Exercise 1, § 13

Let K be a commutative field and A a K-algebra. We call an A-module M absolutely without radical if for every extension L of K, the $A_{(L)}$-module $M_{(L)}$ is without radical. We say that the algebra A is absolutely without radical if the A-module $A_s$ is absolutely without radical.

a) A commutative algebra that is absolutely without radical is separable (V, §15, No. 2, p. 119, Definition 1).

b) Every submodule of a module that is absolutely without radical is absolutely without radical; a direct sum of modules that are absolutely without radical is absolutely without radical.

c) Let M be a finitely generated A-module; then M is absolutely without radical if and only if there exists a perfect field P that is an algebraic extension of K such that $\mathfrak{R}_{A_{(P)}}(M_{(P)}) = 0$.

d) A semisimple A-module M is absolutely without radical if and only if every simple module belonging to the support of M is.

e) Let S be a simple A-module and D its commutant. Prove that the following properties are equivalent:

(i) The A-module S is absolutely without radical.

(ii) The K-algebra D is absolutely without radical.

(iii) The center of D is a separable extension of K.

f ) A semisimple algebra is absolutely without radical if and only if the center of each of its simple components is a separable extension of K.

g) Prove that an A-module that is absolutely without radical and finite-dimensional over K is absolutely semisimple.