Proposition 5

Let S be a simple A-module of finite dimension over K, let D be the opposite field of the commutant of S, and let Z be the center of D. The following properties are equivalent:

(i) The character Tr$_S$ of S is not zero.

(ii) There exists an element $d\in D$ such that Tr$_{D/K}(d)\not= 0$.

(iii) The extension Z of K is separable, and the characteristic $p$ of K does not divide the degree [D : Z].

The right D-vector space S is finite-dimensional. Let $(e_1, . . . , e_n)$ be a basis of S over D, and let $u$ be an element of End$_D(S)$. Let $(d_{ij})$ be the matrix of $u$ with respect to the basis $(e_1, . . . , e_n)$. We have $u(e_j) =\sum^n_{i=1}e_id_{ij}$ for $j\in [1, n]$. Denote by $u_K$ the mapping $u$ viewed as an endomorphism of the K-vector space S and by $(u_{ij})$ the matrix of $u_K$ with respect to the decomposition $(e_iD)$ of the K-vector space S as a direct sum (II, §10, No. 5, p. 346). We have Tr($u_K$) $=\sum_i$ Tr($u_{ii}$). Moreover, $u_{ii}$ is the endomorphism of the K-vector space $e_iD$ defined by $u_{ii}(e_id) =e_id_{ii}d$ for $d\in D$, so its trace is equal to Tr$_{D/K}(d_{ii})$. We have thus proved the equality

(9) Tr($u_K$) $=$ Tr$_{D/K}(\sum_id_{ii})$.

By Burnside’s theorem (VIII, p. 83, Corollary 1 of Proposition 4), the mapping $a\mapsto a_S$ from A to End$_D(S)$ is surjective. The equivalence of properties (i) and (ii) therefore follows from formula (9).

If the extension Z of K is separable, then there exists an element $d\in Z$ such that Tr$_{Z/K}(d)\not= 0$ (V, §8, No. 2, p. 49, Corollary of Proposition 1). Moreover, we have Tr$_{D/K}(d) = [D : Z]$ Tr$_{Z/K}(d)$ (III, §9, No. 4, p. 548, Corollary); consequently, if $p$ does not divide [D : Z], then we have Tr$_{D/K}(d)\not=$ 0. This proves that (iii) implies (ii).

If the extension Z of K is not separable, then we have Tr$_{Z/K}(x) = 0$ for every $x\in Z$, and therefore Tr$_{D/K}(d) =$ Tr$_{Z/K}$(Tr$_{D/Z}(d)$) $= 0$ for every $d\in D$.

Now suppose that $p$ divides the degree [D : Z]. It then divides the reduced degree of D over Z. For every $d\in D$, we have Tr$_{D/Z}(d) = 0$ (VIII, p. 344, Remark), and therefore Tr$_{D/K}(d) =$ Tr$_{Z/K}$(Tr$_{D/Z}(d)$) $= 0$. This completes the proof of the implication (ii) $\Rightarrow$ (iii).