Questions: Nakayama's Lemma

Question 1

Let (R, 𝔪) be a local ring and M a finitely generated R-module. If M/𝔪M = 0, what can you conclude?

• A. M = 0
• B. M is a free module
• C. M ≅ 𝔪M but M need not be zero
• D. Nothing — M/𝔪M = 0 carries no information about M

Answer: (A)

M/𝔪M = 0 means 𝔪M = M. By Nakayama's lemma, if M is finitely generated and 𝔪M = M, then M = 0. This is the 'annihilation' form of Nakayama's lemma. The finite generation hypothesis is essential: the ℚ as a ℤ₍ₚ₎-module satisfies pℚ = ℚ but ℚ ≠ 0 (it is not finitely generated over ℤ₍ₚ₎).

Question 2

Let (R, 𝔪, k) be a local ring and M a finitely generated R-module. If m₁, ..., mₙ ∈ M are elements whose images m̄₁, ..., m̄ₙ form a k-basis of M/𝔪M, then:

• A. m₁, ..., mₙ generate M as an R-module, and n is the minimal number of generators
• B. m₁, ..., mₙ form a basis for M as a free module
• C. m₁, ..., mₙ generate M only if M is torsion-free
• D. m₁, ..., mₙ generate 𝔪M but not necessarily M

Answer: (A)

Let N = Rm₁ + ··· + Rmₙ. Then N + 𝔪M = M (because the images span M/𝔪M), so M/N is killed by 𝔪 (since 𝔪(M/N) = (𝔪M + N)/N = M/N). By Nakayama, M/N = 0, i.e., N = M. The number n = dim_k(M/𝔪M) is the minimal number of generators because any generating set must map surjectively onto M/𝔪M. This is Nakayama's most powerful corollary: linear algebra over k (the residue field) determines the number of generators over R.

Question 3

Nakayama's lemma requires the module to be finitely generated.

• T. True
• F. False

Answer: True

Without finite generation, the lemma fails. The field ℚ, viewed as a module over ℤ₍ₚ₎ (the localization of ℤ at (p)), satisfies pℚ = ℚ. But ℚ ≠ 0. The issue is that ℚ is not finitely generated over ℤ₍ₚ₎ — it requires infinitely many generators. The Cayley-Hamilton-style proof of Nakayama uses a determinant argument that requires a finite generating set.

Question 4

Over a local ring, a finitely generated projective module is free.

• T. True
• F. False

Answer: True

This is a celebrated consequence of Nakayama's lemma. If P is a finitely generated projective module over a local ring (R, 𝔪), then P/𝔪P is a vector space over k = R/𝔪. Choose elements lifting a basis to get a surjection R^n → P. The kernel K satisfies K/𝔪K = 0 (because R^n/𝔪R^n → P/𝔪P is an isomorphism), so K = 0 by Nakayama. Thus P ≅ R^n. This dramatically simplifies the theory of projective modules locally.

Question 5

Prove that if M is finitely generated over a local ring (R, 𝔪) and 𝔪M = M, then M = 0.

Model answer: Let M = Rm₁ + ··· + Rmₙ with n minimal. Since 𝔪M = M, we can write mₙ = a₁m₁ + ··· + aₙmₙ with aᵢ ∈ 𝔪. Then (1 - aₙ)mₙ = a₁m₁ + ··· + aₙ₋₁mₙ₋₁. Since aₙ ∈ 𝔪 and R is local, 1 - aₙ is a unit (it lies outside the unique maximal ideal). So mₙ = (1-aₙ)⁻¹(a₁m₁ + ··· + aₙ₋₁mₙ₋₁), meaning M is generated by m₁, ..., mₙ₋₁ — contradicting minimality of n, unless n = 0, in which case M = 0.

The proof reveals exactly why local rings are essential: the element 1 - aₙ (with aₙ in the maximal ideal) must be a unit. In a non-local ring, 1 - aₙ might be a non-unit sitting in some other maximal ideal, and the argument fails. The local hypothesis converts 'close to the identity' into 'invertible,' which is the geometric intuition: near a point, small perturbations of invertible functions stay invertible.