TITLE: Nilradical of polynomial ring
QUESTION [13 upvotes]: Let $R$ be a commutative ring.  The nilradical $\text{nil}(R)$ is the set of all nilpotent elements, and it is the intersection of all the prime ideals of $R$.  Is the following true in the polynomial ring?
$$\text{nil}(R[X])=(\text{nil}(R))[X]$$
Any polynomial in the right hand side has all its coefficients nilpotent in $R$, hence also in $R[X]$, and therefore it is nilpotent in $R[X]$ as the sum of nilpotent elements.  The other inclusion is not clear to me.  Is there a counterexample?

REPLY [7 votes]: Here is a way to see $\mathrm{nil}(R[X])\subseteq\mathrm{nil}(R)[x]$ without much computation:
Suppose that $f\in R[X]$ is nilpotent. For every prime ideal $\mathfrak{p}$ of $R$ the canonical ring homomorphism $R\to R/\mathfrak{p}$ induces a ring homomorphism $R[X]\to (R/\mathfrak{p})[X]$ that just reduces the coefficients mod $\mathfrak{p}$.
Now the image of $f$ under this map is also nilpotent. But a nilpotent polynomial over the integral domain $R/\mathfrak{p}$ must be zero.  Hence all coefficients of $f$ lie in $\mathfrak{p}$.