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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?

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No, there is no counterexample; we always have $\mathrm{nil}(R[x])=\mathrm{nil}(R)[x]$. This is given as exercise $2.$(ii) in chapter 1 of Atiyah-Macdonald:

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To prove the inclusion you're wondering about, you can proceed by induction on the degree of the nilpotent polynomial:

If $p\in \mathrm{nil}(R[x])$ is of degree 0, then $p=a_0$ for some $a_0\in R$, and clearly $a_0\in\mathrm{nil}(R)$, so the claim is true for polynomials of degree 0. Suppose the claim is true for $p\in\mathrm{nil}(R[x])$ of degree $\leq n.$

Let $p=a_0+\cdots+a_{n+1}x^{n+1}\in\mathrm{nil}(R[x])$ be of degree $n+1$, say with $p^m=0$. For any $r$, the leading coefficient of $p^r$ is always $a_{n+1}^r$; since $p^m=0$, we must have $a_{n+1}^m=0$, so that $a_{n+1}$ is nilpotent, and hence $a_{n+1}x^{n+1}$ is nilpotent. Since $\mathrm{nil}(R[x])$ is an ideal, we have that $p-a_{n+1}x^{n+1}\in\mathrm{nil}(R[x])$, which is of degree $\leq n$, and therefore the inductive hypothesis implies all of its coefficients are nilpotent. Thus, all of the coefficients of $p$ are nilpotent, and by induction, we are done.

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  • $\begingroup$ very nice proof. Thanks. $\endgroup$ – PatrickR Dec 22 '12 at 23:14
  • $\begingroup$ No problem, glad to help! $\endgroup$ – Zev Chonoles Dec 22 '12 at 23:16
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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 $\mathfrak{p}$.

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  • 4
    $\begingroup$ The reverse inclusion can be proved this way as well. Suppose each coefficient $a_i$ is nilpotent in $R$ and let $\mathfrak{q}$ be a prime of $R[x]$. Then $\mathfrak{p}=\mathfrak{q}\cap R$ is a prime of $R$, whence, by assumption, $f\in\mathfrak{p}R[x]\subseteq\mathfrak{q}$ ($\mathfrak{p}R[x]$ consists of the polynomials all of whose coefficients are in $\mathfrak{p}$). Thus $f$ is in $\mathfrak{q}$, and as $\mathfrak{q}$ was arbitrary, $f$ is nilpotent. $\endgroup$ – Keenan Kidwell Jun 25 '14 at 19:53

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