Let $A[x]$ be the algebra of polynomials with coefficients in a $k$-algebra $A$. Assume that, for any simple $A[x]$-module $M$, we have $\text{End}\_{A[x]} M = k$. Does it follow that any element of $J(A)$ is nilpotent?



Does it follow that any element of $J(A)$ is nilpotent?
Let $a \in J(A)$, and suppose for a contradiction that the left ideal $A[x](1 - xa)$ is proper.


We can then choose a maximal left ideal $J$ of $A[x]$ containing $A[x](1-xa)$, so that $M := A[x] / J$ is a simple $A[x]$-module. Let $v := 1 + J$ be the canonical $A[x]$-module generator of $M$; then $(1 - xa)\cdot v = 0$ by construction.


Now $x$ acts on $M$ by $A[x]$-module endomorphisms, so by assumption, there exists $\lambda \in k$ such that $x - \lambda$ kills $M$. Hence


$$(1 - \lambda a) \cdot v = (1 - xa) \cdot v = 0.$$


However, since $a \in J(A)$, $1 - \lambda a$ is a unit in $A$, and it follows that $v = 0$. Hence $M = 0$, contradicting the simplicity of $M$.


So in fact $A[x](1 - xa) = A[x]$, and therefore we can elements $b\_0, b\_1,\ldots, b\_n \in A$ such that


$$(b\_0 + b\_1x + \cdots + b\_nx^n)(1 - xa) = 1.$$


Equating coefficients, we see that $b\_i = a^i$ for all $i = 0, \ldots, n$, and also that $b\_na = 0$. Hence $a^{n+1} = 0$ and $a$ is nilpotent.


