Prime radical of block matrix ring I'm studying the matrix algebra $M$ that consists of all matrices of the form
$$\left(\begin{array}{cc}A&B\\0&C\end{array}\right)$$
where $A,B,C$ are $2\times2$ matrices with coefficients in a fixed field $F$. I want to calculate the prime radical of $M$. Now I know that in this case of a finite-dimensional $F$-algebra, the prime radical is nilpotent and therefore must consist of all nilpotent matrices in $M$. My problem is trying to give an explicit description of these matrices. It is clear to me that if the above matrix is nilpotent, then $A$ and $C$ also need to be nilpotent, but I'm not quite sure how to proceed farther. Any help would be greatly appreciated.
 A: While it certainly consists of nilpotent elements it is not true that it consists of all nilpotent elements in M.  This is also the case for just a square matrix ring over a field: the prime radical is zero despite the existence of nonzero nilpotent elements.
One thing that is going to help you here is that in a finite dimensional algebra like this one, prime ideals are maximal.
Now obviously the matrices with $A,C$ zero form an ideal $I$ that squares to zero, so it is contained in the prime radical .  So to find the maximal ideals one only has to find the maximal ideals of $M/I$.   But these are obvious, because $M/I$ is clearly isomorphic to a product of two full matrix rings.
After describing the two maximal ideals and lifting them to $M$, you will be able to conclude that $I$ is already the prime radical.
The set of nilpotent elements, however, is strictly larger, because it would contain elements of the form $\begin{bmatrix}N_1 & B \\ 0 & N_2\end{bmatrix}$ where $N_1$ and $N_2$ are nilpotent matrices.
