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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.

Katie Dobbs
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    Related to http://math.stackexchange.com/questions/153333/jacobson-radical-of-rx-where-r-is-domain –  Mar 13 '13 at 11:24

2 Answers2

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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 $i\ge 1$.

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I tried in a way. Maybe it's helpful. Suppose nilradical of R[X] = nil(R[X]) and Jacobson of R[X] = J(R[X]). Every maximal is a prime ideal, so nil(R[X]) is a subset of J(R[X]). {As set of prime contains maximal also.} Now check J(R[X]) is a subset of nil(R[X]). let f belongs to J(R[X]). then for every g(x) belongs to R[X], 1-fg is unit in R[X]. Take g(x) = -x. Then 1+xf is a unit in R[X]. So, every coefficient of f is nilpotent. So, f is nilpotent, and f belongs to nilradical of R[X]. Implies J(R[X]) is a subset of nil(R[X]). So, J(R[X])=nil(R[X]) in R[X]. if anything is wrong, please let me know.

Debjit Mallick
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