Questions on the paper "On the finitistic global dimension conjecture for Artin algebras"

Question

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:

1. $[C] = [A] + [B]$ if $C \approx A \oplus B$.
2. $[P] = 0$ 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:

1. $K_0$ 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?
2. 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 $M$.)
3. Does $kM$ in d) mean the direct sum of $k$ many copies of $M$?

Answer 1

1. 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 $\langle \mathrm{add}\,M\rangle$.

2. 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 $\phi(M)=0$.

3. 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 $K_0$.