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(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.

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    $\begingroup$ Have you tried taking $A$ to be a division algebra (not equal to $k$)? $\endgroup$ Oct 5, 2020 at 12:34
  • $\begingroup$ @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? $\endgroup$ Oct 5, 2020 at 13:47
  • $\begingroup$ @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. $\endgroup$ Oct 5, 2020 at 14:19

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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.

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

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