Algebraically Closed Base Fields

In this section, the letter C stands for a field. We assume that, for each element s in S, the characteristic of C does not divide s, so that, according to Theorem 9.1.5(h), OS is semisimple. In addition, we assume C to be algebraically closed. 

From Theorem 9.1.4(i) we know that (CS)( C Z(CS). Thus, we just have to prove that Z(CS) C (CS) . 

Let K be an irreducible submodule of the CS-module CS such that K C Hx. Then, by Theorem 8.5.4(h), Hx = Endc(iL). Thus, by Theorem 9.1.9, there exists an element a in Hx such that, for each element k in K, k(aQ = k. 

Prom Theorem 8.6.4(ii) (together with Theorem 8.5.3(ii) and Theorem 8.5.4(ii)) we conclude that 

the claim follows from Theorem 9.1.4(h) together with Theorem 9.2.1. 

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In the second section, we derive the Schur relations in the case where the characteristic of the underlying base field does not divide any of the integers s with s G S and is algebraically closed. 

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