Finite dimensional division algebras are generated by two elements

Posted: June 20, 2010 in Division Rings, Noncommutative Ring Theory Notes
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The goal is to prove that if D is a division ring which is finite dimensional, as a vector space, over its center F, then there exist two elements a,b \in D such that D=F[a,b]. Here F[a,b] means the F-algebra generated by a,b. We begin with a standard field theory result.

Lemma. (Artin) Let L/F be a field extension with [L:F] < \infty. Then L/F is simple if and only if the number of intermediate fields F \subseteq K \subseteq L is finite.

From now on, D is a noncommutative division algebra with the centerF and [D:F] < \infty.

Theorem. (Herstein and Ramer1972) Let D_0 \neq D be a sub-division algebra of D and L a maximal subfield of D which is a simple extension of F. There exists c \in D such that cLc^{-1} \cap D_0 = F.

Proof. By the lemma there exist only a finite number of intermediate subfields F \subset K \subseteq L, which we’ll call them K_1, \cdots , K_n. Let C_i, \ i \leq n, be the centralizer of K_i in D. Clearly C_i \neq D, for all i. Note that F is an infinite field because otherwise D would be a finite division algebra and thus, by the Wedderburn’s little  theorem, D would have to be a field, which is false because we assumed that D is noncommutative. Therefore D cannot be equal to the union of finitely many of proper F vector subspaces. Hence there exists some x \in D such that

 x \notin \bigcup_{i=1}^n C_i \cup D_0, \ \ \ \ \ \ \ (*).

Now suppose that the claim in the theorem is false. Then for all \alpha \in F: \ (x+\alpha)L(x+\alpha)^{-1} \cap D_0 \neq F, and hence L \cap (x+\alpha)D_0(x+\alpha)^{-1} is one of the K_i. Since F is infinite, the set \{x+\alpha: \ \alpha \in F \} is infinite. But the number of K_i is finite and thus there exist distinct elements \alpha_1, \alpha_2, \alpha_3 \in F such that F \subset M=L \cap (x+\alpha_i)D_0(x+\alpha_i)^{-1}, for all i=1,2,3. Choose a \in M \setminus F. Then there exist d_1,d_2,d_3 \in D_0 such that a=(x+\alpha_i)d_i(x+\alpha_i)^{-1}, for all i=1,2,3. Playing with these relations, with the fact that x \notin D_0, will eventually give us xa=ax. Thus x is in the centralizer of F(a). But F(a)=K_j, for some j \leq n, because a \in L \setminus F. Thus x \in C_j, which is not possible by (*). \ \Box

Recall that D has a maximal subfield L which is a simple extension of F (see the theorem in this post). If L=F(d), then obviously cLc^{-1}=F(cdc^{-1}) is a maximal subfield too. .

Corollary. There exist a,c \in D such that D=F[a,b], where b=cac^{-1}.

Proof. Choose a \in D such that L=F(a) is a maximal subfield of D. By the theorem, there exists c \in D such that cLc^{-1} \cap L = F. Let D_0=F[a,b], where b=cac^{-1}. Since [D:F] < \infty, every subalgebra of D is algebraic over F and hence it is a division ring.  So D_0 is a division ring. Let Z be the center of D_0. Clearly F \subseteq Z. Now let d \in Z. Then, since L and cLc^{-1} are both maximal subfields of D_0 and every maximal subfield of a division algebra contains the center, d \in cLc^{-1} \cap L = F. Thus Z=F. So, since L is a maximal subfield of D_0, we have [D_0:F]=[L:F]^2. But L is also a maximal subfield of D and hence [L:F]^2=[D:F]. Therefore [D_0:F]=[D:F] and so D_0=D. \ \Box

The above corollary is known as “Albert’s theorem”.  For another proof, see Corollary 15.17 in Lam’s book “A first course in noncommutaive ring theory”.


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