Questions: Tor Functors as Derived Tensor Product

Question 1

The tensor product functor − ⊗ B is right-exact but not generally left-exact. What does Tor₁(A, B) measure?

• A. The failure of A ⊗ B to commute with direct sums
• B. The kernel of the map A' ⊗ B → A ⊗ B when 0 → A' → A → A'' → 0 is exact
• C. The failure of Hom(A, −) to be right-exact
• D. The number of non-zero summands in a projective resolution of A

Answer: (B)

When 0 → A' → A → A'' → 0 is exact, tensoring with B gives A' ⊗ B → A ⊗ B → A'' ⊗ B → 0 (right-exactness preserved), but the left map A' ⊗ B → A ⊗ B may fail to be injective. Tor₁(A, B) is exactly the kernel of this map — it measures how much the injection fails. If Tor₁(A, B) = 0 for all B, then − ⊗ A is left-exact (and hence exact), meaning A is flat. Note: Tor measures failure of left-exactness of the tensor product, not Hom — Ext handles the failure of Hom to be right-exact.

Question 2

What is Tor₁(ℤ/6ℤ, ℤ/4ℤ)?

• A. ℤ/24ℤ
• B. ℤ/2ℤ
• C. 0
• D. ℤ/6ℤ ⊕ ℤ/4ℤ

Answer: (B)

For cyclic groups, Tor₁(ℤ/mℤ, ℤ/nℤ) ≅ ℤ/gcd(m,n)ℤ. Here gcd(6,4) = 2, so Tor₁(ℤ/6ℤ, ℤ/4ℤ) ≅ ℤ/2ℤ. This captures 'torsion interaction': the two groups have a common factor of 2, so their Tor₁ is non-trivial. Note that ℤ/6ℤ ⊗ ℤ/4ℤ ≅ ℤ/2ℤ as well — but Tor₁ encodes information about the *failure of exactness* in the resolution, which happens to coincide here. For coprime orders (e.g., ℤ/2ℤ and ℤ/3ℤ), Tor₁ = 0.

Question 3

A module A is flat if and only if Tor_n(A, B) = 0 for all n ≥ 1 and all modules B.

• T. True
• F. False

Answer: True

This is the defining characterization of flatness in terms of Tor. Flatness means tensoring with A preserves exact sequences (− ⊗ A is an exact functor). Tor_n(A, B) measures the failure of exactness at the nth level of a projective resolution tensored with B. If all these failure groups vanish, the tensor product is exact at every level, which is precisely flatness. Free modules and projective modules are flat; flatness is strictly weaker than projectivity (there exist flat modules that are not projective).

Question 4

Tor₀(A, B) is zero whenever A and B have no common torsion elements.

• T. True
• F. False

Answer: False

Tor₀(A, B) = A ⊗ B, always — it recovers the ordinary tensor product and is never 'zero because of no common torsion.' The subscript-zero Tor is just the tensor product itself, regardless of torsion. It is Tor₁ and higher groups that detect torsion interaction. For example, ℤ ⊗ ℤ/nℤ ≅ ℤ/nℤ ≠ 0 even though ℤ is torsion-free.

Question 5

Why is Tor computed using projective resolutions rather than injective resolutions, and what does it mean for a module to be flat in terms of Tor?

Model answer: Tor is the left derived functor of the tensor product. Derived functors of a right-exact functor F are computed by replacing the argument with a projective resolution, applying F, and taking homology — this 'fills in' the left-exactness that F lacks. Injective resolutions are used for right derived functors of left-exact functors (like Hom, which gives Ext). Since − ⊗ B is right-exact (not left-exact), its derived functors go to the left, requiring projective resolutions. A module A is flat precisely when Tor_n(A, B) = 0 for all n ≥ 1 and all B, meaning tensoring with A is already exact and no derived correction is needed.

The left/right distinction is fundamental: left derived functors correct for failure of left-exactness (Tor from tensor), right derived functors correct for failure of right-exactness (Ext from Hom). Using an injective resolution for Tor would produce incorrect (and in fact trivially zero) answers in many cases because projective modules are exactly the ones for which tensoring is exact — they are the 'acyclic objects' for the tensor product functor.