I'm reading Dave Benson's book "Representations and Cohomology," Volume I, and I'm trying to understand the following discussion on page $32$ in which he introduces the notion of a minimal projective resolution. Let $\Lambda$ be a ring and $M$ a left $\Lambda$-module. The book reads:
We write $\tilde{\Omega}^n(M)$ for $\ker(\partial_{n-1})$ in a projective resolution of $M$. Note that by Schanuel's lemma, if $\tilde{\Omega}^n(M)'$ is similarly defined using another projective resolution of $M$ then there are projective modules $P$ and $P'$ with $\tilde{\Omega}^n(M)\oplus P'\cong\tilde{\Omega}^n(M)'\oplus P$. If $M$ is finitely generated and the Krull-Schmidt theorem holds for finitely generated $\Lambda$-modules then there is a unique minimal resolution of $M$, and we write $\Omega^n(M)$ for $\ker(\partial_{n-1})$ in this particular resolution.
For reference, the Krull-Schmidt theorem holds for a module $M$ if it is a finite direct sum of indecomposable modules, unique up to isomorphism and ordering of summands. I understand the discussion above except for the phrase "If $M$ is finitely generated and the Krull-Schmidt theorem holds for finitely generated $\Lambda$-modules then there is a unique minimal resolution of $M$." Benson does not define what a minimal resolution is, and it is not clear to me why he introduces the notion in the middle of this discussion. I have many related questions:
- What is a minimal projective resolution, why does it exist, and why is it unique?
- Why do we need the Krull-Schmidt theorem to hold for there to exist a (unique) minimal resolution?
- Is the initial part of this discussion useful in showing there is a unique minimal resolution, or is Benson introducing the notion of a minimal resolution here to give a definition of $\Omega^n(M)$, that is, is the point of this discussion minimal resolutions or $\Omega^n(M)$?
Best Answer
I've learned the answer to my question, which I write here for future questioners:
First, we assume that $\Lambda$ is Artinian, so that any finitely generated $\Lambda$-module has a composition series, and also any submodule of a finitely generated module is itself finitely generated.
We first show that any finitely generated module has a unique projective cover, constructed as follows. Let $M$ be finitely generated, so there is some free, hence projective, module $P_1$ such that $f:P_1\twoheadrightarrow M$. Suppose also that $P_1$ is minimal with respect to its decomposition into indecomposable modules (that is, no summand of $P_1$ lies in the kernel of $P_1\twoheadrightarrow M$). Now, suppose that $g:P_2\twoheadrightarrow M$ is another such surjection, with $P_2$ projective and minimal. By projectivity, we must have maps $P_1\to P_2$ and $P_2\to P_1$ (commuting with the appropriate diagrams), which we now show are isomorphisms.
Suppose that $h:P_1\to P_2\to P_1$ is not an isomorphism. Then by Fitting's lemma (here is where we use the fact that $P_1$ has a composition series), there is some $n$ such that $P_1=\mathrm{im}(h^n)\oplus\ker(h^n)$, with $\ker(h^n)\ne 0$. But notice that $f\circ h^n=f$, so that $\ker(h^n)\in\ker(f)$, which contradicts the minimality hypothesis of $P_1$. It follows that $h$ is an isomorphism. Applying the same argument to $P_2\to P_1\to P_2$, we see that $P_1\cong P_2$. Define the projective cover of $M$, $P_M$, to be the unique (up to isomorphism) minimal projective module which surjects onto $M$. Let $\Omega(M)$ be the kernel of the map $P_M\twoheadrightarrow M$.
Now, I define a minimal projective resolution of $M$ to be a projective resolution $P^{\bullet}\xrightarrow{\partial} M$ such that no indecomposable summand of $P_n$ lies in the kernel of $\partial_n$ for all $n\ge 0$.
Begin constructing a projective resolution of $M$ by setting $P_0=P_M$, and $\Omega^1(M)=\Omega(M)$. Having defined $P_{n-1}$ and $\Omega^{n}(M)$, let $\Omega^{n+1}(M)=\Omega(\Omega^n(M))$, and $P_{n}=P_{\Omega^{n}(M)}$. By construction, the resolution is minimal, and because the projective cover of a finitely generated module is unique up to isomorphism, the resolution is unique up to isomorphism. Notice that at each step we use the fact that $\Omega^n(M)$ is finitely generated because it is a submodule of a finitely generated module.
Notice also we've defined $\Omega^n(M)$ differently than Benson has, but the two definitions are equivalent.