Let $I$ be a nilpotent ideal in a ring $R$. It could be easily deduced, by the definition of product of ideals, that the full matrix ring $\mathbb M_n(I)$ for any natural number $n$ is also nilpotent.
Now, is the above statement true if "nilpotent" is replaced by "idempotent" in the sense that $I^2=I$?
If $X=(x_{ij})\in \mathbb M_n(I)$ then $x_{ij}=\sum_t a^{(t)}_{ij}b_{ij}^{(t)}$, a finite sum with $a^{(t)}_{ij}, b_{ij}^{(t)}\in I$. And we search for matrices $A^{(u)}_{kl},B^{(u)}_{kl}$ in $ \mathbb M_n(I)$ with $X=\sum_u A^{(u)}_{kl}B^{(u)}_{kl}$, a finite sum.
Thanks for any help!
If I understand the question correctly we want to know if every $X \in \mathbb{M}_n(I)$ can be written as $X = \sum_{i=1}^n A_i B_i$ with $n \in \mathbb{N}$ and $A_i, B_i \in \mathbb{M}_n(I)$. This can be shown as follows:
For all $1 \leq i,j \leq n$ and $a \in I$ let $E_{ij}(a) \in \mathbb{M}_n(I)$ denote the matrix with $a$ as the $(i,j)$-th entry and all other entries $0$. Because every matrix in $\mathbb{M}_n(I)$ is the sum of such matrices it sufficies to show the statement for these matrices.
Because $I^2 = I$ there exist $a_1, b_1, \dotsc, a_n, b_n \in I$ with $a = \sum_{k=1}^n a_k b_k$. Thus $$ E_{ij}(a) = E_{ij}\left( \sum_{k=1}^n a_k b_k \right) = \sum_{k=1}^n E_{ij}(a_k b_k) = \sum_{k=1}^n E_{i1}(a_k) E_{1j}(b_k). $$