A module M over a ring R has projective dimension pd(M) = 0. What does this tell you about M?
AM has no projective resolution — it is too complicated to be built from projective objects
BM itself is projective — the trivial resolution 0 → M → M → 0 works with no correction steps
CM is the zero module — only the trivial module needs no corrections
DM is a free module of rank 0, meaning M is the zero module
pd(M) = 0 means the shortest projective resolution of M has length 0, i.e., the resolution is 0 → P₀ → M → 0 where P₀ = M itself is projective. No correction steps are needed because M is already projective. This is the definition: projective dimension measures how far M is from being projective, with 0 indicating it is projective outright. A projective module has the lifting property that makes computations with it tractable and is 'as close as possible' to free.
Question 2 Multiple Choice
A Noetherian local ring R is shown to have finite global dimension. What does the Serre–Auslander–Buchsbaum theorem imply about the corresponding geometric object?
AThe spectrum of R is an irreducible variety — it has no components that can be removed
BThe spectrum of R is a smooth variety — it has no singular points
CThe spectrum of R is a compact variety — it has no points at infinity
DThe spectrum of R is a projective variety — it embeds in projective space
The Serre–Auslander–Buchsbaum theorem states that a Noetherian local ring R is regular if and only if it has finite global dimension. Regularity of a local ring corresponds geometrically to smoothness: a point on an algebraic variety is smooth (non-singular) if and only if its local ring is regular. This is a profound connection between homological algebra and algebraic geometry — 'how long do projective resolutions need to be?' is equivalent to 'is this point singular?' Homological dimension is an algebraic measurement of geometric smoothness.
Question 3 True / False
If pd(M) = d, then Extⁿ(M, N) = 0 for all n > d and all modules N, because the projective resolution of M terminates at step d.
TTrue
FFalse
Answer: True
Extⁿ(M, N) is computed by applying Hom(−, N) to a projective resolution of M and taking cohomology. If M has a projective resolution of length d, the resolution terminates at step d — there are no terms beyond Pₐ. Applying Hom and taking cohomology of a complex that is zero beyond step d must give zero cohomology beyond degree d. So Extⁿ(M, N) = 0 for all n > pd(M). This is why homological dimension governs when long exact sequences of Ext groups truncate.
Question 4 True / False
A module with finite projective dimension necessarily has finite injective dimension as well, since both invariants measure the same underlying algebraic complexity.
TTrue
FFalse
Answer: False
Projective and injective dimensions are independent invariants. A module can have finite projective dimension but infinite injective dimension, or vice versa. Over ℤ, the module ℤ/pℤ has projective dimension 1 (it admits the two-step free resolution 0 → ℤ →^p ℤ → ℤ/pℤ → 0) but infinite injective dimension. While both invariants appear when studying derived functors, they capture different structural information — projective dimension measures how M sits over projectives, injective dimension measures how M sits under injectives.
Question 5 Short Answer
Explain what projective dimension measures and why it is described as 'the distance from M to the class of projective objects.'
Think about your answer, then reveal below.
Model answer: Projective dimension pd(M) is the length of the shortest projective resolution of M — the smallest n such that there is an exact sequence 0 → Pₙ → ··· → P₀ → M → 0 with all Pᵢ projective. If M is projective, pd(M) = 0 — zero correction steps needed. If M is not projective but admits a two-step resolution, pd(M) = 1 — one projective correction layer suffices. Each successive projective Pₙ corrects the obstruction left by the previous layer. Higher pd(M) means more layers of non-projective structure that must be resolved before reaching projective building blocks — hence 'distance from projective.'
The geometric intuition comes from algebraic geometry: projective modules correspond to vector bundles over smooth varieties, and projective dimension measures how far M is from being a vector bundle. Over a regular local ring (smooth point), every module has finite projective dimension — the Serre-Auslander-Buchsbaum theorem. Over a singular ring, some modules require arbitrarily long resolutions and global dimension becomes infinite. Projective dimension thus literally measures algebraic distance in the category of modules.