The set of endomorphisms with their image contained in Jacobson radical is nilpotent ideal I have a question:


For $A$ a finite dimensional algebra over an algebraically closed field $\mathbb K$ and a finitely generated $A$-module $M$, let $I(M)=\{f\in End_A(M)\mid Img(f)\subseteq MRad(A)\}$. Then $I(M)$ is a nilpotent ieal of $End_A(M)$.


The only thing that I can say is when $M$ is an indecomposable, which in that case $End_A(M)$ will form a two-sided ideal contained in $Nil(End_A(M)$ and in particular, an endomorphism of $M$ is either nilpotent or an isomorphism. And when its image is in $Rad(M)$ and since $Rad(M)$ is nilpotent so that will be nilpotent.
But in general I do not know how to show it!
Can someone let me know how to show it?
Many thanks for your help!
 A: Don't forget to show (Step 0) that $I(M)$ is an ideal first. But I leave that to you.
Step 1: Show that all $f \in I(M)$ are nilpotent. E.g. use that all $f$ are $A$-linear and that $Rad(A)$ is nilpotent. (Do you know that $Rad(A)$ is nilpotent? If yes, the proof for that probably goes along similar lines as this.)
Step 2: Show that $End_A(M)$ is a finite-dimensional $\Bbb K$-algebra. For the finite dimension, e.g. use that it is contained in $End_\Bbb K (M)$ and that $M$ is a finite-dimensional $\Bbb K$-vector space.
Step 3: Show in general that if $B$ is any finite-dimensional $\Bbb K$-algebra and $J$ an ideal of $B$ such that all elements of $J$ are nilpotent, then $J$ is a nilpotent ideal. E.g. you can use that $J$ is a finite dimensional $\Bbb K$-vector space, choose a basis, and use that in partcular all basis vectors are nilpotent.
Step 4: Apply the result of step 3 to $B=End_A(M)$ and $J=I(M)$.
(Note: Steps 2--4 can be avoided with more theory. Your ring is artinian hence noetherian, so by Levitzky's theorem even one-sided ideals consisting of nilpotent elements will be nilpotent. For a two-sided ideal, being finitely generated is enough, etc. There are many cases where one can conclude that an ideal consisting of nilpotent elements is nilpotent, and probably all of them apply to your ring.)
