Take $R=\mathbb Z$, $M=\mathbb Q/\mathbb Z$, and $k=2$. The group $M$ is nonzero and divisible. Since every element of $M$ has finite order, $M$ is the filtered union of finite cyclic groups. For every $n\geq1$,
$$
(\mathbb Z/n\mathbb Z)\otimes_{\mathbb Z}M
\cong M/nM=0.
$$
Tensor products commute with filtered colimits, so
$$
M\otimes_{\mathbb Z}M=0.
$$

Now suppose a nonzero finitely generated $R$-module $M$ satisfied $M^{\otimes k}=0$. Choose a <maximal ideal> $\mathfrak m$ in the <support of a module>[support] of $M$. The localized module $M_{\mathfrak m}$ is nonzero and finitely generated. By <Nakayama lemma>,
$$
M_{\mathfrak m}/\mathfrak mM_{\mathfrak m}\ne0.
$$
This is a nonzero vector space over the <residue field> $\kappa(\mathfrak m)$, so its $k$-fold tensor power is nonzero. But it is the reduction modulo $\mathfrak m$ of $(M^{\otimes k})_{\mathfrak m}$, a contradiction. Thus a nonzero <tensor-nilpotent module> cannot be finitely generated.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 101 / 2 / iii / Solution

An $R$-module $M$ is tensor-nilpotent when $M^{\otimes k}=0$ for some positive integer $k$. A nonzero finitely generated module cannot be tensor-nilpotent: localize at a maximal ideal in its support and reduce modulo that maximal ideal, obtaining a nonzero tensor power of a nonzero vector space.


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