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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?

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  • $\begingroup$ Is $k$ a field? Does "$k$-algebra" mean "associative $k$-algebra with unit"? Please edit. $\endgroup$
    – YCor
    Oct 29, 2015 at 10:32
  • $\begingroup$ J(A) is the Jacobson Radical of A? $\endgroup$
    – Vincent
    Oct 29, 2015 at 10:55
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    $\begingroup$ Crossposted from MSE: math.stackexchange.com/questions/1502604 $\endgroup$ Oct 29, 2015 at 11:46

1 Answer 1

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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.

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