Question concerning a faithful module over an Artinian ring

Question

Let $A$ be an Artinian ring and $M$ an $A$-module.

Is it true that $M$ is faithful if and only if there is an exact sequence of the form $0\rightarrow A \rightarrow M^r$ for some natural $r$ ?

Thanks for the help.

Answer 1

"$\Leftarrow$" If $a\in A$ annihilates $M$, then $a$ annihilates $M^r$. In particular, $a$ annihilates $A$ and hence $a=0$.

"$\Rightarrow$" If $(x_i)_{i\in I}$ is a generating set for $M$, then there is an injective homomorphism $A\to\prod_{i\in I} Ax_i$ given by $a\mapsto(ax_i)_{i\in I}$, and thus we have an exact sequence $$0\to A\stackrel{f}\to M^I.$$ If $|I|<\infty$ we are done. Otherwise, let $p_i:M^I\to M$ be the canonical projections, and $q_i=p_if$. Since $\cap_{i\in I}\ker p_i=0$ we also have $\cap_{i\in I}\ker q_i=0$. Furthermore, since $A$ is artinian there is a finite subset $J\subset I$ such that $\cap_{j\in J}\ker q_j=0$. Thus the map $\oplus_{j\in J} q_j:A\to M^J$ is an embedding.

Edit. This is an alternative proof which works for non-commutative rings as well. For each $x\in M$ let $\phi_x:A\to M$ be the homomorphism defined by $\phi_x(a)=ax$. Since $M$ is faithful we have $\cap_{x\in M}\ker\phi_x=0$, and since $A$ is Artinian there is a finite subset $\{x_1,\dots,x_n\}$ of $M$ such that $\cap_{i=1}^n\ker\phi_{x_i}=0$. Then the map $\oplus_{i=1}^n\phi_{x_i}:A\to M^n$ is an embedding.