Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hope this question is fine. Nakayama's lemma http://en.wikipedia.org/wiki/Nakayama_lemma#Statement is mentioned in the majority of books on algebraic geometry that treat varieties. So I think, I red the formulation of this lemma at least 20 times (and red the proof maybe around 10 times) in my life. But for some reason I just can not get this lemma, i.e. I have tendency to forget it. Last time this happened just a couple of days ago, in the book of Shafarevich (Basic Algebraic geometry in 1.5.3) this lemma is used to prove that for finite maps between quasiprojective varieties the image of a closed set is closed, and again this lemma sounded as something foreign to me (so again I went through the proof of the lemma)...

Question. Is there a path to get some stable understanding of Nakayama's lemma and its corollaries? I would be especially happy if there were some geometric intuition below this lemma. Or some geometric example. Or maybe there is a nice article of this topic? Some mnemonic rule? (or one just needs to get used to the lemma?)

share|improve this question
1  
When you have seen it used a few tens of times, you'll get so used to it that you will say with confidence that you understand it. –  Mariano Suárez-Alvarez Apr 12 '11 at 19:18
2  
Do take a look at math.stackexchange.com/questions/18902/…, though. –  Mariano Suárez-Alvarez Apr 12 '11 at 19:18
34  
Dear aglearner, congratulations on your candidness! You are in good company: I've heard René Thom, a Fields medalist and one of the most creative geometers of the 20th century, admit publicly in a talk in front of a vast audience that he could never remember this lemma. It is a real mystery to me why the statement is indeed so difficult to remember : the algebraic concepts involved are fairly elementary, the statement is pleasantly general and there is nothing counterintuitive to the result. My solution to the problem was essentially to learn it by heart.... –  Georges Elencwajg Apr 12 '11 at 19:33
add comment

7 Answers 7

It's sort of like the inverse function theorem, and that is why it is so strong. If you have $n$ functions vanishing at the origin of $k^n$ and want to know if they give a local coordinate system, you ask if their differentials are independent at the origin. Or equivalently if their differentials generate the cotangent space at the origin. So in a [not necessarily noetherian, thanks Georges!] local ring $(\mathcal{O},\mathfrak{m})$, Nakayama's lemma says you can detect that elements of the maximal ideal generate that ideal, hence act sort of like coordinate functions, just by knowing their differentials, i.e. their residues in the Zariski cotangent space $\mathfrak{m}/\mathfrak{m}^2$, generate that linear space.

Those versions of the lemma you linked to are almost unrecognizable forms of this simple statement, but that's the way abstract math goes as we know. But the idea is the same, you have a hypotheses about a truncated version of your statement, and you get out the fuller version. The Jacobson radical stuff is there to disguise the fact that it doesn't say much unless you are in a local setting. I.e. in a local ring the Jacobson radical is pretty big and you get a better result. In a polynomial ring with tiny Jacobson radical you get nothing.

share|improve this answer
    
Thank you! I'll try to consider Nakayama's lemma from this standpoint. –  aglearner Apr 12 '11 at 19:38
1  
There are many great insights in this answer, roy! You are perfectly right that Jacobson radicals tend to be small: for example, in a Jacobson (!) ring the Jacobson radical is equal to the nilradical and so is zero if the ring is reduced: as you so aptly write, you get nothing in that case. By the way, the ring $(\mathcal O,\frak m)$ needn't be noetherian: I'm not trying to nitpick but just to emphasize how elementary Nakayama's lemma is. And yet so difficult to remember! –  Georges Elencwajg Apr 12 '11 at 20:41
    
Great answer! thanks! –  SGP Apr 12 '11 at 21:41
15  
As a reference, corollary 14.10, p. 179 of Joe Harris' Algebraic Geometry, proves that a bijection of projective varieties with injective differential at all points is an isomorphism, i.e. a version of the inverse function theorem, as a corollary of nakayama. I myself learned this interpretation of nakayama in 1965, from my algebra teacher, the great Maurice Auslander –  roy smith Apr 12 '11 at 22:21
    
Another good reference for many geometric consequences of Nakayama is Mumford's "red book" chapter III, sections 4,5,10 on non singularity, and etale and smooth morphisms. Nakayama's lemma itself, in several forms, is in section 2. –  roy smith Apr 14 '11 at 15:17
show 2 more comments

Mnemonic: $\quad M=IM \Rightarrow m=im$

The version of Nakayama described: If $I$ is an arbitrary ideal of an arbitrary commutative ring $A$ and if a finitely generated module $M$ satisfies $M=IM$, then there exists $i\in I$ such that for all $m\in M$ we have $m=im$.
Please notice: no noetherian nor local assumption on $A$, no assumption at all on $I$.

share|improve this answer
4  
Very nice mnemonic! And also thanks for sharing the story about Rene Tom. –  aglearner Apr 12 '11 at 22:56
11  
You need to assume that your "arbitrary ring" $A$ is commutative. For a non-commutative counterexample, take $A$ to be any domain with a proper non-zero idempotent ideal $I$ and set $M = I$. Then $M = I = I^2 = IM$ since $I$ is idempotent so if your result were true we would get an element $i \in I$ such that $(1-i)I = 0$; since $I \neq 0$ and $A$ is a domain this forces $1 = i \in I$, contradicting the properness of $I$. For a concrete example, let $\mathfrak{g}$ be a perfect Lie algebra and let $I = \mathfrak{g}A$ be the augmentation ideal of its enveloping algebra $A = U(\mathfrak{g})$. –  Konstantin Ardakov Apr 13 '11 at 8:22
3  
Ah yes, Konstantin, you are right. Like many algebraic geometers I tend to say and write "ring" when I mean "commutative ring". I have now edited my answer in order to prevent any misunderstanding.Thanks for your attention and apologies for my sloppiness. –  Georges Elencwajg Apr 13 '11 at 9:18
    
You're welcome. Glad to be of help. –  Konstantin Ardakov Apr 13 '11 at 10:47
add comment

It's easiest to understand for local rings, so let $R$ be one with residue field $k$. Nakayama's lemma just says that a finitely generated $R$-module is zero if and only if the induced $k$-vector space is. Through the magic of abelian categories, this implies that a map of $R$-modules is surjective if and only if the induced $k$-linear map of $k$-vector spaces is (apply the lemma to its cokernel). This says that I can find generators for an $R$-module by lifting a basis of its associated $k$-vector space (that is, I can test whether a map $R^n \to M$ is surjective by testing it after reducing by $k$).

There are two ways to look at this: one (algebraically), it allows you to consider a lot of $R$-module statements as actually being $k$-linear algebra statements; and two (geometrically), it allows you to transfer information from the fiber of a sheaf at a point to the stalk at that point, and from there, to an open neighborhood.

An example of the first property: suppose you want to prove the Cayley-Hamilton theorem for a linear endomorphism $A$ of some finitely-generated $R$-module: that $A$ satisfies its own characteristic polynomial $p_A$. Note that $p_A$, as an element of $R[t]$, reduces correctly when we pass to $k$, so that $p_A(A)$ vanishes after reducing to $k$ by the Cayley-Hamilton theorem for vector spaces. Therefore, by Nakayama's lemma applied to the image of $p_A(A)$, it vanishes over $R$ as well.

An example of the second property: suppose $R$ is noetherian and I have a flat $R$-module $M$, and I choose a basis for its reduction to $k$, giving a presentation $R^n \to M \to 0$ (it is surjective by the lemma applied to the cokernel, as explained before). This turns into a short exact sequence $0 \to K \to R^n \to M \to 0$ in which $K$ is finitely generated (since $R$ is noetherian) and since $M$ is flat, it remains exact after reducing to $k$, where the kernel $K$ vanishes. Conclusion: $M$ is free over $R$. The geometric interpretation of this is that flat, coherent sheaves over a noetherian scheme (if you're reading Shafarevich, your schemes are varieties and are always noetherian) are vector bundles.

share|improve this answer
    
great answer with nice examples! –  SGP Apr 12 '11 at 21:40
    
I don't understand the last example. if you reduce that s.e.s. to $k$, you get an exact sequence $0\to 0\to k^n\to M\otimes k\to 0$. But we knew that already and you don't need flatness of M for that. How does it follow that in the original sequence $K$ vanishes? –  Toink May 16 '13 at 18:53
3  
@Toink: Because by flatness we know that the sequence you wrote is actually of the form $0 \to K \otimes k \to R^n \otimes k \to M \otimes k \to 0$. Therefore $K \otimes k = 0$, so by Nakayama $K = 0$. Without flatness we'd only have $K \otimes k \to R^n \otimes k \to M \otimes k \to 0$, and thus $K \otimes k \twoheadrightarrow \operatorname{ker}(R^n \otimes k \to M \otimes k) = 0$, which is exactly zero information about $K$ :) –  Ryan Reich May 16 '13 at 19:00
    
oh nice, thanks –  Toink May 16 '13 at 19:07
add comment

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 $M=0$.

share|improve this answer
1  
Second fact: no, it's just that if you have a finitely generated $M$, then $V(\operatorname{ann} M) \subset \operatorname{supp} M$, which you can prove straight from the definition of localization at a prime. –  Ryan Reich Apr 12 '11 at 22:06
    
Also, I like this intuition. –  Ryan Reich Apr 12 '11 at 22:07
    
Could you briefly describe or give a reference to the Rees construction? It doesn't have e.g. a Wikipedia article. –  Qiaochu Yuan Apr 13 '11 at 6:36
    
Qiaochu; I added the proof that Filtered=>Graded. I didn't realize use the Rees algebra, instead I used the associated graded. –  Greg Muller Apr 13 '11 at 16:05
    
The Rees algebra itself has a wikipedia article (which only covers the case of $I$-adic filtrations), and Chapter 5 of Eisenbud has a similar $I$-adic focus. –  Greg Muller Apr 13 '11 at 16:08
show 1 more comment

For me the Nakayama lemma (even though maybe not in its strongest form) simply says that:

If $\mathcal{F}$ is a coherent sheaf over the (locally noetherian) scheme $X$, then the dimension of the fiber of $\mathcal{F}$ at a closed point $x\in X$ is equal to the rank of the stalk, and a basis of the fiber lifts to a system of generators of the stalk.

share|improve this answer
add comment

I usually find the statement of Nakayama's Lemma easy to remember because of its proof, which is really nothing more than the definition of the Jacobson radical plus the existence of maximal left ideals in a ring.

Every non-zero finitely generated module $M$ admits a non-zero cyclic quotient module, which in turn (by a Zornication) admits a non-zero simple quotient module. So we can find a submodule $N$ of $M$ with $M/N$ simple. But now $J . (M/N) = 0$ since the Jacobson radical $J$ of the ring kills every simple module, so $JM \leq N < M$ which says that $JM$ is a proper submodule of $M$.

Note that this general form of the Lemma doesn't need any complicated determinant-type arguments. In the commutative case, other forms of the Lemma can easily be obtained from this general form "$JM < M$" by considering localisation.

share|improve this answer
1  
That's the way I like to think of it too: a finitely generated module over any ring (with identity) has a simple quotient. –  user1421 Apr 27 '11 at 21:29
    
+1 for the Zornication! –  ACL Jan 12 '13 at 10:49
    
By the way, if you're ready to zornicate/zornify, rather observe that the set of proper submodules of a non-zero finitely generated module is inductive, hence has a maximal element. –  ACL Jan 12 '13 at 10:50
1  
Do you have a published reference for this proof? –  ACL Jan 12 '13 at 10:50
add comment

As one says, proofs are useful not only to certify the truth of a statement, but also to remember and understand the statement better. But in this second function, proofs should not be understood only as "proof of the statement". Proofs that use the statement, instead, are also a very useful way to remember it.

There was a time where I had difficulty to remember Nakayama. This time ended when I learnt the following basic result: Let $M$ be a finitely generated module over a local domain $A$ of maximal ideal $m$, residue field $A/m=k$, fraction field $K$. To such an $M$ one can attach two finite-dimensional vector spaces, one, $M \otimes_A K$ over $K$, the other, $M \otimes_A k = M/mM$ over $k$. Then one always $\dim_K M \otimes_A K \leq \dim_k M \otimes_A k$ with equality if and only if $M$ is free.

I find this result much more striking and easy to remember than Nakayama itself. Yet it is essentially equivalent to it. Here is the proof: Take $e_1, \dots, e_n$ be a basis of $M \otimes_A k$. Lift this in a family $f_1,\dots,f_n$ of $A$. By Nakayama, $f_1, \dots, f_n$ generates $M$ as an $A$-module, hence $M \otimes_A K$ as a $K$-vector space, hence the stated inequality. If furthermore is is an equality, then $f_1,\dots,f_n$ is a basis of $M \otimes_A K$, hence $K$-free, hence $A$-free, hence an $A$-basis of $M$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.