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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.

  1. A local ring that is absolutely flat is a field - proof given here.
  2. Every set of $n$ - generators for a free module of rank $n$ is a basis - proof given here.
  3. 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.
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5 Answers5

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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 $f$.

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Georges Elencwajg
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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 :)

rschwieb
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    You get even more than this, right? ($I$ is generated by an idempotent, and you get a decomposition of the ring.) There's also something about flatness. – Potato Jul 03 '13 at 19:55
  • Dear @Potato Sure, the idempotent gives rise to a ring decomposition. I'm not sure what you have in mind for flatness: but $I$ is certainly be a projective $R$ module, being a summand of $R$. These implications seem rather disconnected from Nakayama's lemma in my mind though, and seem a little mundane compared to the implication above. If you had something different in mind about flatness, please let me know... – rschwieb Jul 03 '13 at 23:42
  • You get that $A/I$ is flat over $A$. And in fact these properties are equivalent. (But agree it doesn't have much to do with Nakayama.) – Potato Jul 03 '13 at 23:44
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    Dear @Potato Well... $A/I$ is even projective over $A$, since $A/I$ is isomorphic to a complement of $I$ in $A$ (which is a summand of $A$.) – rschwieb Jul 03 '13 at 23:47
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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 $\mathscr{F}\otimes\mathscr{G}=\mathcal{O}_X$.

Jared
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  • Dear Jared, do you have a reference for a proof of the above result? Thanks. –  Jun 30 '13 at 04:16
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    This doesn't have anything to do with noetherian schemes, it holds for locally ringed spaces in general. See http://mathoverflow.net/questions/33489/justification-of-the-term-invertible-sheaf. There you also see how Nakayama is used. – Martin Brandenburg Jun 30 '13 at 09:45
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You might be interested in this. It contains some applications of Nakayama's Lemma in Commutative Algebra and Algebraic Geometry.

Sai
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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)

Louis
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