R with different Jacobson radical and nilradical (both non-zero) I have seen a post in MathStack with $R=\mathbb{Z}[x]$, $p=(x^2+1)$ and $m=(x^2+1,2)$.
I can not understand this example. Can you help me to understand it? Or may be giving any other example?
I think I have not seen this kind of theory in any lecture and that is why I am not able to understand they way he/she build it.
Nilradical is zero here?
 A: $\mathbb Z[x]$ is a domain and its Jacobson radical and nilradical are both zero. I'm not sure what the ideals are supposed to mean because you didn't give any other context as to the example you are alluding to. You should at the very least link to the place you found it.
There is a fairly obvious strategy though: take a domain with prime ideals $P, Q$ such that $\{0\}\neq P\subsetneq Q\lhd R$, and then 


*

*Take $R'=R_Q$, which means $R$ localized at $Q$. This is now a local domain with maximal ideal corresponding to $Q$ and a nontrivial prime ideal between $Q$ and zero

*$R''=R'/(P')^2$, where $P'$ is the ideal of $R'$ corresponding to $P$.  Now in $R''$, the nilradical is smaller than $P'$, but since $R'$ is local, $R''$ is local, and its Jacobson radical is still the maximal ideal.
The example you're suggesting could be modified to do that, but $(x^2+1,2)$ would not work since $(x^2+1,2)$ is not maximal. It'd work with $Q=(x^2+1,3)$, though.
A: To solve this I we will use two Lemmas.

Lemma 1: $x \in J(A) \Leftrightarrow 1-xy$ is unit in $A$ for all $y\in A$.
Proof 1: $\boxed{\Rightarrow}$ Suppose $1-xy$ is not a unit. We know that all non-units of A belongs to some maximal ideal $m$; but $x\in J(A) \subseteq m$, hence $xy\in m$ and therefore $1\in m$, which is absurd.
$\boxed{\Leftarrow}$ Suppose $x \notin m$ for some maximal ideal $m$. Then $m$ and $x$ generate the unit ideal $(1)$, so that we have $u+xy=1$ for some $u\in m$ and some $y \in A$. Hence $1-xy \in m$ and is therefore not a unit.

and

Lemma 2:$f$ is unit in $A[[x]] \Leftrightarrow a_0$ is unit in $A$.
$\boxed{\Rightarrow}$ If $g =\sum_m b_m x^m$ is an inverse of $f$, then $fg=0$ implies $a_0 b_0 =1$ so that $a_0$ is unit.
$\boxed{\Leftarrow}$ Supposing $a_0 $ is unit, we construct an inverse $g=\sum_m b_m x^m$ to f. Let $b_0 = a_0^{-1}$. We want $fg=\sum_j c_j x^j =1$, so for $j \geq 1$ we want $c_j = \sum_{n=0}^j a_n b_{j-n} =0$. Now suppose we have found satisfactory coefficients $b_j$ for $j \leq k$. We need $c_{k+1} = a_0 b_{k+1} + \sum_{n=1}^{k+1} a_n b_{k+1-n} = 0$; but we can solve this to find the solution $b_{k+1} = -a_0^{-1} (\sum_{n=1}^{k+1} a_n b_{k+1-n}) $. Since we can do this for all $k$, we have constructed an invers to $f$.

Now consider $R=(\mathbb{Z}/4\mathbb{Z}) [[x]]=\mathbb{Z}_4[[x]]$.
First of all, we will see which is the Jacobson radical $J(\mathbb{Z}_4[[x]])$.
By Lemma 1, $f \in J(\mathbb{Z}_4[[x]]) \Leftrightarrow 1-fg$ is unit in $\mathbb{Z}_4[[x]]$ for all $g \in \mathbb{Z}_4[[x]]$.  By Lemma 2, $1-fg = \sum c_j x^j$ is unit in $\mathbb{Z}_4[[x]]$ iff $c_0$ is unit in $\mathbb{Z}_4$.\
Consider $f=a_0 +a_1 x + a_2 x^2 +...$ and $g= b_0 + b_1x +b_2 x^2 +...$ with $a_i, b_j \in \mathbb{Z}_4$. That means $c_0 = 1- a_0b_0$. Moreover, we know that $[1], [3]$ are the units of $\mathbb{Z}_4$.

*

*$[1] = [1] - a_0 b_0 \Leftrightarrow a_0b_0= [0] \Leftrightarrow a_0= [0]$ since we need the equality for all $g \in \mathbb{Z}_4[[x]]$.

*$[3] = [1]- a_0 b_0 \Leftrightarrow a_0 b_0 = [-2]$ which it is equal to $a_0b_0 = [2]$ since we are in $\mathbb{Z}_4$. But there is no $a_0$ which satisfies this expression for any $b_0$.

Hence, $f\in J(\mathbb{Z}_4[[x]])$ if and only if $a_0 =[0]$. Hence  $$J(\mathbb{Z}_4[[x]])=\{ \sum_{n=0}^{\infty} a_n x^n = f\in \mathbb{Z}_4[[x]] \hspace{0,1cm}|\hspace{0,1cm} a_0 = [0]\}$$
Once we know $J(\mathbb{Z}_4[[x]]) \neq 0$, we need to see that $N(\mathbb{Z}_4[[x]]) \neq 0$ and $N(\mathbb{Z}_4[[x]]) \neq J(\mathbb{Z}_4[[x]])$.
To see that $N(\mathbb{Z}_4[[x]]) \neq 0$ we will consider $f(x)=2x^2+2$. Since $(f(x))^2 = (2x^2 +2)^2 = 4 ( x^4 +2x^2 +1 ) = 0$ in $\mathbb{Z}_4$, $f \in N(\mathbb{Z}_4[[x]])$. This is because $\exists N$ positive integer such that $f^N = 0$.\
Moreover, $f(x)=(2x^2+2)\notin J(\mathbb{Z}_4[[x]])$ since $a_0 = [2]$.
Clearly, in $\mathbb{Z}_4[[x]]$, $J(\mathbb{Z}_4[[x]]) \neq N(\mathbb{Z}_4[[x]])$ and they are both not zero.
Remark: A similar easy process can be done with any $\mathbb{Z}_n$, with $n$ not prime.
