This is proven by observing that any nil ideal is contained in the Jacobson radical of the ring, and since the Jacobson radical is a nilpotent ideal ( due to the artinian hypothesis ), the result follows.
12.
Finally, I do think there might be a problem in Semisimple algebra because there it is claimed that the radical is the unique nilpotent ideal containing all nilpotent elements of the algebra, and that finitedimensional algebras are semisimple iff the radical is trivial.
13.
Now suppose A has a non-zero radical, so it has a non-zero nilpotent ideal I . Let x be a non-zero element of this ideal, so y = x . x ^ T is also in I, and is symmetric and non-zero ( to show y is non-zero use the fact that if a linear map has matrix x with respect to an orthonormal basis, then its adjoint has matrix x ^ T ).
