4. Central Simple Algebras
4.6 Brauer Group. The Frobenius Theorem
In Sect. 4.3, we have observed that the class of central simple algebras is closed with respect to the tensor multiplication. The ground field K plays the role of identity since A IS> K ~ A for any algebra A. Finally, Theorem 4.3.1 shows that the opposite algebra A ° is an inverse of the algebra A in the sense of this operation. All this allows us to define a group structure on the set of isomorphism classes of the central division algebras in the following way.
We fix a representative in each isomorphism class of the central division algebras. IT Dl and D2 are such representatives, then Dl IS> D2 is a central simple algebra, and therefore isomorphic to Mn(D), where D is a central division algebra. Put D = Dl D2 . It follows from Sect. 4.2 that Dl D2 = D2Dl and D1(D2D3 ) = (D1D2)D3 • Furthermore DK = KD = D, and by Theorem 4.3.1, DDo
=
DOD=
K. In this way, our set of central division4.6 Brauer Group. The Frobenius Theorem 79 algebras forms a commutative group. This group is called the Brauer group of the field K and is denoted by Br (K).
If L is an extension of the field K, then for every central division algebra D also the L-algebra DL is central simple and thus isomorphic to Mn(D'), where D' is a central division algebra over L. It is easy to verify that assigning to D the division algebra D', we obtain a group homomorphism Br (K) -+
Br (L). The kernel of this homomorphism consists of those division algebras for which L is a splitting field. This subgroup of the Brauer group is denoted by Br (L j K). Theorem 5.1 shows that every element of the Brauer group belongs to some subgroup of the form Br (Lj K) and thus Br (K) = U Br (Lj K).
A concrete calculation of the Brauer group is, as a rule, rather complex, and the structure of this group is known only for some fields K. We shall limit ourselves to the most simple cases: the field of real numbers and the finite fields (see Chap. 5).
Of course, if K is an algebraically closed field, then there are no central division algebras (in fact, no division algebras) different from K, and thus the Brauer group is trivial.
Over the field IR of real numbers, there is at least one proper central field, viz. the quaternion algebra lH. A remarkable result asserts that this is the only central division algebra over the field IR.
Theorem 4.6.1 (Frobellius). The only finite dimensional division algebras over the field IR of real numbers are the field IR itself, the field 0:: of complex numbers and the quaternion algebra lH.
Proof. First, let L be a finite extension of the field IR, a an element of L and ma(x) the minimal polynomial ofthe element a over the field IR (see Sect. 1.2).
Since ma(x) is irreducible, it is either linear (and then a E IR) or quadratic of the form x 2
+
2px+
q, where p2 < q. In the second case, the element a+
p is a root of the polynomial x 2+
q - p2. Thus,R
is a root of the polynomialq _ p2
x2
+
1. Therefore the subfield IR[aJ is isomorphic to 0::. Since 0:: is algebraically closed, L ~ 0::.Thus, the finite extensions of the field IR are either IR itself or 0::. Therefore 0:: is a splitting field of any central division algebra Dover IR. Let D
I-
IR,tP
= [D : IRJ and L a maximal subfield of D. Since LI-
IR, necessarily L ~ 0::, and by Theorem 4.5.1, d = [0::: IRJ = 2, i.e. [D: IRJ = 4.Denote by i an element of the subfield L such that i2
=
-1 (the image of the element i E <C in t.he isomorphism 0:: ~ L). The complex conjugation determines an automorphism of the field L in which i is mapped to -i. By Corollary 4.4.2, there is a non-zero element j in D such that jij-l = -i, i. e... ..
J~ = -~J.
Since j and i do not commute, j (j. L and thus 1, i, j are linearly independent. Besides, j 2i
=
-jij=
ij2, i.e.P
E GD(L) = L. ThusP =
0:+
{3i with 0:, {3 E IR. ButP
must commute with j, and thus j(o:+
(3i) = o:j+
{3ji = (0: - {3i)j = (0:+
{3i)j and therefore {3 =o.
It follows80 4. Central Simple Algebras
that
p =
a E IR. Clearly, a<
0 (otherwise a=
"{2 and (j - "{)(j+ "{) =
0which is impossible). Replacing j by jj..,r-::a, we may assume that
p
=-1.Thus, we have already identified two elements i and j such that i2 = j2 = -1 andji
=
-ij. Write k=
ij. Then k2=
ijij=
_i2p =
-1; ik=
i2j=
-j;ki
=
iji=
_i2 j=
j. Similarly, j k=
-kj=
i. In other words, the elements i, j, k have the same multiplication table as the canonic basis of the quaternions IH. Therefore, there is a homomorphismf :
IH -? D which is a monomorphism because IH is a division algebra. But [IH : IR] = [D : IR], and thereforef
is anisomorphism, as required. 0
Corollary 4.6.2. Br(IR) = Br(<CjIR) is the cyclic group of order two.
Exercises to Chapter 4
1. Let D be a finite dimensional division algebra over
fl.:.
Prove that two D- bimodules are isomorphic if and only if the corresponding self-representations are similar (i. e. differ by an inner automorphism of the algebra M,,( D)).2. Let Sand T be finite sets with quasi-order relations -+, A and B the correspond- ing minimal algebras over a field 1( (see Exercises 8-10 to Chap. 3). Introduce a quasi-order on the Cartesian product S X T by (s,t) -+ (s',t') if 8 -+ 8' and t -+ t'. Prove that A®B is the minimal algebra corresponding to this quasi-order relation on the Cartesian product S X T.
3. For the IR-algebra 4:!, prove that 4:! ® 4:! ~ 4:! $ 4:!. This example shows that the tensor product of simple algebras need not be a simple algebra.
4. A linear transformation 8 : A -+ A is said to be a derivation on the algebra A if, for arbitrary elements a, b E A, 8( ab)
=
a8( b)+
8( a )b.a) Show that the map 8", , where x is a fixed element of A, given by the formula 8",a
=
ax - xa is a derivation on the algebra A. This derivation is called inner.b) If 8 is a derivation on the algebra A, prove that the map T : A -+ M2(A) given by the formula
is an algebra homomorphism.
c) Prove that every derivation on a central simple algebra is inner. (Hint: Use Exercise b) and the Skolem-Noether theorem.)
5. Let A be a simple algebra, B its central simple subalgebra and B'
=
CA(B).Prove that:
a) B' is a simple algebra;
b) [B: 1(][B' : 1(] = [A: 1(];
c) if B' ~ Mm(D), then A ® BO ~ M,,(D), where m divides n.
Give an example in which CA(B')
=I
B.Exercises to Chapter 4 81 6. Consider the algebra D over the field <Q of rational numbers with basis {1, i,j, k}
and multiplication table
i ] k i -1 k - ]
j -k -2 2i k ] -2i -2 (verify that, indeed, it is an algebra).
a) Prove that D is a central division algebra.
b) Verify that L1
=
<Q[i] and L2=
<Q(j] are non-isomorphic maximal subfields of D.7. Prove that if D1 and D2 are two central division algebras over J( such that [D1 : J(] and [D2 : J(] are relatively prime, then D1 0 D2 is a division algebra.
(Hint: Assuming that D1 0 D2 ~ Mn( D), calculate D1 0 D2 0 D2 in two different ways and deduce that n divides [D2 : J(].)
8. (Dickson's theorem) Prove that two elements of a central division algebra are conjugate if and only if they have the same minimal polynomials.
9. (Hilbert division algebra) Let L be a field and r.p an automorphism of L. Consider
co
the "power series" of the form
L
aiti, where ai E L, t a symbol (variable)i;}>-oo
and i
»
-00 indicates, as usual, that there is only a finite number of powers with negative exponents. Addition of the series is given simply byL
co aiti+
i>-oo
00 00
L
biti= L
(ai+
bi)ti and multiplication by the rules ta=
r.p(a)t (a E L)i>-oo i>-oo
and a(
f:
aiti)= . f:
(aa;)ti.'»-00 .»-co
a) Verify that the set of the series with the above operations of addition and multiplication forms a division algebra L[t, r.p]. This division algebra is called the Hilbert division algebra.
b) Let J(
=
{a ELI r.p( a)=
a} and n be the order of the automorphism r.p, i. e. the least natural number for which r.pn is the identity automorphism, or 00 if such a natural number does not exist. Prove that the center of the Hilbert division algebra is K[tnD if n =I- 00 and K otherwise.c) Construct an example of a central infinite dimensional division algebra.
(Hint: Take for L the field of rational functions J((x).)