# Prove or give a counter-example about a statement on faithful simple left modules over a finite-dimensional algebra.

(My post in short) Prove or give a counter-example for this statement: (Let $$A$$ be a finite-dimensional algebra over $$k$$, and let $$M$$ be a simple left $$A$$-module. Furthermore, suppose that the action of $$A$$ over $$M$$ is faithful, and let $$\{m_1, m_2, \cdots, m_n \}$$ be a $$k$$-basis for $$M$$. Then for any nontrivail proper set $$\{0\}\neq I \subsetneq \{m_1, m_2, \cdots, m_n \}$$, we have $${\rm Ann}(I)\neq \{0\}$$.

There is no need to read the following texts.

Some unimportant details. (There is no need to read the following texts, you can ignore it)

I was reading the proof of this proposition:

Proposition: Let $$A$$ be a finite-dimensional algebra over $$k$$, and let $$M$$ be a simple left $$A$$-module. Furthermore, suppose that the action of $$A$$ over $$M$$ is faithful. Then $$M$$ is isomorphic to a left ideal of $$A$$, as a left $$A$$-module. Also, there exists an integer $$n$$ such that $$A$$ is isomorphic to $$M^n$$, as a left module.

During the proof, I had $$10$$ questions. I solved $$6$$ of them, so it remained $$4$$ of them. I think I can solve one of them myself, and I should think about another one. The last two are not needed for the proof, but if the second one is true, then the proof would be changed to a much more straightforward and constructive proof.

I will review the situation here: Let $$A$$ be a finite-dimensional algebra over $$k$$, and let $$M$$ be a simple left $$A$$-module. Furthermore, suppose that the action of $$A$$ over $$M$$ is faithful. Then $$M$$ has a finite dimension over $$k$$, as a $$k$$-vector space. (Let $$0\neq m \in M$$, and consider the $$A$$-submodule generated by $$m$$. Since $$M$$ is simple, this submodule is equal to $$M$$. So $$M$$ is generated by $$m$$, over $$A$$, as $$A$$-module.) Let $$\{m_1, m_2, \cdots, m_n \}$$ be a $$k$$-basis for $$M$$.

1. Do the dimension of $$M$$, as a $$k$$-vector space, equals the dimension of $$A$$, as a $$k$$-vector space? I can see that $$\dim_k(M) \leq \dim_k(A)$$, but I can not show the reverse inequality. (This question is not very important for me)

Clearly $${\rm Ann}(\{m_1, m_2, \cdots, m_n \})={\rm Ann}(M)=\{0\}$$, because $$A$$ acts faithfuly on $$M$$.

1. (Main question) Prove or give a counter-example for this statement: (Let $$A$$ be a finite-dimensional algebra over $$k$$, and let $$M$$ be a simple left $$A$$-module. Furthermore, suppose that the action of $$A$$ over $$M$$ is faithful, and let $$\{m_1, m_2, \cdots, m_n \}$$ be a $$k$$-basis for $$M$$.) Then for any nontrivial proper set $$\{0\} \neq I \subsetneq \{m_1, m_2, \cdots, m_n \}$$, we have $${\rm Ann}(I)\neq \{0\}$$.

Note that if $$I \subseteq J$$, then $${\rm Ann}(J) \subseteq {\rm Ann}(I)$$. Therefore if this statement is true, then it suffices to show that $${\rm Ann}(M\backslash \{m_i\})\neq \{0\}$$, for any $$1\leq i \leq n$$. Also if there exists a counter-example, we can find it amont these maximal proper subset $$M\backslash \{m_i\}$$'s.

• Have you tried taking $A$ to be a division algebra (not equal to $k$)? Oct 5, 2020 at 12:34
• @AndrewHubery Nice idea. If $A$ is a division algebra, then for any $0\neq a \in A$, ${\rm Ann}(a)=0$. So it remains to shows that there is a simple left $A$-module $M$, of $\dim_kM \geq 2$, on which $A$ acts faithfully. I think if $\dim_kA \geq 2$, then $M=A$ is a good choice. But why is this a simple left $A$-module? Oct 5, 2020 at 13:47
• @AndrewHubery I think I get the point: The only left $A$-modules of $A$ are $\{0\}$ and $A$. So it is simple. Thank you so much. Oct 5, 2020 at 14:19

## 1 Answer

Consider $$A=M=\mathbb C$$ and $$k=\mathbb R$$. Any annihilator of a nonzero element of $$M$$ in $$\mathbb C$$ is a proper right ideal, but there is only one.

• I think this is a special case of the comment by @AndrewHubery I appreciate both of you. Oct 5, 2020 at 13:51