14
$\begingroup$

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.

$\endgroup$
9
$\begingroup$

Use the following elementary facts about polynomials:

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 $i\ge 1$.

$\endgroup$
  • $\begingroup$ See here for a much simpler approach, a la Euclid. $\endgroup$ – Bill Dubuque Jul 9 '14 at 13:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.