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There are the following Nakayama style lemmata:

  1. (the classical Nakayama lemma) Let $R$ be a commutative ring with $1$ and $M$ a finitely generated $R$-module. If $m_1, \ldots, m_n$ generate $M$ modulo $I$, where $I \subset \mathrm{Jac}(R)$, then they generate $M$.

  2. (the graded Nakayama lemma) See How to memorise (understand) Nakayama's lemma and its corollaries?.

  3. (the filtered Nakayama lemma) See How to memorise (understand) Nakayama's lemma and its corollaries?.

  4. (the topological Nakayama lemma, see [Neukirch, Schmidt, Wingberg], Cohomology of Number Fields, (5.2.18)): Let $\mathcal{O}$ be a commutative local ring complete in the $\mathfrak{m}$-adic topology with finite residue field of characteristic $p$. Assume $G$ is a pro-$p$-group and $M$ is a compact $\mathcal{O}[[G]]$-module. If $M/\mathfrak{m}$ is a finitely generated $\mathcal{O}[[G]]$-module, so is $M$.

  5. (Burnside's basis theorem, see also Let $G$ be a group such that its Frattini subgroup $\Phi(G)$ is finitely generated. Then a subset of $G$ generates $G$ iff it generates it modulo $\Phi(G)$.


Now my question is: Is there a common categorical version, like there is a categorical generalisation of Baer's criterion (In a suitable abelian catgory, an object $I$ is injective iff for all subobjects $U$ of a generator $G$ and morphisms $U \to I$ there is a lift $G \to I$)?

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In what you say about Baer's criterion it seems they don't need a notion of an ideal in a category. I imagine that you won't be so lucky with Nakayama. This MO question about how to do ideals in general categories might be useful:… –  David White May 20 '12 at 23:59

2 Answers 2

up vote 42 down vote accepted

Let me describe a common generalization of Nakayama's lemmas and Burnside's basis theorem which may shed some light here. Let $G$ be a group and $P$ a set of endomorphisms of $G$. A $P$-subgroup will be a subgroup of $G$ which is closed under acting by elements of $P$. We'll call $\mathbb{Fr}_P(G)$ the "$P$-Frattini subgroup of $G$", defined as the intersection of all maximal $P$-subgroups. Note the following special cases:

  • When $P$ is empty then $\mathbb{Fr}_P(G)$ is the Frattini subgroup of $G$.
  • When $R$ is a ring, $G$ its additive group and $P$ its multiplicative semigroup then $\mathbb{Fr}_P(G)$ is $J(R)$, the Jacobson radical of $R$.
  • When $P$ is as above, and $G$ is an $R$-module then $\mathbb{Fr}_P(G)$ contains $J(R)G$.

Let's denote the smallest $P$-subgroup containing a set $S$ by $\langle S\rangle_P$, and call an element $x\in G$, a non-generator if $G=\langle S,x \rangle_P$ always implies $G=\langle S\rangle_P$. We have the following theorem:

The set of non-generators of $G$ is precisely $\mathbb{Fr}_P(G)$.

By taking $P$ empty we obtain Burnside's basis theorem. By taking $G$ to be an $R$-module and $P$ to be the multiplicative semigroup of $R$ we recover Nakayama's lemma. If $R$ is a graded ring and we take $P$ to be the semigroup of elements of positive degree and $G$ to be a graded $R$-module, we recover the graded version of Nakayama's lemma, something similar should hold for the filtered version. Surely someone has taken up this point of view (Which I learned from Gruenberg's "Cohomological topics in group theory") to define a Frattini object for a large class of categories?

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Very interesting! I would never have guessed that there is a common generalization. –  Martin Brandenburg May 21 '12 at 6:25

Let me build a little on Gjergji's answer, using his terminology, in order to get closer to how I have usually heard Nakayama's lemma stated. Suppose, in addition, that $\mathbb{Fr}_p(G)$ is finitely generated and let $S \subseteq G$. Then, if the image of $S$ $P$-generates $G/\mathbb{Fr}_p(G)$, then $S$ $P$-generates $G$. (Hmm, it's not clear to me that $\mathbb{Fr}_p(G)$ is always normal in $\mathbb{Fr}_p(G)$, although it is in all of the examples so far. So we might need to add that as a hypothesis.)

Proof: Let $x_1$, $x_2$, ..., $x_r$ $P$-generate $\mathbb{Fr}_p(G)$. Then $G=\langle S, x_1, x_2, \ldots, x_r \rangle_P$ (exercise!). Repeatedly using the definition of a nongenerator, $G = \langle S \rangle_P$. $\square$

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Is it really necessary to assume $Fr_P(G)$ to be finitely generated? It really suffices to assume that $G$ is finitely generated (mostly to ensure that there are enough maximal subgroups). Then there is a slightly more general version of non-generator-ness that says $G=\langle S,X\rangle_P \wedge X\subseteq Fr_P(G) \implies G=\langle S\rangle_P$. –  Johannes Hahn Jan 11 '14 at 15:15

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