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3 added 600 characters in body

Concerning

When not,

are the centers of the division rings Galois over $k$

A finite dim'l $k$-algebra $A$ is split provided $\operatorname{End}_A(S) = k$ for every simple $A$-module $S$.

[This terminology is consistent with that used in other contexts -- $A$ is split just in case the reductive quotient of the "unit group" $A^\times$ is a split reductive algebraic group over $k$.]

Suppose that $k$ is perfect and that $A_\ell = A \otimes_k \ell$ is a split $\ell$-algebra for a finite, separable extension $\ell \supset k$. Then we have the following:

$(*)$ If $S$ is a simple $A$-module, the center $Z$ of the division $k$-algebra $\operatorname{End}_A(S)$ is a subfield of $\ell$ (indeed, \ell$. To see this, I claim first that$\operatorname{End}_A(S) \otimes_k \ell$is a split semisimple$\ell$-algebra. Indeed, since$k \subset \ell$is separableand since , the$A A_\ell$-module$S \otimes_k \ell$ell = S_\ell$ is splitsemisimple, say $\operatorname{End}_A(S) S_\ell = \otimes_k bigoplus_i S_i$ as $A_\ell$-module, where $S_i$ is the $T_i$-isotypic component of $S_\ell$ and where $T_i$ are distinct simple $A_\ell$-modules. Since by assumption $\operatorname{End}_{A_\ell}(T_i) = \ell$, we see that $$\operatorname{End}_A(S) \otimes_k \ell \simeq \operatorname{End}_{A_\ell}(S_\ell) = \prod_i \operatorname{End}_{A_\ell}(S_i)$$ is a product of full matrix algebras over $\ell$. Now observe that the center of a split semisimple $\ell$-algebra whose center is thus a a split commutative etale $\ell$-algbra $\ell \times \cdots \times \ell$). ell$. Assertion$(*)$now follows. Apply this to$A = kG$for a finite group$G$. By Torsten's comment (following Emerton's answer) we may suppose$k$to be perfect. Then$A \otimes_k \ell$is split for an Abelian extension$\ell$of$k$-- a suitable$\ell$can be obtained by adjoining to$k$enough roots of unity. It follows that$Z$is always Galois over$k$(for$A = kG$). 2 added 17 characters in body Concerning When not, are the centers of the division rings Galois over$k$A finite dim'l$k$-algebra$A$is split provided$\operatorname{End}_k(S) \operatorname{End}_A(S) = k$for every simple$A$-module$S$. [This terminology is consistent with that used in other contexts --$A$is split just in case the reductive quotient of the "unit group"$A^\times$is a split reductive algebraic group over$k$.] Suppose that$k$is perfect and that$A \otimes_k \ell$is a split$\ell$-algebra for a finite, separable extension$\ell \supset k$. If$S$is a simple$A$-module, the center$Z$of the division$k$-algebra$\operatorname{End}_A(S)$is a subfield of$\ell$(indeed, since$k \subset \ell$is separable and since$A \otimes_k \ell$is split,$\operatorname{End}_A(S) \otimes_k \ell$is a split semisimple$\ell$-algebra whose center is thus a split commutative etale$\ell$-algbra$\ell \times \cdots \times \ell$). Apply this to$A = kG$for a finite group$G$. By Torsten's comment (following Emerton's answer) we may suppose$k$to be perfect. Then$A \otimes_k \ell$is split for an Abelian extension$\ell$of$k$-- a suitable$\ell$can be obtained by adjoining to$k$enough roots of unity. It follows that$Z$is always Galois over$k$(for$A = kG$). 1 Concerning When not, are the centers of the division rings Galois over$k$A finite dim'l$k$-algebra$A$is split provided$\operatorname{End}_k(S) = k$for every simple$A$-module$S$. [This terminology is consistent with that used in other contexts --$A$is split just in case the reductive quotient of the "unit group"$A^\times$is a split reductive algebraic group over$k$.] Suppose that$k$is perfect and that$A \otimes_k \ell$is split for a finite, separable extension$\ell \supset k$. If$S$is a simple$A$-module, the center$Z$of the division$k$-algebra$\operatorname{End}_A(S)$is a subfield of$\ell$(indeed, since$k \subset \ell$is separable and since$A \otimes_k \ell$is split,$\operatorname{End}_A(S) \otimes_k \ell$is a split semisimple$\ell$-algebra whose center is thus a split commutative etale$\ell$-algbra$\ell \times \cdots \times \ell$). Apply this to$A = kG$for a finite group$G$. By Torsten's comment (following Emerton's answer) we may suppose$k$to be perfect. Then$A \otimes_k \ell$is split for an Abelian extension$\ell$of$k$-- a suitable$\ell$can be obtained by adjoining to$k$enough roots of unity. It follows that$Z$is always Galois over$k$(for$A = kG\$).