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Nakayama's lemma is as follows:

Let $A$ be a ring, and $\frak{a}$ an ideal such that $\frak{a}$ is contained in every maximal ideal. Let $M$ be a finitely generated $A$-module. Then if $\frak{a}$$M=M$, we have that $M = 0$.

Most proofs of this result that I've seen in books use some non-trivial linear algebra results (like Cramer's rule), and I had come to believe that these were certainly necessary. However, in Lang's Algebraic Number Theory book, I came across a quick proof using only the definitions and induction. I felt initially like something must be wrong--I thought perhaps the proof is simpler because Lang is assuming throughout that all rings are integral domains, but he doesn't use this in the proof he gives, as far as I can see.

Here is the proof, verbatim: We do induction on the number of generators of $M$. Say M is generated by $w_1, \cdots, w_m$. There exists an expression $$w_1 = a_1w_1 + \cdots + a_mw_m$$ with $a_i \in \frak{a}$. Hence $$(1-a_1)w_1 = a_2w_2 + \cdots +a_mw_m$$ If $(1-a_1)$ is not a unit in A, then it is contained in a maximal ideal $\frak{p}$. Since $a_1 \in \frak{p}$ by hypothesis, we have a contradiction. Hence $1-a_1$ is a unit, and dividing by it shows that $M$ can be generated by $m-1$ elements, thereby concluding the proof.

Is the fact that $A$ is assumed to be a domain being smuggled in here in some way that I missed? Or is this really an elementary proof of Nakayama's lemma, in full generality?

YCor
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Phillip Williams
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    Is Cramer's rule really non-elementary? In any case, this is a complete proof. It also appears in many other places (eg it is the "second proof" of Nakayama's lemma in Atiyah-MacDonald). – Andy Putman Oct 11 '10 at 21:15
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    The Cramer's rule proof does give a little more. If we drop the assumption that $\mathfrak{a}$ is contained in the Jacobson radical it shows that $\mathfrak{a}M=M$ implies that $M$ is annihilated by a ring element congruent to $1$ modulo $\mathfrak{a}$. – Robin Chapman Oct 11 '10 at 21:19
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    I was always confused that people use the Cramer's rule proof too. Not only does this proof not use that $A$ is a domain, but it doesn't even use that $A$ is commutative! – Anton Geraschenko Oct 11 '10 at 21:19
  • I know that this this question is ten years old (...) but if the OP is still on MO, I wanted to point out that the strong form of Nakayama's lemma is actually a consequence of the weak form (in particular, there is a proof of the strong form, which does not involve determinants). See exercise 2, chap. 3, in Atiyah-Macdonald's book. – Damian Rössler Jan 04 '21 at 16:17
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    Complementing @DamianRössler's comment, and for those who don't have A–M at hand: for every ideal $I\subseteq A$, let $S=1+I$, then the localization $A\to S^{-1}A$ pushes $I$ into a Jacobson radical in a universal fashion (more precisely, it is the initial $A$-algebra where the image of $I$ lies in the Jacobson radical). Geometrically, it is the Zariski localization along the closed subscheme $V(I)\subseteq\operatorname{Spec}(A)$. This allows us to deduce the strong form of Nakayama's lemma from the weak version. – Z. M Jan 11 '22 at 08:43

2 Answers2

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There are various forms of the Nakayama lemma. Here is a rather general one; note that it does not involve maximal ideals and is a constructive theorem (Atiyah-MacDonald, Commutative Algebra, Prop. 2.4 ff).

Let $M$ be a finitely generated $A$-module, $\mathfrak{a} \subseteq A$ be an ideal and $\phi \in End_A(M)$ such that $\phi(M) \subseteq \mathfrak{a} M$. Then there is an equation of the form $\phi^n + r_1 \phi^{n-1} + ... + r_n = 0$, where the $r_i$ are in $\mathfrak{a}$.

The proof uses the equality $adj(X) \cdot X = \text{det}(X)$ for quadratic matrices over a ring. I call this an elementary linear algebra fact. Of course, there you only prove it for fields but using function fields implies the result for general rings. If we take $\phi=\text{id}_M$, we get the following form:

Let $M$ be a finitely generated $A$-module and let $\mathfrak{a} \subseteq A$ be an ideal such that $\mathfrak{a} M = M$. Then there exists some $r \in A$ such that $rM = 0$ and $r \equiv 1$ mod $\mathfrak{a}$.

In particular, we get:

Let $M$ be a finitely generated $A$-module and let $\mathfrak{a} \subseteq A$ be an ideal such that $\mathfrak{a} M = M$ and $\mathfrak{a}$ lies in every maximal ideal of $A$. Then $M=0$.

Observe that this argument uses Zorn's lemma (namely that every non-unit is contained in a maximal ideal) and is thus nonconstructive. Which is of course not surprising since without Zorn's lemma it is consistent that there are nontrivial rings without any maximal ideals at all. This should convince you that the first form of the Nakayama lemma is the most easy and elementary one. The last form has another short proof, which is standard and given in the question above.

Here is another short well-known proof for the last form, which also works if $A$ is noncommutative (then we have to replace "maximal ideal" by "maximal left ideal"): Assume $M \neq 0$. Since $M$ is finitely generated, an application of Zorn's lemma shows that $M$ has a maximal proper submodule $N$. Then $M/N$ is simple, thus isomorphic to $A/\mathfrak{m}$ for some maximal left ideal $\mathfrak{m}$. Then $N = \mathfrak{m} M = M$, contradiction.

By the way, I don't know if the first form is true if $A$ is noncommutative. The theory of determinants is not really prosperous over noncommutative rings. Hints?

In many texts about algebraic geometry only the last form of the Nakayama lemma is needed. But the first one is stronger and is used in many results in commutative algebra.

Martin Brandenburg
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    Very nice, Martin. – Hailong Dao Oct 12 '10 at 00:20
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    The first form is not true if $A$ is noncommutative. To see an example, note that we can use the first form of the result to prove the following fact: If $R$ is a ring such that the first form holds for $R[X]$, then any finitely generated $R$ module is Hopfian (i.e. any epic map is injective). This follows from applying the result given above to $R[X]$, with $X$ acting as your surjective map. A corollary of this fact is that rings for which the first form hold satisfy the Invariant Basis Number property - $R^{n}\cong R^{m}$ iff $m=n$. Tragically, there are rings which do not satisfy IBN. – Rishi Vyas Oct 12 '10 at 19:13
  • Atiyah's proof of Prop 2.4 is seriously lacking: in order to show that the polynomial has coefficients in $\mathfrak{a}$, he picks a generating set ${x_i}$ for $M$ over $A$, and then writes $\phi$ as a matrix $(a_{ij})$ with respect to these generators. He then assumes that the the entries of the matrix lie in $\mathfrak{a}$, presumably because the images $\sum_j a_{ij} x_j$ are in $\mathfrak{a}M$. But this assumption is false for an arbitrary generating set.

    The proof doesn't appear easily salvageable: if you want to use a generating set, it will require great care to choose it properly.

     – Tim Campion Oct 15 '10 at 00:39
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    @Tim: ? Since $M=\mathfrak{a} M$, the coefficients may be chosen in $\mathfrak{a}$. There is no lack at all. – Martin Brandenburg Oct 15 '10 at 05:38
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    Well, for the application to Nakayama's lemma, I suppose that $M = \mathfrak{a} M$. But the proposition statement needs to be amended to include that hypothesis: the proposition does not apply to all ideals $\mathfrak{a}$ as Atiyah's statement would have it. – Tim Campion Oct 15 '10 at 16:00
  • Even using the fact that $\mathfrak{a}M = M$, there's a little more to be said: first you pick a generating set ${x_i}$, then you write that basis as ${\alpha_i y_i}$ with $\alpha_i \in \mathfrak{a}$. Then you can write a matrix $\phi(y_i) = \sum f_{ij} x_j = \sum f_{ij} \alpha_j y_j$, which gives you a matrix relative to the ${y_i}$ generating set with coefficients in $\mathfrak{a}$. – Tim Campion Oct 15 '10 at 16:05
  • Sorry, "basis" should be "generating set", of course. – Tim Campion Oct 15 '10 at 16:09
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    I don't understand your problem in Atiyah's proof. – Martin Brandenburg Oct 17 '10 at 14:59
  • @MartinBrandenburg I think what TimCampion is saying is that there is a change of base matrix needed in order to write the matrix of $\phi$ so that all the entries of $\phi$ are elements of $\mathfrak a$. However, Atiyah & MacDonald do not acknowledge this. – user5826 Feb 09 '21 at 18:16
  • @TimCampion I know this is very old, but I do not understand your objection. Sure, if you write the $x_i$ in terms of a new basis $y_i$ you need a change of basis, but why would you do that? Every element of $aM$ can be written as $\sum_i a_i x_i$ with $a_i$ in $a$... – Denis Nardin Oct 09 '21 at 14:18
  • @DenisNardin I am feeling too lazy to actually look into this now, but let's just say that without even looking at it, I'd say it's at least 10x more likely that the proof is fine and $Tim_{2010}$ was silly than the other way around :). – Tim Campion Oct 11 '21 at 12:56
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I think the following proof is valid and avoids both determinants and maximal ideals. The cost is induction over all $A$-modules generated by $n$ elements.

Nakayama. Let $J$ be the Jacobson radical. Let $M$ be a finitely generated $A$-module satisfying $M=JM$. Then $M=0$.

Proof. Induction on size of generating set. If $M$ is generated by zero elements then it is zero. Assume the assertion holds for modules generated by $n-1$ elements and let $M$ be generated by $n$ elements $m_1,\dots ,m_n$. Then $m_n=\sum _{i=1}^n \varepsilon _im_i$ with $\varepsilon _i\in J$ so $(1-\varepsilon _n)m_n=\sum_{i=1}^{n-1}m_i$. Dividing by the unit $1-\varepsilon _n$ we obtain $m_n$ as a linear combination of $m_1,\dots ,m_{n-1}$. Proceeding in this fashion we find $M=0$.

Arrow
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  • And this proof works for noncommutative rings with the element-based characterization of Jacobson radical as in https://qchu.wordpress.com/2012/05/30/the-jacobson-radical/, i.e. the axiom of choice is not needed. – Z. M Jan 11 '22 at 08:36