Proposition

In the ring $A[x]$, the Jacobson radical is equal to the nilradical.

Solution

Since every maximal ideal is prime, $N(A[x]) \subset J(A[x])$. Let $f \in J(A[x])$. By Proposition 1.9[Atiyah], $1 - xf$ is a unit in $A[x]$. As we showed previously, this implies $a_0, \cdots, a_n \in A$ are nilpotent. Moreover, the same post implies that $a_0 + a_1x + \cdots + a_nx^n \in A[x]$ is nilpotent. Therefore, $f \in N(A[x])$.