About faithful module Let $K$ be a field and $A$ be an algebra over  $K$($A$ not neccssarily commutative algebra). Let $M$ be an $A$-module and it's called a faithful $A$-module if $\operatorname{Ann}_{A}(M)=0$. There is a question about faithful module:

Let $M$ be a finitely generated $A$-module, then $M$ is faithful if and only if there exist an injective $A$-module homomorphism $f:A \rightarrow {\oplus}^{n}_{i=1}M \quad$ for some $n < \infty$.

From RHS to LHS is easy, but from LHS to RHS I have no idea to finish it. And I think maybe the algebra should add more conditions?
Thanks for the help.
 A: Firstly, the last homomorphism would be determined by $1\mapsto (x_1,\dots,x_n)$ for some $x_i\in M$.  In order to be injective, it would have to be the case that $\bigcap ann(x_i)=\{0\}$.  So as you can see, success depends on finding this set of $x_i$'s.
If $A$ were commutative, we would be fine: we could just select a generating set $m_1,\ldots m_n$ of $M$, and it would necessarily mean $ann(M)=\bigcap ann(m_iR)=\bigcap ann(m_i)=\{0\}$.
In the noncommutative case you still have $\bigcap ann( m_iR)=\{0\}$, but it is not so clear why there should also be a finite set of point annihilators that intersects to zero.
One way to do it is to require $A$ to be right Artinian, because then $A$ is also "right finitely cogenerated" meaning that any collection of submodules which intersects to zero necessarily has a finite subcollection that intersects to zero. Applying this to $\{ann(x)\mid x\in m_iR\}$ would extract a finite subset of $x_{ij}$'s for $m_i$, and you do it for all $i$ to get a larger (but finite) collection of $x_{ij}$'s, say $N$ of them.
Then you would be able to conclude that $1\mapsto (\ldots, x_{ij},\ldots)$ would determine a map into $\oplus_{i=1}^N M_i$ that is injective.
A non-Artinian counterexample
Let $A$ be the algebra of linear transformations of a countable dimensional vectors space $V_F$. Fix a basis $b_1, b_2,\ldots $ and consider the idempotent element $e$ which sends $b_1\mapsto b_1$ and $b_i\mapsto 0$ otherwise.
I claim that $eA$ is a faithful $A$ module. There is only one notrivial ideal that could be the annihilator: the transformations of finite rank.  However, you can easily see that it is impossible for $eRa=\{0\}$ for a nonzero finite-rank transformation $a$. No matter what nonzero $a$ is picked, you can pick a $g$ that maps a nonzero element in $Im(a)$ to $b_1$, and then $ega\neq 0$. So, $M=eA$ is faithful, and actually it is a simple module.  Then you're never going to fit a copy of $A$ into $\oplus_{i=1}^N M$ because the latter will be semisimple and $A$ isn't.
