# Jacobson radical equal to nilradical in $R[X]$

Let $R$ be a non-zero commutative ring with identity. Let $\textrm{nilrad}(R)$ be the nilradical of $R$, which can be characterised either as the intersection of all prime ideals of $R$, or as the ideal of nilpotent elements. Let $J(R)$ be the Jacobson radical of $R$, which can be characterised either as the intersection of all maximal ideals of $R$ or as the ideal of elements $x\in R$ with the property that $1-xy$ is a unit for all $y\in R.$

In general, the nilradical of R is contained in the Jacobson radical. I want to show that the reverse inclusion holds in the polynomial ring $R[X].$ Can someone please give me a hint for this problem? Thank you.

• – user26857 Mar 13 '13 at 11:24

Let $f\in R[X]$, $f=a_0+a_1X+\cdots+a_nX^n$. Then
$(1)$ $f$ is nilpotent iff $a_i$ is nilpotent for all $i\ge 0$;
$(2)$ $f$ is invertible iff $a_0$ is invertible and $a_i$ is nilpotent for all $i\ge 1$.