Questions: Exact Sequences of Modules

Question 1

Given a short exact sequence 0 → A →^f B →^g C → 0 of R-modules, which of the following must be true?

• A. B ≅ A ⊕ C as R-modules
• B. f is injective and g is surjective
• C. A is isomorphic to C
• D. B is a free module

Answer: (B)

Exactness at A (with 0 → A) means ker(f) = im(0 → A) = 0, so f is injective. Exactness at C (with C → 0) means im(g) = ker(C → 0) = C, so g is surjective. Exactness at B means im(f) = ker(g). The sequence does NOT imply B ≅ A ⊕ C — that holds only when the sequence splits. For example, 0 → ℤ →^{×2} ℤ → ℤ/2ℤ → 0 does not split: ℤ is not isomorphic to ℤ ⊕ ℤ/2ℤ.

Question 2

The sequence 0 → 2ℤ → ℤ → ℤ/2ℤ → 0 is a short exact sequence of ℤ-modules. Does it split?

• A. Yes — there is a ℤ-module homomorphism ℤ/2ℤ → ℤ that is a section of the quotient map
• B. No — any ℤ-module homomorphism ℤ/2ℤ → ℤ must be zero, since ℤ is torsion-free, so no section exists
• C. Yes — because ℤ is a PID, all short exact sequences of ℤ-modules split
• D. No — because 2ℤ ≅ ℤ is free, the sequence must split by the splitting lemma

Answer: (B)

A splitting would require a homomorphism s: ℤ/2ℤ → ℤ with the projection composed with s being the identity on ℤ/2ℤ. But any homomorphism from ℤ/2ℤ to ℤ maps the generator 1̄ to some n ∈ ℤ satisfying 2n = 0, hence n = 0. So the only homomorphism is the zero map, which is not a section. The sequence does not split: ℤ ≇ 2ℤ ⊕ ℤ/2ℤ = ℤ ⊕ ℤ/2ℤ.

Question 3

If 0 → A → B → C → 0 is a short exact sequence and C is a free module, then the sequence splits.

• T. True
• F. False

Answer: True

If C is free with basis {cᵢ}, choose any preimage bᵢ ∈ B of each cᵢ under the surjection g: B → C. The map s: C → B defined by s(cᵢ) = bᵢ (extended linearly) is a section: g ∘ s = id_C. So the sequence splits, and B ≅ A ⊕ C. This is why free modules have the 'projective' property — maps to them can always be lifted. The previous example showed that when C = ℤ/2ℤ (not free), splitting can fail.

Question 4

Every short exact sequence of vector spaces over a field splits.

• T. True
• F. False

Answer: True

Over a field, every module (vector space) is free — it has a basis. By the previous result, if C is free then 0 → A → B → C → 0 splits. Since every vector space is free, every short exact sequence of vector spaces splits, and B ≅ A ⊕ C. This is equivalent to the rank-nullity theorem: dim(B) = dim(A) + dim(C). The failure of splitting over general rings is what makes module theory richer and harder than linear algebra.

Question 5

Explain the relationship between exact sequences and the isomorphism theorems from abstract algebra.

Model answer: The first isomorphism theorem says that for a homomorphism f: M → N, we have M/ker(f) ≅ im(f). This is encoded by the exact sequence 0 → ker(f) → M → im(f) → 0. More generally, a short exact sequence 0 → A →^f B →^g C → 0 says f is injective (A ≅ im(f) = ker(g)) and C ≅ B/im(f) ≅ B/A. So exact sequences package the isomorphism theorems into a compact, composable notation that generalizes to longer sequences and functorial constructions.

The power of exact sequences over raw isomorphism theorems is composability. You can splice exact sequences together, apply functors to them (localization, tensor product, Hom), and track how exactness is preserved or broken. When a functor fails to preserve exactness, the 'defect' is measured by derived functors — Ext and Tor — which form the basis of homological algebra.