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Let $A = M_2(\mathbb{C})$ and let $M = \mathbb{C}^2$ with its usual $A$-module structure. How do I show that $M$ is a faithful $A$-module? I understand that this amounts to showing that its annihilator is just the zero ideal, but I am not sure what do from there. Any help would be appreciated, thanks.

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    $\begingroup$ Faithful means that the mapping is injective, which is clear for the identity mapping. $\endgroup$ May 22, 2015 at 9:54
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    $\begingroup$ If $X \in M_2(\mathbb{C})$ is a nonzero matrix, can you find a vector $v \in \mathbb{C}^2$ such that $Xv \neq 0$? It helps to put abstract definitions in simple words, sometimes... $\endgroup$ May 22, 2015 at 10:33

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So the question has an answer:

Suppose $M$ is a matrix whose first column is somewhere nonzero. Then if $v=(1,0)^t$ then $Mv\neq 0$ because it is equal to the first column of $M$. A similar argument works if the second column is nonzero with the vector $(0,1)^t$. Thus no nonzero matrix annihilates the module.

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