The following is taken from: $\textit{Partial Differential Control Theory Vol 1: Mathematical tools}$ by J F. Pommaret
$\color{Green}{Background:}$
$\textbf{Definition 1.50.}$ $M$ is call a $\textit{free module}$ if it has a $\textit{basis},$ that is a system of generators linearly independent over $A.$ When $M$ is free, the number of generators in a basis is called the $\textit{rank}$ of $M$ and does not depend on the basis if it is finite. Indeed, if $\{x_i\}$ and $\{y_j\}$ are two bases and $y_j=\sum {a^i}_jx_i,$ then $\text{det}({a^i}_j)$ must be invertible in $A.$ If $M$ is free of rank $n,$ then $M\cong A^n.$
$\textbf{Lemma 1.54.}$ In a short exact sequence of modules:
$$0\to M'\xrightarrow{f}M\xrightarrow{g}M''\to 0$$
the module $M$ is noetherian if and only if the momdules $M'$ and $M''$ are noetherian.
$\textbf{Proposition 1.55.}$ If $A$ is a noetherian ring and $M$ is a finitely generated $A-$module, then $M$ is a noetherian module.
$\textit{Proof.}$ Applying the preceding lemma (Lemma 1.54) to the short exact sequence $0\to A^{r-1}\to A^r\to A\to 0$ where the right epimorphism is the projection onto one factor, we deduce by induction that $A''$ is a noetherian module. Now, if $M$ is generated by $\{x_1,\ldots,x_n\}$ then there is an epimorphism $A^n\to M:(1,0,\ldots,0)\to x_1,\ldots,(0,\ldots,0,1)\to x_n.$ According to the preceding lemma, $M$ is a noetherian module too.
$\color{Red}{Questions:}$
I have questions about the short exact sequence $A^n\to M:(1,0,\ldots,0)\to x_1,\ldots,(0,\ldots,0,1)\to x_n.$ in the proof of proposition 1.55 quoted above.
I know the notation $A^n$ means that the vector space/module has $n$ number of basis elements. But for $A^n\to M:(1,0,\ldots,0)\to x_1,\ldots,(0,\ldots,0,1)\to x_n,$ $(1,0,\ldots,0)$ should be an $n-$tuple element, but then what does the notation $x_1,\ldots,(0,\ldots,0,1)\to x_n$ mean? Does it mean something like:
$(1,0,\ldots,0)\to\begin{pmatrix}x_1 \\\ x_2 \\\ \vdots \\\ (0,\ldots,0,1) \end{pmatrix}\to x_n?$ So $(1,0,\ldots,0)$ becomes $(0,\ldots,0,1)$ and $(0,\ldots,0,1)$ is in the $n-$th place in $\begin{pmatrix}x_1 \\\ x_2 \\\ \vdots \\\ (0,\ldots,0,1) \end{pmatrix}$
Thank you in advance
No. It is simpler than that. First, $A^n$ is just the direct sum of $n$-copies of $A$, so elements are really $n$-tuples of elements in $A$. $A^n$ is, of course, a free $A$-module and the standard basis is $\{e_1,\ldots,e_m\}$ where $e_i$ is the element that has only zeroes except for a 1 in the $i$-th coordinate.
Second, to give a morphism of $A$-modules $F\to M$ where $F$ is a free $A$-module, it suffices to specify the images of the elements of a basis of $F$, and in fact they can be anything we want on $M$! (it suffices to specify it on a basis because it generates $F$ and we can choose anything on $M$ because it is linearly independent).
So what the book is saying is: Construct a morphism $A^n\to M$. Why? Well, by what I said, it suffices to choose images for the standard basis $$(1,0,\ldots,0)\,,\,(0,1,\ldots,0)\,,\,\ldots\,,\;(0,0,\ldots,1).$$
And he chooses $x_1$ for $(1,0,\ldots,0)$, he chooses $x_2$ for $(0,1,\ldots,0)$ and so on.