Contradictions about simple modules of matrix rings

Question

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?

Answer 1

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.

Answer 2

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 $AM=\pmatrix{1 & 0\cr 0 & 0}\neq 0$.