a categorical Nakayama lemma? - MathOverflow

a categorical Nakayama lemma?

2012-05-20T17:41:24Z

There are the following Nakayama style lemmata:

(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 $M$.
(the graded Nakayama lemma) See http://mathoverflow.net/questions/61446/how-to-memorise-understand-nakayamas-lemma-and-its-corollaries.
(the filtered Nakayama lemma) See http://mathoverflow.net/questions/61446/how-to-memorise-understand-nakayamas-lemma-and-its-corollaries.
(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 $M$.
(Burnside's basis theorem, see also http://groupprops.subwiki.org/wiki/Burnside%27s_basis_theorem) 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 $\Phi(G)$.

[tbc]

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 $G \to I$)?

Answer by Gjergji Zaimi for a categorical Nakayama lemma?

2012-05-21T01:37:39Z

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 Jacobdon 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?

Answer by David Speyer for a categorical Nakayama lemma?

2012-05-21T11:58:21Z

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, $G = \langle S \rangle_P$. $\square$