$m ∈ M$ such that the image in $M_{\mathfrak{p}}/\mathfrak{p}M_{\mathfrak{p}}$ is not zero Let $M$ be a finitely generated $A$-module. For every prime ideal $\mathfrak{p}\in \operatorname{Spec}{A}$, define
$$μ(\mathfrak{p},M)=\dim_{k(\mathfrak{p})}(M_{\mathfrak{p}}/\mathfrak{p}M_{\mathfrak{p}}),$$
where $k(\mathfrak{p}) = A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$
Let $S$ be a finite set of prime ideals of $A$ such that $M_{\mathfrak{p}} \ne 0$ for every $\mathfrak{p} \in S$.
My text reads the following equivalence is clear,

For an element $m \in M$, the image of $m$ in $M_{\mathfrak{p}}/\mathfrak{p}M_{\mathfrak{p}}$ is not zero for all $\mathfrak{p} \in S$ if and only if $μ({\mathfrak{p} }, M/Am)=μ({\mathfrak{p} }, M)-1$ for all $\mathfrak{p}\in S$.

But why? Do I need to calculate $$μ({\mathfrak{p}}, M/Am)$$ directly? Thank you for your help.
lem1.2 of this page 'in other words' the last paragraph of statement lem1.2 is my question.
https://www2.math.upenn.edu/~chai/602-3_2010-11/notes/forster-swan_notes.pdf
 A: Hints: $M_{\mathfrak{p}}/\mathfrak{p}M_{\mathfrak{p}}$ is a module over $\kappa(\mathfrak{p})$, i.e. a $\kappa(\mathfrak{p})$-vector space.
The image of an element $m\in M$ in this vector space is non-zero if and only if it spans a one dimensional subspace. The claim follows then by playing around with the isomorphism theorems $(A/B)/(C/B)\cong A/C$ and $(T+S)/T\cong S/(S\cap T)$ and using the fact that localization is exact and thus commutes with quotients.
Edit: (more detailed answer)
Let us introduce some notation:

*

*$\langle m\rangle:=Am$ the $A$-submodule of $M$ generated by $m$.

*$\langle \frac{m}{1}\rangle$ the $A_{\mathfrak{p}}$-submodule of $M_{\mathfrak{p}}$ generated by the image $\frac{m}{1}$ of $m$ in $M_{\mathfrak{p}}$.

By definition we have
$$ \mu(\mathfrak{p},M/\langle m\rangle)=\dim_{\kappa(\mathfrak{p})}\left(\frac{(M/\langle m\rangle)_{\mathfrak{p}}}{\mathfrak{p}(M/\langle m\rangle )_{\mathfrak{p}}}\right). $$
We want this to be equal to $\dim_{\kappa(\mathfrak{p})}(M_{\mathfrak{p}}/\mathfrak{p}M_{\mathfrak{p}})-1$.
We are assuming that $m$ has non-zero image in the $\kappa(\mathfrak{p})$-vector space $M_{\mathfrak{p}}/\mathfrak{p}M_{\mathfrak{p}}$, which as we already argued in the hints and comments implies that it spans a linear subpsace of dimension $1$.
This linear subspace is the quotient $\langle \frac{m}{1}\rangle /(\mathfrak{p}M_{\mathfrak{p}}\cap \langle \frac{m}{1}\rangle)$.
So it suffices to show that there is an isomorphism
$$ \frac{(M/\langle m\rangle)_{\mathfrak{p}}}{\mathfrak{p}(M/\langle m\rangle )_{\mathfrak{p}}}\cong \frac{M_{\mathfrak{p}}/\mathfrak{p}M_{\mathfrak{p}}}{\langle \frac{m}{1}\rangle /(\mathfrak{p}M_{\mathfrak{p}}\cap \langle \frac{m}{1}\rangle)}. $$
Using the isomorphism theorem $(T+S)/T\cong S/(S\cap T)$ we may rewrite the right-hand side as
$$\frac{M_{\mathfrak{p}}/\mathfrak{p}M_{\mathfrak{p}}}{(\mathfrak{p}M_{\mathfrak{p}}+\langle \frac{m}{1}\rangle)/\mathfrak{p}M_{\mathfrak{p}}}, $$
and after applying the isomorphism theorem $(A/B)/(C/B)\cong A/C$ we may rewrite this last expression as
$$ \frac{M_{\mathfrak{p}}}{\mathfrak{p}M_{\mathfrak{p}}+\langle \frac{m}{1}\rangle}. $$
There are at least two ways to arrive at this expression from the left-hand side.
The two ways that I have now in mind use as a first step that localization commutes with quotients, one of them going "outward" and the other one going "inward".
Going "inward" and using the isomorphism theorems as before we have
$$ \frac{(M/\langle m\rangle)_{\mathfrak{p}}}{\mathfrak{p}(M/\langle m\rangle )_{\mathfrak{p}}}\cong \frac{M_{\mathfrak{p}}/\langle \frac{m}{1}\rangle}{\mathfrak{p}M_{\mathfrak{p}}/(\mathfrak{p}M_{\mathfrak{p}}\cap\langle \frac{m}{1}\rangle)}\cong \frac{M_{\mathfrak{p}}/\langle \frac{m}{1}\rangle}{(\mathfrak{p}M_{\mathfrak{p}}+\langle\frac{m}{1}\rangle)/\langle\frac{m}{1}\rangle}\cong \frac{M_{\mathfrak{p}}}{\mathfrak{p}M_{\mathfrak{p}}+\langle \frac{m}{1}\rangle}. $$
And going "outward" and using the isomorphism theorems as before we have
$$ \frac{(M/\langle m\rangle)_{\mathfrak{p}}}{\mathfrak{p}(M/\langle m\rangle )_{\mathfrak{p}}}\cong \left(\frac{M/\langle m\rangle}{\mathfrak{p}M/(\mathfrak{p}M\cap \langle m\rangle)}\right)_{\mathfrak{p}}\cong \left(\frac{M/\langle m\rangle}{(\mathfrak{p}M+\langle m\rangle)/\langle m\rangle}\right)_{\mathfrak{p}}\cong \left(\frac{M}{\mathfrak{p}M+\langle m\rangle}\right)_{\mathfrak{p}}\cong \frac{M_{\mathfrak{p}}}{\mathfrak{p}M_{\mathfrak{p}}+\langle \frac{m}{1}\rangle}. $$
