Nilradical of a noncommutative Ring I know in a commutative ring, the $\text{Nil}(R)$ is the ideal consisting of the Nilpotent elements of the ring.  
$$
\text{Nil}(R) = 
\Bigg{\{}  r \in R‎ \mid \forall n \in \mathbb{N} ‎r^n=0  \Bigg{\}}‎. 
$$
And I know it's not true for any ring which is not commutative.
I have proved the statement for the case of commutative rings; 
Could any one give a counter example of (“ In any Ring the set of all Nil(R) is an ideal”)?
 A: In the non-commutative ring case the definition of the nilradical being the ideal consisting of nilpotent elements does not work in general.
Take the ring $R=M_2(\mathbb{Z})$. The nilpotent elements do not form an ideal in $R$, since in general the sum of two nilpotent elements is not nilpotent. As an example, take the two nilpotent elements
$$x = \left(\begin{array}{cc}
0 & 1\\
0 & 0
\end{array}\right)\quad\text{and}\quad y = \left(\begin{array}{cc}
0 & 0\\
1 & 0
\end{array}\right).$$ Their sum $x+y$ is not nilpotent.
Reference: Nilradical of a ring
A: Take any field $F$ and consider the full matrix ring $R=M_2(F)$.  It's well known that this ring only has trivial ideals.
Obviously there are elements which are nilpotent, like $\begin{bmatrix}0&1\\0&0\end{bmatrix}$, so if the nilpotent elements formed an ideal, it would have to be all of $R$. But that's absurd since there are non-nilpotent elements (like the identity.)
Dietrich gave an example of the sum of two nilpotents not being nilpotent. Let me also point out that the product of a nilpotent element with another element may not be nilpotent either, so that the set of nilpotents does not absorb like an ideal is required to do.
For example:
$$
\begin{bmatrix}0&1\\0&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}=\begin{bmatrix}1&0\\0&0\end{bmatrix}
$$
The first matrix is nilpotent, the second is a unit, and the last one is an idempotent (and not nilpotent.)
This works for any field and any of its full matrix rings.
