Minimal free resolution I'm studying on the book "Cohen-Macaulay rings" of Bruns-Herzog 
(Here's a link and an image of the page in question for those unable to use Google Books.)
At page 17 it talks about minimal free resolution, but it doesn't give a proper definition (or I'm misunderstanding the one it gives), could you give me a definition?
And if $(R,\mathfrak{m},k)$ is a Noetherian local ring, $M$ a finite $R$-module and
$F.:\cdots\rightarrow F_n\rightarrow F_{n-1}\rightarrow\cdots\rightarrow F_1\rightarrow F_0\rightarrow 0$  
a finite free resolution of $M$. The it is minimal if and only if $\varphi_i(F_i)\subset\mathfrak{m}F_{i-1}$ for all $i\geq1$. Why?
 A: This answer is a slight reformulation of jspecter's; perhaps it will help.

A minimal free resolution is one in which each free module has the minimal number of generators.  (Hence the name.)
Here is how you make it:
Start with $M$, f.g. over $R$.  Its minimal number of generators is $\dim M/\mathfrak m M$.  So we can take a free module $F_0$ with this number of generators and a surjection $F_0 \to M$ (but we can't do this with any 
free module of smaller rank).
If this map is an isomorphism, we're done.
Otherwise, note that since $F_0/\mathfrak m F_0 \to M/\mathfrak m M$ is
a surjection between $R/\mathfrak m$-vector spaces of the same dimension,
it is an isomorphism, and so the kernel of the surjection $F_0 \to M$
is contained in $\mathfrak m F_0$.  Now apply the same process that we
used to construct $F_0 \to M$ to this kernel,
to obtain a map of free modules $F_1 \to F_0$ whose cokernel is $M$,
now with both $F_0$ and $F_1$ being free on the minimal possible number
of generators.
The same argument as above will show that the kernel of $F_1 \to F_0$
is contained in $\mathfrak m F_1$.  
We now continue inductively, and so produce a minimal free resolution of $M$.
As a biproduct of the construction, we find that each map $F_i \to F_{i-1}$ has image lying in $\mathfrak m F_{i-1}$.  Equivalently, each map $F_i\to F_{i-1}$ reduces to the zero map modulo $\mathfrak m$.
Now, as jspecter points out,
there is a converse to the preceding remark: any free resolution with the property that $F_{i+1} \to F_{i}$ has image lying in $\mathfrak m F_{i}$ for every $i \geq 0$ (or equivalently, with the property that the maps $F_{i+1} \to F_{i}$ reduce to $0$ mod $\mathfrak m$ for $i \geq 0$) is a minimal free resolution, in the sense that the $i$th stage (for $i \geq 0$), the number of generators of $F_i$
is equal to the minimal number of generators of the kernel of the map
$F_{i-1} \to F_{i-2}$.  (Here we agree that $F_{-1} = M$ and that $F_{-2} = 0$.)
(In jspecter's answer, things are phrased in terms of 
of cokernels rather than kernels.  But we are saying the same thing: since the $F_i$ form a complex,
the cokernel of $F_i \to F_{i-1}$ is the same as the kernel of $F_{i-1} \to F_{i-2}$.  I have given a formulation in terms of kernels just because I find it slightly more intuitive.)  
As jspecter also says,
in the book you are reading, the definition of minimal resolution is almost surely the one I give at the beginning of this answer.  The book is then using the preceding remark and its converse to give the alternative characterization of minimal resolutions that you asked about.
A: I can't see that book online, but let me paraphrase the definition from Eisenbud's book on commutative algebra. See Chapter 19 page 473-477 for details. 
Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak{m},$ then
Definition: A free resolution of a $R$-module $M$ is a complex 
$$\mathcal{F}: ...\rightarrow F_i \rightarrow F_{i-1} \rightarrow ... \rightarrow F_1 \rightarrow  F_0$$ 
with trivial homology such that $\mathbf{\text{coker}}(F_1 \rightarrow F_0) \cong M$ and each $F_i$ is a free $R$-module. 
He then defines a minimal resolution as follows:
Definition: A complex 
$$\mathcal{F}: ...\rightarrow F_i \rightarrow F_{i-1} \rightarrow ...$$ 
over $(R,\mathfrak{m})$ is minimal if the induced maps in the complex $\mathcal{F}\otimes R/\mathfrak{m}$ are each identically $0$. (Note that this is equivalent to the condition that $Im(F_i \rightarrow F_{i-1}) \subset \mathfrak{m}F_{i-1}$)  
after this he proves the fact that a  free resolution $\mathcal{F}$ is minimal if and only if a basis for $F_{i-1}$ maps into a minimal set of generators for $\mathbf{\text{coker}}(F_i \rightarrow F_{i-1}).$
The proof is via a straightforward appeal to Nakayama's Lemma.
From the way your question is worded, I would guess that your text has taken the latter of these equivalent conditions as the definition of a minimal free resolution and proved it to be equivalent to the former. But I can't be sure without seeing the text.  
