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Let $(A,\mathfrak{m})$ be a local ring. Let us call $\mathfrak{m}$ almost nilpotent if for every sequence $a_1,a_2,\dotsc$ in $\mathfrak{m}$ there is some $n \geq 1$ such that $a_1 \cdot \dotsc \cdot a_n = 0$. Let $M$ be an $A$-module with $M = \mathfrak{m} M$. I would like to prove $M = 0$.

This is an exercise in Lenstra's Galois Theory for Schemes I've been struggling with quite some time. If $M$ is finitely generated, it is trivial (Nakayama). Now let's say $M$ is countably generated, by $m_1,m_2,m_3,\dotsc$. By elimination, we may then assume that $m_i \in \langle m_{i+1},m_{i+2},\dotsc \rangle$. If we had $m_i \in \langle m_{i+1} \rangle$, it would be easy to conclude: Choose $a_i \in \mathfrak{m}$ with $m_i = a_i m_{i+1}$. Now apply the assumption to the sequence $(a_i)$, this shows $m_1 = 0$. Since we could choose $m_1$ arbitrary, $M=0$. However, in general case, the equations become quite horrible and sums, not just products, are involved. If you draw a tree representing the linear combinations, I know that every path must end in a zero eventually, but not that the whole tree must end eventually. My gut feeling is that there may be universal counterexamples ...

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You possibly already know this, but for the benefit of other readers almost nilpotence is also known as T-nilpotence (T for transfinite). It's an important condition in the discussion of perfect rings. – rschwieb Jul 2 '12 at 11:03
I didn't know that, thank you. I already wondered why "almost nilpotent" didn't give appropriate search results. – Martin Brandenburg Jul 2 '12 at 17:54
Fantastic :) I'm glad to be alerted to this connection with Nakayama's lemma. I never use T-nilpotence, I just know that a ring is right perfect iff $R/rad(R)$ is semisimple and $rad(R)$ is right T-nilpotent. So here we are talking about perfect local ring. – rschwieb Jul 2 '12 at 17:56
up vote 3 down vote accepted

Lemma 28.3 from Anderson & Fuller, Rings and Categories of Modules, solves your question completely.

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Thank you. D'oh it is so simple! – Martin Brandenburg Jul 3 '12 at 14:51

König's Lemma states that a tree with infinitely many nodes and in which each node has only finitely many children, has an infinite branch.

In the tree that represents the linear combinations: cut off every branch as soon as the corresponding product of coefficients is zero. As the tree splits only finitely in every node, the tree must by König's Lemma be finite if $\mathfrak{m}$ is almost nilpotent.

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In fact I already thought of König's Lemma, but was not convinced that it is useful. And I am still not convinced. There are lots of infinite branches, I don't need König's lemma for that. Why does this have anything to do with finiteness? – Martin Brandenburg Jul 2 '12 at 17:51
Cut off as soon as the coeffients multiply to 0. An infinite branch in this new tree corresponds to a sequence violating the almost nilpotent property. – Gido Scharfenberger-Fabian Jul 5 '12 at 9:22

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