Contradictions about simple modules of matrix rings Let $k$ be a field and $R := M_2(k)$ be the ring of $2 \times 2$ matrices over $k$.
Let $S$ be a simple left $R$-module.
Then, matrices
$ A = 
\begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix}$ and $ B=
\begin{pmatrix}
0 & 0 \\
1 & 0
\end{pmatrix}
$
are nilpotent.
So, they are annihilaters of $S$.
Moreover, $ AB = \begin{pmatrix}
1 & 0 \\
0 & 0
\end{pmatrix},BA= \begin{pmatrix}
0 & 0 \\
0 & 1
\end{pmatrix}
$.
Thus, all elements in $R$ annihilate $S$.
So, $S$ is the zero module.
However, it is well-known that the simple module of $R$ is isomorphic to $k^{\oplus 2}$.
This fact contradicts the above argument.
Where is the mistake in the above argument?
 A: In a commutative ring, nilpotent elements annihilate every simple module, because any nilpotent element belongs to the Jacobson radical.
This is generally false for noncommutative rings. For instance, the Jacobson radical of $M_2(k)$ is the zero ideal, which in particular means that for every nilpotent element $A\in M_2(k)$ there exists a simple module $S$ which is not annihilated by $A$.
What's a possible obstruction? An element $r\in R$ belongs to the Jacobson radical if and only if $1-rs$ is invertible for every $s\in R$. When the ring is commutative and $r$ is nilpotent, then $rs$ is nilpotent as well; moreover $1-r$ is invertible whenever $r$ is nilpotent. However, for noncommutative rings, even if $r$ is nilpotent, it's not generally true that $rs$ is nilpotent as well.
Consider
$$
B\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} =
\begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix}
$$
and the characteristic polynomial is $X^2-X$, so this matrix is not nilpotent.
A: Nilpotent elements of $R$ do not necessarily annihilate $S$.
Assume for example that $S=\{ \pmatrix{a & 0 \cr b & 0}, a,b\in k\}$. This is a simple left $R$-module.
Then $M=\pmatrix{ 0 & 0 \cr 1& 0}\in S$ but $AM=\pmatrix{1 & 0\cr 0 & 0}\neq 0$.
