Suppose we want to prove a stronger form of Nakayama's Lemma, which goes as follows: Let $M$ be a finitely-generated $A$-module, and let $B$ be a (non-unital) subring of $A$ with $BM=M$ (here, $BM$ denotes the set of $B$-linear combinations of $M$). Then there exists $b\in B$ with $(1+b)M=0$. This Nakayama's Lemma is used in the answer here, motivated by an attempt to prove the Cayley-Hamilton theorem avoiding determinants. How can we prove this? Each of my attempts have led to the negative requirement that $AB\subseteq B$, i.e., that $B$ be an ideal, i.e., that $B$ be a submodule of $A$. Does anyone have any pointers or proofs?

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    $\begingroup$ I've added a comment in the linked thread which corrects the proof. $\endgroup$
    – user26857
    Jan 18, 2017 at 10:07

1 Answer 1


Actually, this statement is not true (I thought I had a proof, but it turns out it used the assumption that $B$ is an ideal in a sneaky way). The counterexample I gave to your previous question gives a counterexample here as well: let $A=\mathbb{Z}_{(2)}$ (or actually just $A=\mathbb{Z}[1/3]$ would work) and $B=3\mathbb{Z}$. Then $BM=M$ for any $M$, since the element $3\in B$ is a unit in $A$. But for instance, if $M=A$, then there is no element of the form $1+b$ that annihilates $M$, since the only element of $A$ that annihilates $M$ is $0$ and $-1\not\in B$.


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