Proposition 1

Let K be a commutative field and A a K-algebra.

a) Every direct sum of absolutely semisimple A-modules is an absolutely semisimple A-module. Every submodule and quotient module of an absolutely semisimple module is absolutely semisimple.

b) Let M be an A-module and let L be an extension of the field K. The A-module M is absolutely semisimple if and only if the $A_{(L)}$-module $M_{(L)}$ is absolutely semisimple.

Assertion a) follows from the analogous assertion for semisimple modules (VIII, p. 56, Corollaries 1 and 3).

Suppose that the A-module M is absolutely semisimple, and let $L'$ be an extension of L. As an $A_{(L')}$-module, $L'\otimes_LM_{(L)}$ is isomorphic to $M_{(L')}$ (II, §5, No. 1, p. 273, Proposition 2); it is therefore a semisimple module. This proves that $M_{(L)}$ is absolutely semisimple.

Conversely, suppose that $M_{(L)}$ is absolutely semisimple. Let $L'$ be an extension of K. There exists a composite extension $(\Omega , u, v)$ of L and $L'$ (V, §2, No. 4, p. 13, Corollary); we identify L and $L'$ with subextensions of Ω. The $A_{(\Omega )}$-module $M_{(\Omega )}$ is isomorphic to $(M_{(L)})_{(\Omega )}$; it is therefore semisimple. But $M_{(\Omega )}$ is also isomorphic to $(M_{(L')})_{(\Omega )}$, and Proposition 8, a) of VIII, p. 222 implies that $M_{(L')}$ is semisimple. So M is absolutely semisimple.