Let $A$ be a ring and $M$ a finite $A$-module; for $p \in \text{Spec} \space A$, write $\mathcal{K}(\mathfrak{p})$ for the residue field of $A_\mathfrak{p}$, and let $\mu (\mathfrak{p}, M)$ denote the dimension over $\mathcal{K}(\mathfrak{p})$ of the vector space $M \otimes \mathcal{K}(\mathfrak{p}) = M_\mathfrak{p}/\mathfrak{p}M_\mathfrak{p}$ (in the usual sense of linear algebra). This is the cardinality of a minimal basis of the $A_\mathfrak{p}$-module $M_\mathfrak{p}$. Hence, if $\mathfrak{p} \supset \mathfrak{p'}$, then $\mu(\mathfrak{p}, M) \geq \mu(\mathfrak{p'}, M)$. (Matsumura p.35)
There is something that confuses me with the last sentence. Let us suppose that $M$ has the minimal basis $\{x_1, x_2, x_3\}$ over $A$. Let $\mathfrak{p}$ be a prime ideal of $A$. Then any element of $M_\mathfrak{p}$ is of the form $\frac{m}{s}$ with $m \in M$ and $s \in A-\mathfrak{p}$, right? So $\frac{m}{s} = \frac{a_1}{s}\frac{x_1}{1}+\frac{a_2}{s}\frac{x_2}{1}+\frac{a_3}{s}\frac{x_3}{1}$. So any localization of $M$ is generated by three elements. So don't we need to have $\mu(\mathfrak{p}, M) = \mu(\mathfrak{p'}, M)$?
For $\mathfrak{p} \in \text{Spec} \space A$ and $x \in M$, we will say that $x$ is basic at $\mathfrak{p}$ if $x$ has non-zero image in $M \otimes \mathcal{K}(\mathfrak{p})$. It is easy to see that this condition is equivalent to $\mu(\mathfrak{p}, M/Ax) = \mu(\mathfrak{p},M)-1$. (Matsumura p.36)
If $\mu(\mathfrak{p}, M/Ax) = \mu(\mathfrak{p},M)-1$ holds, then the minimal generating set for $(M/Ax)_\mathfrak{p} = M_\mathfrak{p}/(Ax)_\mathfrak{p}$ is less than the minimal generating set for $M_\mathfrak{p}$, so this of course tells us that $x$ is nonzero in $M_\mathfrak{p}$. But how does that imply that it is nonzero in $M_\mathfrak{p}/\mathfrak{p}M_\mathfrak{p}$?
For the converse, suppose that $x$ has a nonzero image in $M_\mathfrak{p}/\mathfrak{p}M_\mathfrak{p}$. Then $x \not\in \mathfrak{p}M_\mathfrak{p}$...and I'm kind of stuck here...
You should think of $\mu(\mathfrak p,M)$ as it is: $\dim_{k(\mathfrak p)}M_{\mathfrak p}/\mathfrak pM_{\mathfrak p}$.
If $V$ is a $k$-vector space, and $x\in V$, then $\dim V/\langle x\rangle=\dim V-1$ iff $x\ne 0$.
If $\mathfrak p'\subset \mathfrak p$, then there is a homomorphism $R_{\mathfrak p}\to R_{\mathfrak p'}$, $a/s\mapsto a/s$. If $M_{\mathfrak p}$ is generated (over $R_{\mathfrak p}$) by $x_1/1,\dots,x_n/1$, then $M_{\mathfrak p'}$ is generated by $x_1/1,\dots,x_n/1$ (over $R_{\mathfrak p'}$).