Questions: Modules over Rings

Question 1

Which of the following is true about modules over ℤ?

• A. ℤ-modules are the same thing as abelian groups, with scalar multiplication given by repeated addition
• B. ℤ-modules always have a basis, just like vector spaces over a field
• C. Every ℤ-module is finitely generated
• D. ℤ-modules cannot have torsion elements

Answer: (A)

A ℤ-module is precisely an abelian group: the scalar multiplication n·m (for n ∈ ℤ, m ∈ M) is just adding m to itself n times (or its inverse for negative n). This identification means that every theorem about ℤ-modules is a theorem about abelian groups, and vice versa. Not every abelian group has a basis (ℤ/2ℤ does not — it has torsion), not every abelian group is finitely generated (ℚ is not), and torsion is ubiquitous (every finite abelian group is torsion).

Question 2

A student says: 'Every finitely generated module over a PID is free, just like every finite-dimensional vector space has a basis.' What correction is needed?

• A. The student is completely correct — finitely generated modules over PIDs are always free
• B. Finitely generated modules over a PID decompose as a direct sum of a free part and a torsion part; only the torsion-free ones are free
• C. Modules over PIDs never have bases because PIDs are not fields
• D. The classification only works for principal ideal domains that are also Euclidean domains

Answer: (B)

The structure theorem for finitely generated modules over a PID says M ≅ R^r ⊕ R/(d₁) ⊕ ··· ⊕ R/(dₖ) where d₁ | d₂ | ··· | dₖ. The free part R^r has a basis, but the torsion summands R/(dᵢ) do not — they have elements m with dᵢ·m = 0. For example, ℤ/6ℤ is a finitely generated ℤ-module that is not free. The module is free if and only if the torsion part is trivial.

Question 3

Every vector space over a field is a free module.

• T. True
• F. False

Answer: True

A vector space is a module over a field, and having a basis is exactly what it means to be a free module. Every vector space has a basis (by Zorn's lemma in general, or by elementary arguments in finite dimensions), so every vector space is free. This is the key property that fails for modules over general rings: the existence of a basis is guaranteed over fields but not over arbitrary commutative rings.

Question 4

If M is a finitely generated module over a Noetherian ring R, then every submodule of M is also finitely generated.

• T. True
• F. False

Answer: True

This is a fundamental property of Noetherian rings, extending the definition from ideals to modules. An ideal of R is a submodule of R viewed as a module over itself, so 'every ideal is finitely generated' is the special case M = R. The general statement follows by induction on the number of generators of M: if M = Rm₁ + ··· + Rmₙ, consider the submodule N ∩ Rm₁ and the image of N in M/Rm₁, both of which are finitely generated by induction.

Question 5

What is a torsion element in a module, and why does torsion represent a fundamentally new phenomenon compared to vector spaces?

Model answer: An element m ∈ M is a torsion element if there exists a nonzero r ∈ R with rm = 0. In a vector space over a field, if λv = 0 with λ ≠ 0, then v = λ⁻¹(λv) = 0 — so the only torsion element is zero. Over a general ring, nonzero scalars need not be invertible, so rm = 0 can hold with both r ≠ 0 and m ≠ 0. For example, in the ℤ-module ℤ/6ℤ, the element 2̄ satisfies 3·2̄ = 6̄ = 0̄, so 2̄ is a nonzero torsion element.

Torsion is the module-theoretic manifestation of zero divisors in the ring acting on the module. It prevents the module from embedding into a free module in a dimension-preserving way and is the primary obstruction to modules behaving like vector spaces. The structure theorem for finitely generated modules over a PID cleanly separates torsion from free behavior, but over more general rings, torsion can be far more complex.