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Suppose $V\neq0$ is a representation of an algebra A.

Definition: $v\in V$ is cyclic if and only if it generates $V$, thus $Av=V$. If a representation has a cyclic vector we call the representation cyclic.

I proved the following: $V$ irreducible iff all non-zero vectors of $V$ are cyclic (this follows immediatly from definitions).

But now i want to prove next result: $V$ is cyclic iff it is isomorphic to $A/I$ where $I$ is a left-ideal of $A$. But how to do this, i can not imagine an left ideal such that the representations $V$ and $A/I$ are isomorphic. We have thus to construct an intertwiner, but how?

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Hint 1 Look at the homomorphism $A\to V, a\mapsto av$ for some cyclic vector $v$. Then use the isomorphism theorem.

Hint 2 For the converse consider $1+I \in A/I$.

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Can you give a hint for the other direction of the proof? So prove that if $V$ is isomorphic to $A/I$ for some left ideal $I$, then $V$ is cyclic. – Badshah Sep 21 '14 at 19:10
@Badshah I edited to include another hint. – Julian Kuelshammer Sep 21 '14 at 19:16
Reading your hint I believe the argument should be as follows: I only have to show that $A/I$ is a cyclic representation of $A$, which should be trivial if you take $1+I\in A/I$ and if the representation is just left multiplication. But can there be no other type of representation of $A$ by $A/I$, more explicitly: not just left or right multiplication? – Badshah Sep 21 '14 at 19:20
@Badshah Of course, there could be other module structures on the vector space $A/I$. But if one writes $A/I$ for a left ideal $I$ it is implicitly assumed that the action is given by left multiplication. The right multiplication does not give you a left module in general. – Julian Kuelshammer Sep 21 '14 at 19:25
But how can we know that if we have this representation $V$ that then the representation $A/I$ of $A$ is just left multiplication? I mean, the representation $V$ can be anything, so why is then the representation $A/I$ just left multiplication? – Badshah Sep 21 '14 at 19:31

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