Questions: Tensor Products of Modules

Question 1

Which of the following correctly describes ℤ/2ℤ ⊗_ℤ ℤ/3ℤ?

• A. ℤ/6ℤ, by the Chinese Remainder Theorem
• B. ℤ/2ℤ ⊕ ℤ/3ℤ
• C. The zero module, because 2 and 3 are coprime
• D. ℤ/2ℤ, because ℤ/3ℤ has no 2-torsion

Answer: (C)

For any a̅ ⊗ b̅ ∈ ℤ/2ℤ ⊗_ℤ ℤ/3ℤ, we have a̅ ⊗ b̅ = a̅ ⊗ 3b̅ = 3(a̅ ⊗ b̅) (since 3b̅ = 0̅ in ℤ/3ℤ) and also a̅ ⊗ b̅ = 2(a̅ ⊗ b̅)·something. More directly: 1̅ ⊗ b̅ = 1̅ ⊗ (3·1̅)b̅... The cleanest argument: a̅ ⊗ b̅ = a̅ ⊗ 1̅ · b = a̅ · 1 ⊗ b̅, and since 2(a̅ ⊗ b̅) = 0 and 3(a̅ ⊗ b̅) = 0, we get 1·(a̅ ⊗ b̅) = (3-2)(a̅ ⊗ b̅) = 0. So every pure tensor is zero, and the entire module is zero. This is the tensor product's way of saying 'ℤ/2ℤ and ℤ/3ℤ have no common structure to combine.'

Question 2

For an R-module M and an ideal I ⊆ R, which of the following is the correct relationship between M ⊗_R R/I and M/IM?

• A. They are not generally related
• B. M ⊗_R R/I ≅ M/IM — tensoring with R/I is the same as 'reducing modulo I'
• C. M ⊗_R R/I ≅ IM
• D. M ⊗_R R/I ≅ Hom_R(R/I, M)

Answer: (B)

This is one of the most useful computational identities for tensor products. Applying − ⊗_R R/I to M is equivalent to quotienting by IM. For example, M ⊗_ℤ ℤ/nℤ ≅ M/nM for any abelian group M. This identity says that tensoring with R/I performs 'reduction modulo I' — it is the algebraic version of restricting to a closed subvariety in geometry. The proof uses the right exactness of tensor product applied to 0 → I → R → R/I → 0.

Question 3

The tensor product functor − ⊗_R N is right exact: it preserves surjections and cokernels.

• T. True
• F. False

Answer: True

If A → B → C → 0 is exact, then A ⊗_R N → B ⊗_R N → C ⊗_R N → 0 is exact. This is a fundamental property of tensor products. The critical limitation is that tensor products are NOT left exact in general: the map A ⊗ N → B ⊗ N need not be injective even when A → B is injective. The failure of left exactness is measured by the Tor functor, and modules for which tensoring is exact are called flat.

Question 4

Tensor products commute with direct sums: M ⊗_R (⊕ᵢ Nᵢ) ≅ ⊕ᵢ (M ⊗_R Nᵢ).

• T. True
• F. False

Answer: True

This follows from the universal property: bilinear maps from M × (⊕Nᵢ) correspond to families of bilinear maps from each M × Nᵢ. Concretely, R^n ⊗_R M ≅ M^n (tensoring with a free module of rank n gives n copies). This is one of the key computational tools: to compute M ⊗ N, present N as a quotient of a free module (via generators and relations) and use right exactness and the direct sum formula.

Question 5

Why is the tensor product ℤ/2ℤ ⊗_ℤ ℤ/3ℤ zero, and what general principle does this illustrate?

Model answer: For any element a̅ ⊗ b̅, we compute: a̅ ⊗ b̅ = 1·(a̅ ⊗ b̅) = (3 - 2)(a̅ ⊗ b̅) = 3(a̅ ⊗ b̅) - 2(a̅ ⊗ b̅). But 2(a̅ ⊗ b̅) = (2a̅) ⊗ b̅ = 0̅ ⊗ b̅ = 0, and 3(a̅ ⊗ b̅) = a̅ ⊗ (3b̅) = a̅ ⊗ 0̅ = 0. So a̅ ⊗ b̅ = 0 for every pure tensor, hence M ⊗ N = 0. The general principle: ℤ/mℤ ⊗_ℤ ℤ/nℤ ≅ ℤ/gcd(m,n)ℤ. When gcd(m,n) = 1, the tensor product is zero. Modules annihilated by coprime integers have 'no common information' to combine.

This illustrates how tensor products detect common structure. The tensor product of two modules is zero when their annihilators generate the whole ring. Geometrically, this corresponds to two subvarieties with empty intersection: their structure sheaves have zero tensor product. The formula ℤ/mℤ ⊗ ℤ/nℤ ≅ ℤ/gcd(m,n)ℤ is a clean example of the general identity M ⊗_R R/I ≅ M/IM.