Proposition 3

Let A be a central simple K-algebra of finite degree, and let L be a semisimple commutative subalgebra of A. The following properties are equivalent:

(i) The algebra L is a maximal commutative subalgebra of A.

(ii) The left L-module A is free, of dimension [L : K].

(iii) We have $[A : K] = [L : K]^2$.

Suppose, moreover, that A is the algebra End$_K(V)$, where V is a vector space of nonzero finite dimension over K. Then the previous properties are also equivalent to the following:

(iv) V is a free L-module of dimension 1.

A) Suppose that A is of the form End$_K$(V), where V is a vector space of nonzero finite dimension over the field K. We will establish the equivalence of conditions (i) through (iv) following the diagram

(i) $=\Rightarrow$ (iv) $=\Rightarrow$ (ii) $=\Rightarrow$ (iii) $=\Rightarrow$ (i) .

Since L is a semisimple commutative algebra of finite degree over K, we can identify it with a finite product $\prod_{i\in I}L_i$ of extensions of K of finite degree (VIII, p. 137, Proposition 3). For every $i\in I$, let $V_i$ be the isotypical component of type $L_i$ of the L-module V; it is a vector space of nonzero finite dimension over $L_i$ because V is a faithful L-module (VIII, p. 144, Corollary). We can then identify V with $\prod_{i\in I}V_i$. Under these conditions, the commutant $L'$ of L in A, which is simply the algebra End$_L$(V), can be identified with the product $\prod_{i\in I}$ End$_{L_i}(V_i)$.

Suppose that L is a maximal commutative subalgebra of End$_K(V)$. By Lemma 3, a), we have $L = L'$, so $L'$ is commutative and we have dim$_{L_i}(V_i) = 1$ for every $i\in I$. So (i) implies (iv).

Suppose that the L-module V is free of dimension 1. Let $(e_1, . . . , e_r)$ be a basis of V over K. The mapping $a\mapsto (ae_1, . . . , ae_r)$ is an isomorphism of left A-modules and therefore of L-modules from A to $V^r$. Consequently, A is a free left L-module of dimension $r$, and we have $r=$ dim$_K(V) = [L : K]$. So (iv) implies (ii).

It is clear that (i) implies (iii).

Finally, suppose that we have $[A : K] = [L : K]^2$, in other words, dim$_K(V) = [L : K]$. We then have $\sum_i$ dim$_{L_i}(V_i)[L_i: K] =\sum_i[L_i: K]$, so that $V_i$ has dimension 1 over $L_i$ for every $i$. We then have End$_{L_i}(V_i) = L_i$ for every $i$, and so $L'= L$. By Lemma 3, a), L is a maximal commutative subalgebra of A. So (iii) implies (i).

B) Let us continue with the general case. By Theorem 1 of VIII, p. 252, there exists a separable extension $K'$ of K of finite degree such that the $K'$-algebra $A_{(K')}$ is isomorphic to an algebra End$_{K'}(V')$, where $V'$ is a vector space of nonzero finite dimension over $K'$. Then the $K'$-algebra $L_{(K')}$ is commutative and semisimple (VIII, p. 222, Corollary 2). By the first part of the proof, the following properties are equivalent:

(i$'$) The algebra $L_{(K')}$ is a maximal commutative subalgebra of $A_{(K')}$.

(ii$'$) The left $L_{(K')}$-module $A_{(K')}$ is free, of dimension $[L_{(K')}: K']$.

(iii$'$) We have $[A_{(K')}: K'] = [L_{(K')}: K']^2$.

By Lemma 3, b), properties (i) and (i’) are equivalent.

Set $n= [L : K]$; then $n= [L_{(K')}: K']$. Property (ii) means that the left L-modules A and $L^n$ are isomorphic; by Theorem 3 of VIII, p. 37, this is equivalent to the $L_{(K')}$-modules $A_{(K')}$ and $(L_{(K')})^n$ being isomorphic. The equivalence of (ii) and (ii$'$) follows.

Finally, we have $[A : K] = [A_{(K')}: K']$ and $[L : K] = [L_{(K')}: K']$, which give the equivalence of properties (iii) and (iii$'$). We have thus proved the equivalence of properties (i), (ii), and (iii).