The Ext Functor

Executive summary

Balance: either variable may be resolved

Extⁿ(C, A) can be computed by taking a projective resolution of C and applying Hom(−, A), or by taking an injective resolution of A and applying Hom(C, −). The two agree, and for n = 1 both agree with the group of extensions under the Baer sum. That triple agreement — the balance of Ext — is what makes the functor computable in practice, because one variable is usually far easier to resolve than the other.

Learning objectives

Compute Ext from a projective resolution.
Compute Ext from an injective resolution.
State the balance theorem and why it matters.
Apply the standard vanishing criteria.
Interpret Ext¹ as classifying extensions.

Section 01The three definitions

Via extensionsYoneda / Baer

Ext¹(C, A) = equivalence classes of extensions under the Baer sum. Needs no resolutions; extends to Extⁿ via n-fold extensions.

Via projectivesResolve C

Take P_• ↠ C projective, apply Hom(−, A), take cohomology of the resulting cochain complex.

AlgorithmExt from a projective resolutionin: modules C, A → out: Extⁿ(C, A) for all n

Choose a projective resolution … → P₁ → P₀ → C → 0.
Delete C to obtain the deleted resolution P_•.
Apply Hom(−, A), giving a cochain complex 0 → Hom(P₀, A) → Hom(P₁, A) → … Contravariance reverses the arrows.
Extⁿ(C, A) is the n-th cohomology of that complex.
Ext⁰(C, A) = Hom(C, A), recovering the original functor in degree 0.

The balance theorem

Resolving C by projectives and resolving A by injectives give canonically isomorphic answers. The proof compares both against a double complex built from a projective resolution of C and an injective resolution of A simultaneously. In practice this means: resolve whichever variable is easier.

Section 02Vanishing criteria

When Ext vanishes
Condition	Consequence
C projective	Extⁿ(C, A) = 0 for all n ≥ 1 and all A
A injective	Extⁿ(C, A) = 0 for all n ≥ 1 and all C
proj dim C ≤ d	Extⁿ(C, −) = 0 for n > d
inj dim A ≤ d	Extⁿ(−, A) = 0 for n > d
Λ a PID	Extⁿ = 0 for n ≥ 2 — global dimension 1
Λ semisimple	Extⁿ = 0 for n ≥ 1 — every sequence splits

Ext<sup>1</sup> = 0 means every extension splits

This is the working interpretation. Semisimplicity of a ring is exactly the statement that Ext¹ vanishes identically, and Maschke's theorem — that k[G] is semisimple when the characteristic does not divide |G| — is the reason ordinary representation theory has no higher cohomology.

Section 03A worked computation

Compute Ext^*(ℤ/mℤ, A) over ℤ. The resolution is short:

0 → ℤ →^×m ℤ → ℤ/mℤ → 0

Applying Hom(−, A) gives A →^×m A, so

Ext⁰ = A[m], Ext¹ = A/mA, Extⁿ = 0 for n ≥ 2

where A[m] is the m-torsion. So Ext¹(ℤ/m, ℤ) = ℤ/mℤ, matching the extension count: the m extensions of ℤ/m by ℤ correspond to the m elements of that group.

The pattern to remember

Ext¹ against ℤ converts a torsion module into its dual torsion module. This single computation drives the universal coefficient theorem in topology and the classification of finitely generated abelian group extensions.

ReferenceFrequently asked questions

Why does the resolution not matter?

Because any two projective resolutions of the same module are chain homotopy equivalent, by the comparison theorem, and homotopic chain maps induce equal maps on homology. The resulting isomorphism is canonical, so the answer is well defined.

Is Ext<sup>1</sup> always the extension group?

Yes, in any abelian category with enough projectives or enough injectives. Without either, the Yoneda definition by extensions still works and is taken as the definition — which is one reason it is worth knowing independently.

What if the category has neither enough projectives nor injectives?

Yoneda Ext remains defined and is a group. Higher Ext is defined by n-fold extensions modulo an equivalence relation. It agrees with the derived-functor definition whenever both exist.

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