Why do nil ideals annihilate simple modules?

Question

A nil ideal $N$ of a ring $R$ is defined as follows:

$(N,+)$ is a subgroup of $(R,+)$

$\forall x \in N, \forall r \in R :\quad x \cdot r \in N$

$\forall x \in N, \forall r \in R : \quad r \cdot x \in N$

$\forall x \in N, \exists n\in \mathbb{Z}: x^n=0$

For any element $a\in R$ and $e\in N$, $ae$ is nilpotent: $(ae)^n=a^ne^n=a^n.0=0$

Hence, $eR$ is a nilpotent ideal and hence a nil ideal. My book says that hence, it is a subset of the Jacobson radical (the set of elements which annihilate simple $R$-modules), but I can't really see how that follows.

NOTE: $R$ is assumed to be Artinian, in this case (i.e. it satisfies the descending chain condition on ideals)

Answer 1

if the $R$-module $M$ is simple then for $a \in R$ we have $aM=0$ or $aM=M$. suppose $aM \ne 0$ then $aM=M$. if $a \ne 0$ is nilpotent, then $\exists n \gt 1$ such that $a^n=0$. then $$ a^nM=0 $$ this requires $$ a(a^{n-1}M) =0 $$ but now $a^{n-1}M=M \Rightarrow aM=0 \Rightarrow a^{n-1}M=0$, a contradiction.

so $a^{n-1}M=0$. if $n-1 \gt 1$ we may repeat the argument until we arrive at $a^1M =0$

Answer 2

The nilideal is contained in the nilradical, since $1-ax$ is a unit whenever $x$ is nilpotent. The nilradical is the intersection of all the kernels of irreducible representations.