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Let $A$ be a commutative ring with identity and $N(A[x]):=$ the nilradical of $A[x]$, $J(A[x]):=$ the Jacobson radical of $A[x]$. Then we prove $N(A[x])=J(A[x])$.

First, it's obvious that $N(A[x])\subset{J(A[x])}$. So just need to prove $J(A[x])\subset{N(A[x])}$.

Second, use the following lemma.
[Lemma]Let $f=a_{0}+a_{1}X+\cdots+a_{n}X^n\in{A[X]}$. Then
$(1)$ $f$ is nilpotent iff $a_{i}$ is nilpotent for all $i\ge{0}$;
$(2)$ $f$ is a unit in $A[x]$ iff $a_{0}$ is a unit in $A$ and $a_{i}$ is nilpotent for all $i\ge{1}$.

Let $f=a_{0}+a_{1}X+\cdots+a_{n}X^n\in{J(A[x])}$. The fact is equal to say that for any $g\in{A[x]}$, $1+fg$ is a unit in $A[x]$. So $a_{0}\in{J(A)}$ and $a_{i}$ is nilpotent for all $i\ge{1}$. Now we just need to prove $a_{0}\in{N(A)}$. That is to say we must prove $J(A)\subset{N(A)}$.

This can't be always true for any ring $A$ that $J(A)={N(A)}$?

I don't know where I am wrong and if someone can make an example about $J(A)$ doesn't equal to ${N(A)}$, I will be very thankful.

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Let $g = x$ and look at the degree $1$ term of $1 + fg$.

P.s. It's true that $J(A) = N(A)$ is not true in general. I don't know of an example off hand, you can probably google for one. But $J(A) = N(A)$ is true when $A$ is a finitely generated algebra over a field, so such examples aren't usually very nice.

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    $\begingroup$ Every local integral domain (which is not a field) satisfies $J(A)\ne N(A)$. For instance, take $A=K[[X]]$. The funny thing is that you have a post about this. $\endgroup$ – user26857 Dec 17 '13 at 8:37
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    $\begingroup$ Ha, I feel like I'm forgetting more math then I'm learning these days... $\endgroup$ – Jim Dec 17 '13 at 17:09

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