$A$ noetherian ring, $M$ finite $A$-module. Why is this an exact sequence?

Question

So I read that under this conditions, for some $p$ (which I believe to be prime but I am not entirely sure) we can generate the following short exact sequence: $$0\longrightarrow\ker\beta\xrightarrow{f}A^p\xrightarrow{\beta}M\longrightarrow0$$ So I guess this comes from the fact that if $M=\langle\{m_i\}_{i=1}^k\rangle$ (since it is a finite module), since $A$ is noetherian (and so is $A^p$) we can send the ${e_i}$ that generate $A^p$ to the $m_i$ that generate $M$. However I don't really understand why the $p$ is needed, because we can also find some $e'_i$ that generate $A$. Why can't it always be $p=1$?

Answer 1

To have the sequence exact at $M$ one requires $\beta$ to be surjective.

If $M$ is generated by $m_1,\dots,m_k$ then we can set $p=k$ (not necessarily prime) and send the basis $e_1,\dots,e_k$ of $A^k$ to $m_1,\dots,m_k$, respectively.
This ensures $\beta$ is surjective.