I'm trying to understand the proof of Proposition 2 from here (page 3).
1 $\implies $ As far as I understand, the map $R^n\rightarrow M$ is defined by $e_i\mapsto z_i$. First, what does it mean mathematically that the composition $R^n\rightarrow M\rightarrow M/\mathscr{m}M$ "factors through" the surjection $R^n\rightarrow k^n$? Furhter, how does the desired result follow from the fact that we have a surjection $k^n\rightarrow M/\mathscr{m}M$? Where does this argument use the hypothesis?
1 $\leftarrow$ Why $M=N+\mathscr{m}M$? Again, how does it use the hypothesis?
In statement 2, I'm only interested in the first equivalence (not involving tensor products).
2 $\implies$ Generally, if a set $B=(b_1,\dots,b_n)$ doesn't form a basis for $V$, then either $B$ is dependent or $SpanB\ne V$. The proof claims that in the case being considered, the former is the case, i.e., it is said that the $\overline{z_i}$ are dependent. How do we know this? Maybe they are independent but don't span the space.
2 $\leftarrow$ How to prove this implication without knowing the notion of tensor product?
Let's start with 1:
$(\Longrightarrow)$ Let's denote the surjective composition $R^n \twoheadrightarrow M \twoheadrightarrow M/\mathfrak mM$ by $\theta:R^n \twoheadrightarrow M/\mathfrak mM$. Let's denote the surjective map $R^n \twoheadrightarrow k^n$ given by reducing each coordinate modulo $\mathfrak m$ by $\pi:R^n \twoheadrightarrow k^n$. To say that $\theta$ factors through $\pi$ is to say that there is a unique map $\bar\theta: k^n \to M/\mathfrak mM$ such that $\theta = \bar\theta \circ \pi$. Moreover, $\bar\theta$ must be surjective. The map $\bar \theta$ sends $(a_1,\dots,a_n)$ to the linear combination $\sum_{i=1}^n a_i \bar{z_i}$, where $\bar{z_i}$ is the image of $z_i$ in $M/\mathfrak mM$, so the fact that $\bar\theta$ is a surjection implies the result.
$(\Longleftarrow)$ To say that the $\bar{z_i}$ generate $M/\mathfrak mM$ is to say that, for any $m\in M$, we may write $\bar m = \sum_{i=1}^n \bar{a_i} \bar{z_i}$ for some $a_i\in R$ with $\bar{a_i}$ meaning the image in $R/\mathfrak m = k$. If we lift all the $a_i$ to $R$ and all the other elements to $M$, then we still get a relation $$ m = m' + \sum_{i=1}^n a_iz_i,$$ for some $m' \in \mathfrak mM$. We have shown that any element of $M$ can be written as the sum of an element of $\mathfrak mM$ and an element of $N$, i.e. $M = N + \mathfrak mM$.
Now let's move on to 2:
$(\Longrightarrow)$ You're right; a priori we don't know whether the $\bar{z_i}$ are dependent or don't span. However, we know that the $z_i$ span $M$, so we may use part 1 to conclude that the $\bar{z_i}$ span $M/\mathfrak mM$, and thus they must be dependent if they are not a basis. Do you understand the rest of this implication?
$(\Longleftarrow)$ We are assuming that the $\bar{z_i}$ form a basis for $M/\mathfrak mM$, and this implies that no proper subset of the $\bar{z_i}$ spans $M/\mathfrak mM$. If the $z_i$ were not a minimal generating set, then a proper subset of them would span $M$, and then by part 1, a proper subset of the $\bar{z_i}$ would span $M/\mathfrak mM$, which we have seen already is not possible.