1 ) and ind(D2 ) are coprime, then D1 ⊗ D2 is division. Proof. Suppose D1 ⊗ D2 ' Mm (E) for some central division algebra E over F . We need to show that m = 1. It follows from [DK94, Th. 4.3.1] that D1op ⊗ D1 ' Mn (F ) where n = ind(D1 )2 . Then Mn (D2 ) ' D1op ⊗ D1 ⊗ D2 ' D1op ⊗ Mm (E) ' Mm (D1op ⊗ E). Then m|n = ind(D1 )2 . Similarly, m|n = ind(D2 )2 . Since ind(D1 ) and ind(D2 ) are coprime, m = 1.



Theorem 4.12. Let D be a central division algebra over a field F . Suppose ind(D) = pe11 · · · pekk for distinct primes pi . Then D ' D1 ⊗ · · · ⊗ Dk for central division algebras Di over F such that ind(Di ) = pei i . Proof. By Theorem 4.7, per(D)|pe11 · · · pekk . By Theorem 4.5, [D] = [A1 ] · · · [Ak ] for some central simple algebras Ai over F such that per(Ai )|pei i . By Corollary 4.10, ind(Ai ) is a positive power of pi if per(Ai ) is so. Let Di be the underlying central division algebra of Ai . Then ind(Di ) = ind(Ai ) are coprime to each other. By Lemma 4.11, D1 ⊗ · · · ⊗ Dk is division. Since D is Brauer equivalent to D1 ⊗ · · · ⊗ Dk , by Review 2.2, D ' D1 ⊗ · · · ⊗ Dk and hence ind(Di ) = pei i .



18

ZHENGYAO WU (吴正尧)

Remark 4.13. What is known: Gauss If F is an algebraically closed field, then per(A) = ind(A) = 1. Tsen(曾炯之) If F is of transcendence degree 1 over an algebraically closed field, then per(A) = ind(A). de Jong-Starr-Lieblich If F is of transcendence degree 2 over an algebraically closed field, then ind(A)| per(A)2 . Wedderburn If F is a finite field, then per(A) = ind(A) = 1. Brauer-Hasse-Noether If F is of transcendence degree 1 over a finite field, then per(A) = ind(A). Albert-Brauer-Hasse-Noether If F is a p-adic field or a number field, then per(A) = ind(A). Together with Grunwald-Wang(王湘浩) theorem, every central simple algebra over F is cyclic i.e. (K, G, f ) where G is a cyclic group. Saltman If F is of transcendence degree 1 over a p-adic field, then ind(A)| per(A)2 . References [DK94] Yurij A. Drozd and Vladimir V. Kirichenko, Finite-dimensional algebras, SpringerVerlag, Berlin, 1994, Translated from the 1980 Russian original and with an appendix by Vlastimil Dlab. MR 1284468 →5, →6, →11, →14, →16, →17 [Her68] I. N. Herstein, Noncommutative rings, The Carus Mathematical Monographs, No. 15, Published by The Mathematical Association of America; distributed by John Wiley & Sons, Inc., New York, 1968. MR 0227205 →1 [Sch85] Winfried Scharlau, Quadratic and Hermitian forms, Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], vol. 270, SpringerVerlag, Berlin, 1985. MR 770063 →11 Department of Mathematics, Shantou University, 243 Daxue Road, Shantou, Guangdong, China 515063 E-mail address: [email protected]