Questions: Ext Functors as Derived Hom

Question 1

What does it mean when Ext¹(A, B) = 0 for two modules A and B?

• A. There are no short exact sequences involving A and B
• B. The only short exact sequence 0 → B → E → A → 0 (up to equivalence) is the split one, so E ≅ A ⊕ B
• C. Hom(A, B) = 0, meaning there are no morphisms from A to B
• D. A and B are both injective modules

Answer: (B)

Ext¹(A, B) classifies equivalence classes of extensions of A by B — short exact sequences 0 → B → E → A → 0. The zero element of Ext¹ corresponds to the split extension, where E ≅ A ⊕ B. If Ext¹(A, B) = 0, this is the only class: every extension splits, and A and B cannot be non-trivially glued together. This does not say anything directly about Hom(A, B) or the injectivity of A or B.

Question 2

How is Ext^n(A, B) computed?

• A. By applying Hom(−, B) to a projective resolution of A and taking cohomology
• B. By applying Hom(A, −) to an injective resolution of B and taking cohomology
• C. By taking the n-th homology of the chain complex A → A → A → ⋯
• D. By iterating the Hom functor n times: Hom(A, Hom(A, ⋯ Hom(A, B) ⋯))

Answer: (B)

Ext^n(A, B) is defined as the n-th cohomology of the cochain complex obtained by applying Hom(A, −) to an injective resolution 0 → B → I⁰ → I¹ → I² → ⋯ of B (after dropping B). The Hom functor is left exact but not right exact, so cohomology at each step measures the failure of exactness. Notably, Ext can also be computed via a projective resolution of A (applying Hom(−, B)), and both methods give the same result — which is why Ext^n(A, B) is well-defined as an invariant of the pair (A, B).

Question 3

Ext⁰(A, B) is a new invariant that captures information not already contained in Hom(A, B).

• T. True
• F. False

Answer: False

Ext⁰(A, B) = Hom(A, B). It is the zeroth cohomology of the complex obtained by applying Hom(A, −) to the injective resolution, and left exactness of Hom means no information is lost at the zeroth step. The higher Ext groups Ext^n for n ≥ 1 are genuinely new invariants measuring the failure of exactness at subsequent steps — they capture information that Hom alone does not see.

Question 4

The Ext groups Ext^n(A, B) are well-defined invariants of the pair (A, B), independent of the choice of injective resolution of B used to compute them.

• T. True
• F. False

Answer: True

This independence is fundamental to Ext being a useful invariant rather than an artifact of construction. Any two injective resolutions of B are related by a chain map (unique up to chain homotopy), and chain homotopic maps induce the same maps on cohomology. Therefore, the cohomology groups you compute are the same regardless of which injective resolution you choose. This is the essential content of the derived functor framework: the construction is canonical.

Question 5

Describe the bijection that gives Ext¹(A, B) its geometric meaning as a classifier of extensions.

Model answer: There is a natural bijection between elements of Ext¹(A, B) and equivalence classes of short exact sequences 0 → B → E → A → 0, where two extensions are equivalent if there is an isomorphism E → E' making both triangles (involving B and A) commute. The zero element of Ext¹(A, B) corresponds to the split extension B ⊕ A. A nonzero element corresponds to a genuinely non-split extension — a module E that contains B as a submodule with quotient A but is not isomorphic to their direct sum. For example, Ext¹_ℤ(ℤ/2, ℤ) ≅ ℤ/2, reflecting the two extensions: the split one ℤ ⊕ ℤ/2, and the non-split one ℤ (via 0 → ℤ →×2 ℤ → ℤ/2 → 0).

The bijection makes Ext¹ a computable algebraic invariant for a fundamentally geometric question: in how many essentially different ways can A and B be 'glued together' into a single module? This connection between algebraic computation (derived functors, injective resolutions) and structural classification (extensions) is the hallmark of homological algebra and the reason Ext appears throughout algebraic topology, group cohomology, and algebraic geometry.