Let $A$ be a commutative associative ring. Show that $\mathfrak{Nil}(A[x])=\mathfrak{Jac}(A[x])$.
Since $\mathfrak{Nil}(A[x])=\bigcap\limits_{\mathfrak p\triangleleft A[x]}\mathfrak{p}$, it's contained in $\bigcap\limits_{\mathfrak m\triangleleft A[x]}\mathfrak{m}=\mathfrak{Jac}(A[x]).$
I don't know how to show the inverse inclusion. Let $f=a_0+a_1x+\dots+a_nx^n\in\mathfrak{Jac}(A[x]).$ That means that $1-fg$ is unit for all $g\in A[x]$ . In particular, $1-f$ is unit or, equivalently, $a_1,\dots,a_n$ are nilpotent and $1-a_0$ is unit. It would be enough to demonstrate that $a_0$ is nilpotent too but I have no idea how to do that. Can you please help me?
UPD: I've seen the answer to "Jacobson radical equal to nilradical in R[X]" but it contains nothing new for me. I have got a problem with the last step.
