Questions on the paper "On the finitistic global dimension conjecture for Artin algebras" I am reading the paper On the finitistic global dimension conjecture for Artin algebras.
On page two, there are two lemmas:

Lemma 0.1 (Fitting’s Lemma)
a)
  Let $M$ be a f.g. module over a Noetherian ring $R$ and let $f \colon M \to M$ be an endomorphism of $M$.
  Then for any submodules $X$ of $M$ there is an integer $n$ so that $f$ sends $f^m(X)$ isomorphically onto $f^{m+1}(X)$ for all $m \geq n$.
  Let $\eta_f(X)$ denote the smallest such value of $n \geq 0$.
b)
  If $Y$ is a submodule of $X$ then $\eta_f(Y) \leq \eta_f(X)$.
c)
  If $R$ is an Artin algebra and $X = M$ there is a direct sum decomposition $X = Y \oplus Z$ so that $Z = \ker f^m$ and $Y = \operatorname{im} f^m$ for all $m \geq \eta_f(X)$.
Let $K_0$ be the abelian group generated by all symbols $[M]$, where $M$ is a f.g. $\Lambda$-module, modulo the relations:
  
  
*
  
*$[C] = [A] + [B]$ if $C \approx A \oplus B$.
  
*$[P] = 0$ if $P$ is projective.
  
  
  Then $K_0$ is the free abelian group generated by the isomorphism classes of indecomposable f.g. nonprojective $\Lambda$-modules.
  For any f.g. $\Lambda$-module $M$ let $L[M] = [\Omega M]$ where $\Omega M$ is the first syzygy of $M$.
  Since $\Omega$ commutes with direct sums and takes projective modules to zero this given a homomorphism $L \colon K_0 \to K_0$.
  For every f.g. $\Lambda$-module $M$ let $\langle \operatorname{add} M \rangle$ denote the subgroup of $K_0$ generated by all the indecomposable summands of $M$.
  Let
  $$
  \phi(M) := \eta_L \langle \operatorname{add} M \rangle.
$$
Lemma 0.2
a)
  If $M$ has finite projective dimension than $\phi(M) = \operatorname{pd} M$.
b)
  If $M$ is indecomposable with $\operatorname{pd} M = \infty$ then $\phi(M) = 0$.
c)
  $\phi(A) \leq \phi(A \oplus B)$.
d)
  $\phi(kM) = \phi(M)$ if $k \geq 1$.
(Original picture of the above text here.)

I am confused at some places:


*

*$K_0$ is the free ablian group, so it can be seen as a $\mathbb{Z}$-module and $\mathbb{Z}$ is a Notherian ring. But here how to make sure $K_0$ is a finitely generated $\mathbb{Z}$-module to use Lemma 0.1?

*How to prove a), b) in Lemma 0.2? (I don't know why there are connections between $\phi(M)$ and the projective dimension of $M$.)

*Does $kM$ in d) mean the direct sum of $k$ many copies of $M$?

 A: *

*It is not really necessary in 0.1 a) to assume that $M$ is finitely generated, one only needs that $X$ is a finitely generated (hence Noetherian) submodule of $M$. That is clearly the case for $\langle \mathrm{add}\,M\rangle$.

*To show 0.2 a): 
For an (indecomposable) direct summand $N \leq_{\oplus} M$, we have $\mathrm{pd}\,N \leq \mathrm{pd}\,M$.
Consider a finite projective resolution 
$$0\longrightarrow P_n \stackrel{p_n}\longrightarrow P_{n-1}\stackrel{p_{n-1}}\longrightarrow \dots \longrightarrow P_1 \stackrel{p_1}\longrightarrow P_0 \stackrel{p_0}\longrightarrow N \longrightarrow 0$$
where $n=\mathrm{pd}\,N.$ Then $\Omega N=\ker p_0, \;\Omega^2 N=\Omega \Omega N=\ker p_1, \dots, \Omega^n M=\ker p_{n-1}=P_n$. Thus, $L^n([N])=[P_n]=0$. Note that all the syzygies $\Omega N, \;\Omega^2 N, \dots \Omega^{n-1} N$ were non-projective, so $L^i([N])\neq 0$ for $i < n$.
On the other hand, if $M=\bigoplus_{j=1}^mN_j,$ then $\mathrm{pd}\,M=\mathrm{pd}\,N_j$ for some $j$: This is because one can sum up the projective resolutions as above to produce a projective resolution of $M$ (whose length is $\max \{\mathrm{pd}\,N_j\;|\; j=1,2, \dots, m\}$). Altogether, since $[N_1], \dots [N_m]$ are the generators of $\langle\mathrm{add} M\rangle$, we see that $$L^k(\langle\mathrm{add} M\rangle)=0,  \; L^{k-1}(\langle\mathrm{add} M\rangle)\neq 0,$$
where $k=\mathrm{pd}\, M$. This shows that $L:L^{k-1}(\langle\mathrm{add} M\rangle) \rightarrow L^{k}(\langle\mathrm{add} M\rangle)$ is not injective, but after that, $L:L^{t}(\langle\mathrm{add} M\rangle) \rightarrow L^{t+1}(\langle\mathrm{add} M\rangle)$ for $t \geq n$ of course injective is, since it is the zero map between zero modules. 
As for $b)$: Note that by an argument as above, $L^n$ does not ever vanish on $\langle[M] \rangle$. That means that for any $n,$ the map $L: L^n(\langle [M] \rangle) \rightarrow L^{n+1}(\langle [M] \rangle)$ is a nonzero map between cyclic modules inside free Abelian group, i.e. between free Abelian groups of rank one. That automatically means that the map is injective. This means that $\phi(M)=0$.


*If I had to guess, I would say that it is a slight misuse of notation, meaning $M^k$, which corresponds to the class $k[M]$ inside $K_0$.

