Favourite applications of the Nakayama Lemma Inspired by a recent question on the nilradical of an absolutely flat ring, what are some of your favourite applications of the Nakayama Lemma? It would be good if you outlined a proof for the result too. I am also interested to see the Nakayama Lemma prove some facts in Algebraic Geometry if possible. Here are some facts which one can use the Nakayama lemma to prove.


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*A local ring that is absolutely flat is a field - proof given here.

*Every set of $n$  - generators for a free module of rank $n$ is a basis - proof given here.

*For any integral domain $R$ (that is not a field) with fraction field $F$, it is never the case that $F$ is a f.g. $R$ - module. Sketch proof: if $F$ is f.g. as a $R$ - module then certainly it is f.g. as a $R_{\mathfrak{m}}$ module for any maximal ideal $\mathfrak{m}$. Then $\mathfrak{m}_{\mathfrak{m}}F = F$ and so Nakayama's Lemma implies $F = 0$ which is ridiculous.

 A: I never remember where it is used, but I did manage to retain one simple application that follows directly from the statement.

If $I$ is a finitely generated ideal of a ring (with identity) such that $I^2=I$, then $I$ is a ring with identity.

Maybe I am just too surprised by theorems which conclude that a rng has an identity :)
A: Here is a result in Algebraic Geometry that can be an application of Nakayama:

Let $X$ be a noetherian scheme and $\mathscr{F}$ a coherent sheaf.  Then $\mathscr{F}$ is invertible if and only if there exists some coherent sheaf $\mathscr{G}$ such that $\mathscr{F}\otimes\mathscr{G}=\mathcal{O}_X$.

A: Let $A$ be a noetherian ring. $\mathfrak{a} \subset A$ an ideal and $M$ be a finitely generated $A$-Module.
Let $N = \bigcap_{n \in \mathbb{N}} \mathfrak{a}^nM$.
Then there exists $x \in A$ such that $x \equiv 1 \mod\mathfrak{a}$ with $x \cdot N = (0)$.
In particular, if $x \in rad(A)$, then it is $N=(0)$.
Corollar Let $A$ be noetherian integral domain. $\mathfrak{a} \subset A$ an ideal with $\mathfrak{a} \neq A$. Then it holds
$\bigcap_{n \in \mathbb{N}} \mathfrak{a}^n = (0)$
(proof by Artin-Rees Lemma and Nakayama Lemma)
A: Let $A$ be a commutative ring and $M$ a finitely generated $A$-module.    Then every surjective endomorphism $f:M\to M$ is injective .
 This result (due to Vasconcelos) surprizingly holds in complete generality, without any noetherianness assumption, and crucially uses Nakayama in its proof: 
The trick is to consider  $M$  also as an $A[X]$-module via the multiplication $P(X)\cdot m=P(f)(m)$, so that for example $(X^3-X)\cdot m=f^3(m)-f(m)$.
The surjectivity asssumption translates into $M=IM$, where $I$ is the ideal $I=(X)\subset A[X]$.
Nakayama then says that for some $i=Q(X)X\in I$ we have $m=i\cdot m$ for all $m\in M$.
[Needless to say, since $M$  is finitely generated over $A$, it is a fortiori  finitely generated over $A[X]$ so that Nakayama may legitimately be invoked.]
And now the injectivity of $f$ follows: if $f(m)=0$ we have successively $$m=i\cdot m=Q(X)X\cdot m=Q(f)(f(m))=Q(f)(0)=0$$ so that $m=0$, which proves the injectivity of $f$.
A: You might be interested in this.
It contains some applications of Nakayama's Lemma in Commutative Algebra and Algebraic Geometry.
