## 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 center$F$ 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”.