it suffices to exhibit just ONE counter-example, so let's pick an easy A to work with:
let A = Z+Z, the direct sum of the additive group (Z,+), where Z is the integers we know and love.
let f(m,n) = (n,0). first, i show that f is indeed in End(A):
f((m,n)+(m',n')) = f((m+m',n+n')) = (n+n',0) = (n,0) + (n',0) = f(m,n) + f(m',n').
next i show that f is nilpotent:
f2((m,n)) = f(f((m,n))) = f((n,0)) = (0,0), so f2 is the zero-map.
now, let g(m,n) = (0,m). prove that g is likewise a nilpotent endomorphism of Z+Z.
next, show f+g is NOT nilpotent (where (f+g)(m,n) = f((m,n)) + g((m,n)).
for the first problem, consider (1A - f)(1A + f + f2 +....+ fn-1).
Thank you very much
Excuse me please clarify the problem(1A - f)is an invertible endomorphism (has inverse)
Have you done what Deveno suggested? What is (1A - f)(1A + f + f2 +....+ fn-1).
Answer the question