Auslander-Buchsbaum formula - proof I have a final exam in commutative algebra and Auslander-Buchsbaum formula is one of the theorems that we have to self-study for the exam. Unfortunately, our course did not cover minimal resolutions and presentations at all. Taking this information into account, I want to write down a self-contained proof of A-B formula that is rigorous on its own without using unproved theorems.
Here you can find the proof I'm trying to rewrite (page 91 of the book, 107 of the PDF).

Let $(R,m,k)$ be a commutative local ring. if $M$ is a non-zero finitely generated $R$-module of finite projective dimension, then
  $$\rm{pd}_R(M) + \rm{depth_R(M)} = \rm{depth(R)}$$

The hardest part is that I need to expand on how the minimal presentation is obtained without assuming anything about minimal resolutions. So here is what I understand and what I don't understand:
Given $M$, since it is finitely generated and $R$ is a local ring, using Nakayama's lemma we can show that any minimal generating set for $M$ induces a basis on $M/mM$ (as a k-vector space) and any basis of $M/mM$ lifts to a minimal generating set for $M$. 
If $\dim_k(M/mM) = r$, we get a homomorphism $\psi: R^{r} \to M$ mapping $e_i$ to $m_i$ where $m_i$'s are generators of $M$. The kernel of this map is a sub-module of $R^{r}$ and since $R$ is commutative, it must be of the form $R^{s}$ (Is this right)? Therefore we get the s.e.s. 
$$ 0 \to R^s \to R^r \to M \to 0$$
The map $\varphi: R^s \to R^r$ can be represented by a $r\times s$ matrix with entries in $R$, but it claims that its entries are in $m$. Why?
Why is the map $\rm{Ext_R^i(k,\varphi)}$ represented by the matrix of $\varphi$ reduced modulo $m$?
 A: Ok, I will take the chance to enlighten you a little bit.
First of all, if you are really willing to fully understand why $\operatorname{Ext}^i(k,R^s) \to \operatorname{Ext}^i(k,R^r)$ is the zero map, you should not care that much about the explanations given in textbooks. These short explanations actually phrase the following: Checking that this map is zero is a messy thing, which I have done at some time - or at least some reliable guy has claimed to have done - but I will not spend 3 pages on my book to explain it. The reader might check it himself.
Speaking for myself, I have checked it at some time. Here is what I have come up with it:
The situation is the following: We are give a map $f:R^s \to R^r$ with $f(R^s) \subset \mathfrak m R^r$. We want to show that the induced map of ext groups $\operatorname{Ext}^i(k,R^s) \to \operatorname{Ext}^i(k,R^r)$ is the zero map.


*

*First of all - since $\operatorname{Hom}$ commutes with finite direct sums in both entries, a map $R^s \to R^r$ is the same as $r \cdot s$ maps $R \to R$ and each such map is just multiplication by an element of $R$. In linear algebra, one gathers these elements in a matrix, but let us forget this for the moment. The assumption $f(R^s) \subset \mathfrak m R^r$ says that each of these maps is multiplication by an element of $\mathfrak m$.

*Furthermore - since $\operatorname{Ext}^i$ also commutes with finite direct sums in both entries - a map $\operatorname{Ext}^i(k,R^s) \to \operatorname{Ext}^i(k,R^r)$ is the same as $r \cdot s$ maps $\operatorname{Ext}^i(k,R) \to \operatorname{Ext}^i(k,R)$ and if we want to show that the map is zero, we have to show that each of this maps is zero.
Summarizing our thoughts, we have reduced to the case $r=s=1$, i.e. we are given a map $R \xrightarrow{\cdot r}R$ with $r \in \mathfrak m$ and we want to show that the induce map $\operatorname{Ext}^i(k,R) \to \operatorname{Ext}^i(k,R)$ is zero.
We split the proofs into parts:


*

*We have the following lemma, which covers the $i=0$-case for the general case of multiplication on a module: 



Let $M$ be a $R$-module and $M \to M$ multiplication by $r \in \mathfrak m$. The induced map $\operatorname{Hom}(k,M) \to \operatorname{Hom}(k,M)$ is zero map.

For the proof, start with a map $k \to M$ and consider the composition $R \twoheadrightarrow k \to M \to M$. A map $R \to M$ is just an element of $M$ and the fact that it factors over $k$ tells us, that this element is annihilated by $\mathfrak m$, i.e. by $r$. Hence the composition is the zero map. By surjectivity of the first map, the composition $k \to M \to M$ is the zero map. This proves the lemma.


*Now we have to look at the construction of the induced map $\operatorname{Ext}^i(k,R) \to \operatorname{Ext}^i(k,R)$. It is constructed as follows: Take an injective resolution $R \to I^0 \to I^1 \to \cdots$. By some fundamental lemma in homological algebra (The Lemma is much more general than our case and it is Lemma 5.2 in Chapter XX of Serge Lang's Algebra. It is fundamental to the construction of derived functors), we can extend a morphism $R \to R$ to a morphism of complexes $I^{(\bullet)} \to I^{(\bullet)}$ and that resulting morphism is unique up to homotopy. In our case, we have a natural choice of such a morphism: Multiplication by $r$ on the complex of course extends multiplication by $r$ on $R$.


Then we apply the Hom-Functor to the complexes to get a morphism of complexes $$\operatorname{Hom}(k,I^{(\bullet)}) \to \operatorname{Hom}(k,I^{(\bullet)})$$
Then we apply the cohomology functor (with the cohomology groups of the complexes being the ext groups by definition) and for each $i$, we get morphisms
$$\operatorname{Ext}^i(k,R) \to \operatorname{Ext}^i(k,R).$$
In particular it suffices to show that the morphism of complexes $$\operatorname{Hom}(k,I^{(\bullet)}) \to \operatorname{Hom}(k,I^{(\bullet)})$$ is the zero morphism, since the zero morphism of complexes gives rise to the zero morphism of cohomology groups. I.e. we have to show that for each $i$
$$\operatorname{Hom}(k,I^i) \to \operatorname{Hom}(k,I^i)$$
is the zero morphism. But this is precisely what we have done in the lemma: Multiplication by $r \in \mathfrak m$ induces the zero morphism after applying $\operatorname{Hom}(k,-)$.
