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?

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.

• Thank you for the detailed explanation. Aug 7, 2022 at 16:09
• Is there any good reference about what is written above ? Aug 8, 2022 at 7:50
• @Walterfield Any book on noncommutative rings, for instance Anderson-Fuller. Aug 8, 2022 at 9:02
• Thank you for the reply. Aug 8, 2022 at 9:32

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

• Thank you for the answer. I think $A^2S= 0$ induces $AS=0$. This is also not correct ? Aug 7, 2022 at 13:58
• My counterexample tells you that the answer is NO. Anyway, $A^2=0$, so basically you are saying that $0=0\Rightarrow AS=0$. It cannot be correct. Aug 7, 2022 at 14:09
• Thank you for the reply. I thought $AS$ is a submodule of $S$. I misunderstood. Aug 7, 2022 at 14:26
• Why I misunderstood seems to be that I didn't pay attention to whether $A$ is a normalizing element (i.e. AS = SA). Aug 7, 2022 at 14:36