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    Nakayama's lemma, in its several forms

    T-DEFNAKA

    For finite modules, generators lift modulo the Jacobson radical; locally, minimal generators are a basis of $M/\mm M$.Suggest a slogan

    (Nakayama's lemma)

    Let \(A\) be a ring, \(I\) an ideal, and \(M\) an \(A\)-module.

    1. If \(M\) is finitely generated and \(M = IM\), there exists \(a \equiv 1 \pmod I\) with \(aM = 0\).

    2. If in addition \(I \subseteq \Jac A\), then \(M = 0\).

    3. If \(I \subseteq \Jac A\), \(M\) is finitely generated and \(N \subseteq M\) is a submodule such that \(N/IN \to M/IM\) is surjective, then \(M = N\).

    4. If \((A,\mm)\) is local, \(M\) is finitely generated, and \(f_1,\ldots,f_n \in M\) have images generating \(M/\mm M\) as an \(A/\mm\)-vector space, then the \(f_i\) generate \(M\) as an \(A\)-module.

    Part (i) is the determinant trick: writing \(m_i = \sum_j a_{ij}m_j\) with \(a_{ij}\in I\) says \(\id - (a_{ij})\) kills \(M\), and multiplying by the adjugate shows \(\det(\id - (a_{ij}))\) does too, an element congruent to \(1\) modulo \(I\). Part (ii) follows because such an \(a\) lies in no maximal ideal, hence is a unit. Part (iii) applies (ii) to the cokernel \(L\) of \(N \injects M\), using right exactness of \(\wait\tensor A/I\) to get \(L = IL\). Part (iv) is (iii) with \(N = \generators{f_1,\ldots,f_n}\) and \(I = \mm\).

    For a finite module \(M\) over a local ring \((A,\mm)\), the images of a minimal generating set form a basis of the vector space \(M/\mm M\). Conversely, lifts of a basis generate by part (iv), and no proper subset generates because its residue classes do not span. Hence the minimal number of generators is \(\dim_{A/\mm}(M/\mm M)\).

    (Constant fiber dimension over a reduced local ring)

    Let \(A\) be a reduced local ring, let \(M\) be a finite \(A\)-module, and let \(r\ge0\) be an integer. For each prime \(\mathfrak p\subseteq A\), let \(k(\mathfrak p)\) be the residue field of \(A_{\mathfrak p}\). If \[ \dim_{k(\mathfrak p)}\bigl(M\otimes_A k(\mathfrak p)\bigr)=r \] for every prime \(\mathfrak p\subseteq A\), then \(M\cong A^r\). This is the local case of the finite locally free criterion; the noetherian sheaf formulation is proved in P-AGH258SEMICONT.

    (Constant fiber dimension over a nonreduced local ring)

    Let \(k\) be a field, \(A=k[\varepsilon]/(\varepsilon^2)\), and \(M=A/(\varepsilon)\). The only prime of \(A\) is \((\varepsilon)\), and \(M\otimes_A k\cong k\), so the fiber dimension is constantly one. The module \(M\) is cyclic but not free: the nonzero element \(\varepsilon\) annihilates \(M\), while every nonzero free \(A\)-module has zero annihilator.

    (Finite generation in Nakayama's lemma)

    Let \(p\) be a prime number, \(A=\ZZ_{(p)}\), and \(\mm=pA\). The \(A\)-module \(M=\QQ\) is nonzero and satisfies \(M=\mm M\), since multiplication by \(p\) is surjective on \(\QQ\). It is not finitely generated: the denominators of any finite set of rational numbers have bounded powers of \(p\), and taking \(A\)-linear combinations preserves that bound, whereas \(\QQ\) contains \(p^{-n}\) for every \(n\ge0\).

    State Nakayama's lemma.
    Over a local ring, how do you produce a minimal generating set of a finitely generated module?
    Where does finite generation get used, and what goes wrong without it?

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