Nakayama's Lemma

Statement

Let $R$ be a commutative ring, $I \subseteq R$ an ideal contained in the Jacobson radical $\operatorname{rad}(R)$ (the intersection of all maximal ideals), and $M$ a finitely generated $R$-module.

Nakayama's Lemma. If $M = IM$ (i.e., $M \le I \cdot M$), then $M = 0$.

Local version. If $(R, \mathfrak{m})$ is a local ring and $M$ is a finitely generated $R$-module with $M = \mathfrak{m}M$, then $M = 0$.

Generator corollary. Elements $m_1, \ldots, m_n \in M$ generate $M$ if and only if their images $\bar{m}_1, \ldots, \bar{m}_n$ generate $M/\mathfrak{m}M$ as an $R/\mathfrak{m}$-vector space.

Visualization

Schematic: why $M = \mathfrak{m}M$ forces $M = 0$ for a local ring.

M  finitely generated by  {m_1, ..., m_n}
          |
          | M = mM means each m_i = sum a_{ij} m_j  with a_{ij} in m
          |
          v
    Matrix equation:  (I - A) · [m_1 ... m_n]^T = 0
    where A = (a_{ij}) has entries in m
          |
          | det(I - A) = 1 - (terms in m) ∈ 1 + m ⊆ R^×  (units in local ring)
          |
          v
    (I - A) is invertible  ⟹  [m_1 ... m_n]^T = 0
          |
          v
    M = 0

Lifting generators example. Let $R = \mathbb{Z}_{(p)}$ (integers localised at $p$), $\mathfrak{m} = (p)$, $M = \mathbb{Z}_{(p)}^2$, and suppose $M/\mathfrak{m}M \cong (\mathbb{Z}/p)^2$ is generated by $\{\bar{e}_1, \bar{e}_2\}$. By Nakayama, $\{e_1, e_2\}$ generate $M$ over $R$.

Step	Content
$M/\mathfrak{m}M$ generators	$\bar{m}_1, \ldots, \bar{m}_k$ (finite, over field $R/\mathfrak{m}$)
Lift to $M$	$m_1, \ldots, m_k \in M$
Let $N = \langle m_1, \ldots, m_k \rangle$	submodule of $M$
$M/N$ satisfies	$M/N = \mathfrak{m} \cdot (M/N)$ (by choice of generators)
Nakayama gives	$M/N = 0$, i.e., $N = M$

Proof Sketch

1. Cayley-Hamilton trick. Since $M = IM$ and $M$ is generated by $m_1, \ldots, m_n$, write $m_i = \sum_j a_{ij} m_j$ with $a_{ij} \in I$. In matrix form: $(A - \mathrm{Id}) \mathbf{m} = 0$ where $A = (a_{ij})$ has entries in $I$.

2. Determinant is a unit. Multiply by the adjugate: $\det(A - \mathrm{Id}) \cdot m_i = 0$ for all $i$. Expanding: $\det(A - \mathrm{Id}) = -1 + (\text{terms in } I) \in -1 + I \subseteq R^\times$ (since $I \subseteq \operatorname{rad}(R)$ and elements of $1 + \operatorname{rad}(R)$ are units).

3. Conclusion. A unit kills all $m_i$, so $m_i = 0$ for all $i$, hence $M = 0$.

Connections

Nakayama's Lemma is the principal tool for working with modules over local rings in algebraic geometry and commutative algebra. The finite generation hypothesis is crucial — compare with Hilbert Basis TheoremHilbert Basis Theorem$$R \text{ Noetherian} \implies R[x] \text{ Noetherian}$$If R is Noetherian then so is R[x]: every ideal in a polynomial ring over a Noetherian ring is finitely generated.Read more →, which ensures it holds for ideals. The Cayley-Hamilton determinant step is a cousin of Cayley–Hamilton TheoremCayley–Hamilton Theorem$$p_A(A) = 0 \;\text{where}\; p_A(\lambda) = \det(\lambda I - A)$$Every square matrix satisfies its own characteristic polynomialRead more → in linear algebra.

-- Mathlib: Submodule.eq_bot_of_le_smul_of_le_jacobson_bot
-- in Mathlib.RingTheory.Nakayama (line 118)

/-- Nakayama's Lemma: if N ≤ I • N, I ≤ jacobson ⊥, and N is finitely generated,
    then N = ⊥ (the zero submodule). -/
example (R M : Type*) [CommRing R] [AddCommGroup M] [Module R M]
    (I : Ideal R) (N : Submodule R M) (hfg : N.FG)
    (hIN : N ≤ I • N) (hI : I ≤ Ideal.jacobson ⊥) : N = ⊥ :=
  Submodule.eq_bot_of_le_smul_of_le_jacobson_bot I N hfg hIN hI