Nakayama's lemma states that given a finitely generated $A$-module $M$, and $J(A)$ the Jacobson radical of $A$, with $I\subseteq J(A)$ some ideal, then if $IM=M$, we have $M=0$.

I've read the proof, and while being relatively simple, it doesn't give much insight on why this lemma should be true, for example - is there some way to see how the fact that $J(A)$ is the intersection of all maximal ideals related to the result?

Any intuition on the conditions and the result would be of great help.

  • $\begingroup$ Have you read the Wikipedia article? $\endgroup$ Jan 25, 2011 at 14:15
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    $\begingroup$ @Qiaochu Yuan - I've looked at it, but there doesn't seem to be any actual intuition given there (at least none that I can see). $\endgroup$
    – Pandora
    Jan 25, 2011 at 14:27
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    $\begingroup$ This question was just referenced in MathOverflow.Net - you might want to check out the answers! mathoverflow.net/questions/61446/… $\endgroup$
    – pageman
    Apr 21, 2011 at 20:51

4 Answers 4


Suppose your module is of finite length. Then you can consider on it the so called radical filtration, which organises the module into an onion-like thing, with elements of the maximal ideal pushing elements of the module farther in from their starting layer to one right below and, moreover, each layer obtained from the one above it in this way.

Now, the condition $\mathfrak m M=M$ tells you that the outermost layer of the module is actually empty: obviously, then, there is not much in the whole thing and $M=0$. We have just discovered Nakayama's lemma!

If your module is arbitrary, exactly the same happens.

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    $\begingroup$ I am talking about the local case, with $I$ equal to the maximal ideal. The general case is just a natural generalization. $\endgroup$ Jan 25, 2011 at 14:51
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    $\begingroup$ Reading a bit about path algebras and their modules, and specially their interpretation as representations of the underlying quiver, IMO, are a great way to develop an intuition about this, for it allows you to draw modules and algebras in a very explicit way, making statements such as Nakayama's pretty much self-evident. $\endgroup$ Jan 25, 2011 at 15:27
  • $\begingroup$ Thank you for your answer! Could you refer me to some source where I could read more about this radical filtration? I can't find anything relevant and it seems like such a background could really help me see things more clearly. $\endgroup$
    – Pandora
    Jan 26, 2011 at 18:53
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    $\begingroup$ One place where you'll find it is in Skowronski+Assem+Simpson book about representation theory of finite dimensional algebras, vol. 1. $\endgroup$ Jan 26, 2011 at 21:05

Here's something that might or might not make sense to you. You know that every ideal $I$ of a commutative ring $R$ gives rise to an $R$-module $R/I$; these are precisely the $R$-modules on one generator. The $R$-modules of the form $R/m$ where $m$ is maximal are special among these. In this language, the elements of the Jacobson radical are precisely the ones that act trivially on all $R$-modules of the form $R/m$. It follows, for example, that if $I$ is in the Jacobson radical, then we cannot have $IM = M$ for any module of the form $R/m^k$.

Nakayama's lemma asserts that a similar statement is true for all finitely generated $R$-modules. This should be reasonable if you are aware of, for example, the classification of finitely generated $R$-modules when $R$ is a PID. The Wikipedia describes a geometric interpretation of this, but I'm not familiar enough with it to say more.

  • $\begingroup$ First, welcome back. This may be a really naive question: The version I have in Reid "Undergrad. Comm. Alg" is stated over a local ring $(A,m)$. I was wondering if this is correct: since $m$ is maximal, it has no units. Thus for $M$ to equal $mM$, I would think $M$ has to be $0$. If you think this is worthy of being a question, I will be happy to post it where you can answer it. Thanks, regards. $\endgroup$
    – user12802
    Dec 21, 2012 at 19:10
  • $\begingroup$ @Andrew: that is a special case of other statements of Nakayama's lemma. See the Wikipedia article. It would be fine to post a separate question if you are still confused. $\endgroup$ Dec 21, 2012 at 22:07

This is probably not the answer you're looking for, but here's another proof:

Suppose $M \neq 0$ and let $u_1,\ldots,u_n$ be a minimal set of generators for $M$. Then $u_n \in IM$ so we may write $u_n=a_1u_1+\ldots+a_nu_n$ with $a_i \in I$. Hence $(1-a_n)u_n=a_1u_1+\ldots+a_{n-1}u_{n-1}$. Since $I$ was contained in the Jacobson radical, it follows that $1-a_n$ is a unit. Hence $u_n$ belongs to the submodule generated by the $u_1,\ldots,u_{n-1}$, contradiction.


There is a more primitive and natural version of Nakayama's lemma:

Let $R$ be a commutative ring, and $M$ be a finitely generated $R$-module and $I\subset R$ be an ideal. If $IM=M$ then $\exists f\in I,\forall m\in M,f\cdot m=m$.

The proof is a bit harder than the above version, but this pretty much says "If $IM=M$ then one of the elements of $I$ stablizes all elements of $M$." And I think this is much more natural than the above version. Assuming this version, then we can see that the condition of "Jacobson ideal" is only used to show that "$1-f$ is a unit" which leads to "$M=0$".

For completion, I give a proof of the above version here:

Proof: Let $m_1,...,m_n$ be a set of generators of $M$. Since $IM=M$, we have a matrix $A'=(a_{ij})\in \textrm{M}_{n\times n}(R)$ such that $A'\underline{m}=\underline{m}$ where $\underline{m}=(m_1,...,m_n)^t$ and $a_{ij}\in I,\forall i,j\leq n$. Now let $A=\textrm{Id}-A'$ so we have $A\underline{m}=\underline{0}$.

Note that $\det A\in R$ is well-defined and there exists a matrix $B\in \textrm{M}_{n\times n}(R)$ s.t. $BA=AB=\det A\cdot \textrm{Id}$, the proof is in Corollary 9.161 of Advanced Modern Algebra by Joseph J. Rotman. Also $\det$ commutes with the quotient map $R\rightarrow R/I$, with $A\equiv \textrm{Id} \ (\textrm{mod }I)$ we can deduce that $\det A\in 1+I$, so $\exists f\in I$, s.t. $\det A=1-f$.

To show $f$ stablizes $M$ it suffices to show that $f\cdot m_i=m_i,\forall i$. $$\underline{0}=BA\underline{m}=\det A\cdot \underline{m}=(1-f)\cdot \underline{m}$$ So $$f\cdot \underline{m}=\underline{m}$$ The result follows. Q.E.D

  • $\begingroup$ I think you want to let $A = \operatorname{Id} - A'$ instead of the other way. Nice writeup though! This is the first time I've understood a presented proof of the lemma. $\endgroup$
    – feralin
    Dec 23, 2019 at 21:25
  • $\begingroup$ @feralin You re right, thx. Fixed now. $\endgroup$
    – Z Wu
    Dec 30, 2019 at 12:47
  • $\begingroup$ Can you elaborate for me what it means to have $A'm = m$? Why does $IM = M$ imply that there's a matrix $A'$ such that $A'm = m$? $\endgroup$ Mar 14, 2021 at 23:03
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    $\begingroup$ @SiddharthBhat Each element $m\in M=IM$ can be written as a finite sum $\sum_j s_j t_j$ where $s_j\in I$ and $t_j\in M$. Expanding $t_j$ as a finite sum of $r_{ij}m_i$'s (since $\{m_i\}$'s are the generators), we eventually have $m=\sum_i a_i m_i$ where $a_i\in I$. For each $m_j$, we get an equaltiy $m_j=\sum_i a_{ij}m_i$ so that we can write them in a matrix form. $\endgroup$
    – Z Wu
    Mar 15, 2021 at 8:06

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