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Let $\mathcal A$ be a Banach algebra over $\mathbb{C}$, $\mathcal X$ a irreducible left $\mathcal A$-module. If $x,y \in \mathcal X$ are linearly independent, there exists an element $a\in\mathcal A$ such that $ax=x$ and $ay=0$.

Is it true? If it is true, how to prove?

Thanks a lot.

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If your irreducible means simple and if the linear independence is taken over $\cal A$ (rather than the base field of the algebra) then it's vacuously true since in simple modules every element generates the whole module so there are no linearly independent elements. Could you please clarify what definition of irreducibility and independence you are working with? –  Marek Sep 25 '13 at 13:57
    
@ Marek:Thank you. Irreducible means simple, independence is taken over the base field of the algebra. –  Zhonghua Wang Sep 26 '13 at 0:48
    
You 're right my example didn't work. –  plusepsilon.de Oct 8 '13 at 9:58

1 Answer 1

This is false. Consider $\mathcal{A} = \mathbb{C}$ as an algebra over $\mathbb{R}$, and let $\mathcal{X} = \mathcal{A}$, viewed as a (left) $\mathcal{A}$-module via left multiplication.

If $M$ is a non-zero submodule of $\mathcal{X}$ and $0 \neq m \in M$ then $a = (a/m)\cdot m \in M$ for all $a \in \mathcal{A}$ so $M = \mathcal{X}$. Hence $\mathcal{X}$ is simple.

Now $1$ and $i$ are elements in $\mathcal{X}$ which are linearly independent over the ground field $\mathbb{R}$, however if $ax = x$ then $a = 1$ and $ay = i \neq 0$.

The same example works whenever your simple module has endomorphism ring $D := End_{\mathcal{A}}(\mathcal{X})$ strictly bigger than the base field of the algebra.

However if you ask the same question with the stronger condition of linear independence over $D$, then the result is true. This follows from the Jacobson Density Theorem.

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Thank you. I forgot to say that $\cal{A}$ is an algebra over $\mathbb{C}$, which I have added. –  Zhonghua Wang Jan 11 at 4:21
    
Then take $\mathcal{A} = \mathcal{X}$ to be the quaternions, $x = 1$ and $y = j$. These are $\mathbb{C}$-linearly independent but since $\mathcal{A}$ is a division ring $ax = x$ forces $a = 1$ so $ay = y \neq 0$. Note that in this example the division ring $D$ is the opposite division ring of the quaternions, so $x,y$ are certainly not $D$-linearly independent. –  Konstantin Ardakov Jan 11 at 10:12

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