Are nilpotent elements in the right singular ideal? Let $F$ be a field. I want to find the right singular ideal $Z_r(R)$ of the ring $$R=\begin{bmatrix}
F[x] & 0 \\
F[x,y]/(x,y^2) & F[y]/(y^2) \\
\end{bmatrix},$$ introduced in Exercise 6, p.36 of Goodearl's book:"Ring Theory, Nonsingular Rings and Modules", and to know whether $Z_r(R)$ contains the nilpotent elements of $R$. 
In the Book, it is claimed that $Z_r(R/Z_r(R))\neq 0$. From the exercise, one infers that $Z_r(R)\neq 0$. Thanks for any cooperation!
 A: No, $Z_r(R)$ doesn't have to (and doesn't, in this case) contain the nilpotent elements. Actually, it is the other way around for a Noetherian ring. That is, $Z_r(R)$ is a nil ideal in any right Noetherian ring. So we at least know that $Z_r(R)$ is certainly contained in $\begin{bmatrix}0&0\\ F[x,y]/(x,y^2)&(y)\end{bmatrix}$.
It is true, however, that the singular ideal contains central nilpotent elements. (This is an easy proof by contradiction.)
The interesting thing to notice is that $\begin{bmatrix}0&0\\F&0\end{bmatrix}$ is a right ideal of $R$.
From there, you can compute that $\begin{bmatrix}0&0\\ 0&y\end{bmatrix}$ is nilpotent, and yet its right annihilator has trivial intersection with $\begin{bmatrix}0&0\\ F&0\end{bmatrix}$, so it is not an element of $Z_r(R)$.
You can further compute that $\begin{bmatrix}0&0\\ F[x,y]/(x,y^2)&0\end{bmatrix}$ is contained in $Z_r(R)$, and with the previous observation, is equal to $Z_r(R)$.
Then $R/Z_r(R)\cong F[x]\times F[y]/(y^2)$, which clearly has nonzero central nilpotent elements, which therefore must be in $Z_r(R/Z_r(R))\neq \{0\}$.
