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?
Best Answer
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.