1.2 Projective Resolutions
Let \(R\) be a ring. Let \(M\) be an \(R\)-module. A projective resolution of \(M\) over \(R\) Is an exact sequence
. That is to say, a projective resolution is a quasi-isomorphism of complexes between the inclusion of \(M\) into chain complexes over \(R\) (i.e. the complex which has \(M\) in degree zero, trivial everywhere else, with trivial maps) and a bounded—below complex whose components are projective \(R\)-modules.
A free resolution of \(M\) over \(R\) is a projective resolution whose components \(P^i\) are free.
The length of a projective resolution \(P_\cdot \) is the highest \(i\) such that \(P^i\) is nonzero. If there exists no such \(i\), the length is infinity
Let \(M\) be an \(R\)-module. Then \(\text{proj dim}M\) is the minimum length of a projective resolution of \(M\). It lives in the set \(\mathbb {N} \cap \infty \).
If there exists a projective resolution of \(M\) with finite length, then \(\text{proj dim}M {\lt} \infty \).
Expand definitions, use the definition of minimum.
Any resolution of \(M\) over \(R\) has length at least that of the minimal resolution.
Expand definitions, use definition of minimum
There exists a free resolution \(F_\cdot \) of \(M\) such that length of \(F_\cdot \) is equal to the projective dimension of \(M\).
One has the generators-relations explicit construction, which should eventually get its own definition. Then one has to show that this is indeed minimal among projective resolutions, which I don’t remember how to do the top of my head.
The global dimension of a ring \(R\) is the supremum over all \(R\)-modules of \(\text{proj dim}M\).
Localization by a multiplicative set is an exact functor.
Let \(M\) be an \(R\)-module. Let \(S\) be a multiplicative set. Let \(\ldots \to A_2 \to A_1 \to A_0 \to M \to 0\) be a free resoltion of \(M\). Then \(\ldots S^{-1}(A_2) \to S^{-1}(A_1) \to S^{-1}(A_0) \to S^{-1}M \to 0\) is a free resolution of \(S^{-1}M\).