Tensor-nilpotent modules Call a module $M$ over a commutative ring $R$ "tensor-nilpotent" if there is a positive integer $n$ for which $M^{\otimes_{R} n}$ (the tensor product of $n$ copies of $M$ over $R$) is the zero module. Note that this definition is very similar to the definition of nilpotent elements in a ring.
Examples of "tensor-nilpotent" modules include:


*

*$S/f(R)$ (the cokernel of $f$ as an $R$-linear map) where $f:R \to S$ is an epimorphism in the category of commutative rings (in this case, $n$ can in fact be chosen to be $2$)

*Any $R$-module $M$ for which the statement "$\forall m \in M \exists r \in R ((rm = 0) \land (\forall n \in M \exists p \in M (rp = n)))$" is true (i.e. any element of $M$ is annihilated by an element of $R$ for which the corresponding endomorphism of $M$ is surjective; again, $n$ can be chosen to be $2$)


*

*As a special case, any divisible torsion $R$-module where $R$ is an integral domain is "tensor-nilpotent".


*Any flat nilpotent ideal (if nonzero, the ring must not be Noetherian; see also this related question)


Question:

Is there a general classification of "tensor-nilpotent" modules (those for which some tensor power vanishes)?

Note that no tensor power of any nonzero finitely generated module can be zero, so "tensor-nilpotent" modules (except for the zero module) must be infinitely generated.
Also, the last lemma in the answer below can easily be generalized to show that if $M$ is a "tensor-nilpotent" $R$-module, then $\mathbf{m}M$ must be equal to $M$ for any maximal ideal $\mathbf{m}$ of $R$. In particular, if $R$ is a PID (principal ideal domain), then an $R$-module is "tensor-nilpotent" if and only if it is a divisible torsion module, and any "tensor-nilpotent" $R$-module must in fact vanish when tensored with itself.
 A: This is not a complete classification over a general ring, but only over a Dedekind domain. I'm not sure if it is reasonable to expect a classification over a general base ring.
Let $R$ be an integral domain with fraction field $K$, then if $M$ is tensor-nilpotent, $K \otimes_RM=0$, because if $K \otimes_R M \neq 0$, then $0 \neq (K \otimes_R M)^{\otimes_K n} =K \otimes_R M^{\otimes_R n}$. This implies that $M$ is a torsion module.  
Now let $R$ be a Dedekind domain.
For a maximal ideal $\mathfrak{m}$ of $R$ and a module $M$, set $M[\mathfrak{m}^\infty]=\{v \in M \mid \exists n \in \Bbb N: \mathfrak{m}^n v =0\}$. Then we have the following result:

Lemma If $M$ is a torsion $R$-module, then $M=\displaystyle \bigoplus M[\mathfrak{m}^\infty]$ where $\mathfrak{m}$ ranges over all maximal ideals of $R$

Proof: Let $v \in M$, $v \neq 0$, then $\mathrm{Ann}_R(v)$ is a non-zero proper ideal of $R$, so we have a factorization $\mathrm{Ann}_R(v)=\mathfrak{m}_1^{n_1} \cdot \ldots \cdot \mathfrak{m}_k^{n_k}$. We $\mathfrak{m}_1^{n_1} + \mathfrak{m}_2^{n_2} \cdot \ldots \cdot \mathfrak{m}_k^{n_k}=(1)$, so let $a_i \in \mathfrak{m}_i^{n_i}$ such that $a_1+a_2 \cdot \ldots \cdot a_n=1$. Then $v=a_1v+a_2 \cdot \ldots \cdot a_nv$. Note that $\mathrm{Ann}_{R}(a_1v) \subset \mathfrak{m}_2^{n_2} \cdot \ldots \cdot \mathfrak{m}_k^{n_k}$ and $\mathrm{Ann}_R(a_2 \cdot \ldots \cdot a_n) \subset \mathfrak{m}_1^{n_1}$, so that $a_2 \cdot \ldots \cdot a_nv \in M[\mathfrak{m}_1^\infty]$. Now inductively apply the same construction to $a_1v$, reducing the number of distinct prime ideals in the factorization of the annihilator at each step. It remains to show that $M[\mathfrak{m}_1^\infty] \cap M[\mathfrak{m}_2^\infty]$ for distinct maximal ideals $\mathfrak{m}_1,\mathfrak{m}_2$. This is because if $v \in M[\mathfrak{m}_1^\infty] \cap M[\mathfrak{m}_2^\infty]$, then $\mathrm{Ann}_r(v)$ contains both $\mathfrak{m}_1^{n_1}$ and $\mathfrak{m}_2^{n_2}$ and hence $\mathfrak{m}_1^{n_1}+\mathfrak{m}_2^{n_2}=(1)$.

Lemma If $M$ is a module over $R$, then $M[\mathfrak{m}_1^\infty] \otimes_R M[\mathfrak{m}_2^\infty]=0$ for distinct maximal ideals $\mathfrak{m}_1$ and $\mathfrak{m}_2$.

Proof: This is essentially the same argument as the last part of the previous proof: any element in $M[\mathfrak{m}_1^\infty] \otimes_R M[\mathfrak{m}_2^\infty]$ is annihilated by some power of $\mathfrak{m}_1$ and by some power of $\mathfrak{m}_2$ and as those powers are comaximal, we obtain that the annihilator contains $1$.

Corollary If $M$ is torsion module, then $M^{\otimes_R n}=\bigoplus M[\mathfrak{m}^\infty]^{\otimes_R n}$. Consequently, $M$ is tensor-nilpotent if and only if $M[\mathfrak{m}^\infty]$ is tensor-nilpotent for all $\mathfrak{m}$.

Proof: Clear.

Lemma $M[\mathfrak{m}^\infty]$ is a module over the local ring $R_\mathfrak{m}$

Let $a \in R \setminus \mathfrak{m}$, then we need to show that $a$ acts as an isomorphism by multiplication on $M[\mathfrak{m}^\infty]$. Let $v \in M[\mathfrak{m}^\infty]$, then $\mathfrak{m}^nv=0$ for some $n$. We have $(a)+\mathfrak{m}^n=(1)$, so choosing $\lambda a \in (a), m \in \mathfrak{m}^n$ with $1=\lambda a+m$, then $v=(\lambda a +m)v=a \lambda v$.This shows that multiplication by $a$ is surjective. It's also injective as $a \notin \mathfrak{m}^n$ for any $n$, so it's not in the annihilator of any element of $M[\mathfrak{m}^\infty]$.
We have thus reduced the classification over Dedekind domains to the classification over discrete valuation rings. Henceforth, let $R$ be a discrete valuation ring and let $\pi \in R$ be a uniformizer and let $K$ be the fraction field of $R$ and let $\kappa=R/(\pi)$.
Note that since $R$ is a DVR, a module $M$ is divisible if and only if $\pi M=M$.

Lemma If $M$ is a tensor-nilpotent module over $R$, then $M$ is divisible.

Proof: Suppose that $\pi M \subsetneq M$, then $\kappa \otimes_R M \neq 0$ is a non-zero vector space over $\kappa$ so that $0 \neq (\kappa \otimes_R M)^{\otimes_{\kappa}n}=\kappa \otimes_R M^{\otimes_R n}$, so $M^{\otimes_R n} \neq 0$.
By Baer's criterion, divsible modules over a PID are injective.
It is a result of Matlis (see for example, Lam - Lectures on Modules and Rings, §3) that over a Noetherian commutative ring $R$, every injective module decomposes uniquely as a direct sum of indecomposable injective modules of the form $E(R/\mathfrak{p})$ for $\mathfrak{p} \in \mathrm{Spec}(R)$, where $E$ denotes the injective hull. Over a DVR $R$ with fraction field $K$ and uniformizer $\pi$, this means that all injective modules are direct sums of $E(R/(0))=K$ and $E(R/(\pi))=K/R$. Clearly $K$ is not torsion, but $K/R$ is torsion and divsible and hence tensor-nilpotent.
Putting all these things together, we obtain the following classification:
If $R$ is a Dedekind domain, then any tensor-nilpotent module is of the form $\displaystyle \bigoplus_{\mathfrak{m} \in \mathrm{Spec}(R) \setminus 0} (K/R_{\mathfrak{m}})^{(I_\mathfrak{m})}$ for uniquely determined cardinal numbers $I_\mathfrak{m}$.
