### The Graded Nakayama's Lemma

My intuition for Nakayama's lemma is rooted in the graded version.

(Graded Nakayama's Lemma)
Let $R$ be a $\mathbb{N}$-graded algebra, and let $R_+$ be the 'irrelevant' ideal of positive degree elements. Let $M$ be a finitely-generated $\mathbb{Z}$-graded $R$-module.
If $I\subseteq R_+$, and $IM=M$ then $M=0$.

I find this version of the lemma very clear and intuitive. A finitely generated $R$-module will be zero in sufficiently low degree. If $M$ is non-zero, then there will be some minimal degree $d$ where $M_d\neq0$. But $R_+$ strictly increases degrees, and so $(R_+ M)_d=0$, and so $IM\neq M$.

In the study of connected graded algebras, the vector space $M/R_+M$ is an extremely useful gadget, which in a natural way parametrizes the generators of $M$. The graded Nakayama's lemma is just the first step along this correspondence.

### Other Nakayama's Lemmas

If you understand the graded Nakayama's lemma, the other version follow rather directly. The filtered version follows from the graded version by passing to the associated graded algebra.

(Filtered Nakayama's Lemma)
Let $R$ be a descending filtered algebra, and let $R_1$ be the ideal of positively filtered elements. Let $M$ be a finitely-generated good-filtered $R$-module so that $\cap M_i=0$.
If $I\subseteq R_1$, and $IM=M$ then $M=0$.

Proof: To see this, let $\overline{R}:=\oplus R_i/R_{i+1}$ be the associated graded algebra, and let $\overline{M}:=\oplus M_i/M_{i+1}$ be the associated graded module (the good-filtered condition on $M$ is exactly that $\overline{M}$ is f.g.). Then $I\subset R_1$ means $\overline{I}\subset \overline{R}_+$, and $\overline{I}\overline{M}=\overline{M}$, and so $\overline{M}=0$. Since $\cap M_i=0$, it follows that $M=0$.

The local Nakayama's Lemma is just a special case of the filtered version, with the $m$-adic filtration.

(Local Nakayama's Lemma)
Let $R$ be a local algebra, and let $m$ be the maximal ideal. Let $M$ be a finitely-generated $R$-module.
If $I\subseteq m$, and $IM=M$ then $M=0$.

Finally, the global Nakayama's lemma follows from the local one. This is because the Jacobson radical is contained in the maximal ideal in every localization, and if every localization of $M$ is zero, then $M$ is zero (uh, does this second fact use Nakayama's Lemma?).

(Global Nakayama's Lemma)
Let $R$ be an algebra, and let $J$ be the Jacobson radical. Let $M$ be a finitely-generated $R$-module.
If $I\subseteq J$, and $IM=M$ then $M=0$.

understandit. $\endgroup$ – Mariano Suárez-Álvarez Apr 12 '11 at 19:18