Zero-divisors of the tensor prodct

Question

Let $R$ be a commutative ring with identity and $M$ an $R$-module. An element $a\in R$ is called a zero-divisor of $M$ if the homothesy map $h_a:M\rightarrow M$, given by $h_a(x):=ax$ for all $x\in M$, is not injective. Then how are the zero divisors of $M$ and $M\otimes_RM$ are related? Note that, zero divisors of $M$ may become non-zero divisors of $M\otimes_RM$. For example, one may consider the subgroup $M$ of $\mathbb Q/\mathbb Z$ consisting of the equivalence classes of all rational numbers whose denominators are powers of $2$. Then $2$ is a zero-divisor of $M$, but $M\otimes_\mathbb ZM=0$. Can we get any positive result if $M$ is assumed to be finitely generated?

Answer 1

There can be many zero-divisors on $M\otimes_R M$, even when $M$ is torsion-free! Let me record the following observations, which are comments really, but I am mentioning this as an answer as it is too long, but also it leads to some very interesting concepts. In what follows, all rings are assumed to be Noetherian and all modules are assumed to be finitely generated.

Observation: Let $(R,\mathfrak m, k)$ be a local ring. Let $M$ be an $R$-module of infinite projective dimension. Then, every element of $\mathfrak m$ is a zero-divisor on $M\otimes_R \Omega^n_R k$ for every $n\ge 1$, i.e., $\mathfrak m \in \text{Ass}(M\otimes_R \Omega^n_R k)$ for every $n\ge 1$.

proof: Let $n\ge 1$. We have an exact sequence $0\to \Omega^n_R k \to F \to \Omega^{n-1}_R k \to 0$ for some free $R$-module $F$. Tensoring with $M$ we see an embedding $0\to \text{Tor}^R_1(\Omega^{n-1}_R k, M)\to M\otimes_R \Omega^n_R k$ . Now, $\text{Tor}^R_1(\Omega^{n-1}_R k, M)\cong \text{Tor}^R_n( k, M)$ is a $k$-vector space, and moreover it is non-zero since $M$ has infinite projective dimension. Thus, we get an embedding $0\to k \to M\otimes_R \Omega^n_R k$, hence $\mathfrak m \in \text{Ass}(M\otimes_R \Omega^n_R k)$, so every element of $\mathfrak m$ is a zero-divisor on $M\otimes_R \Omega^n_R k$ .

Corollary 1: Let $(R,\mathfrak m, k)$ be a local ring of depth $1$ (for e.g. a reduced local ring of dimension $1$). Let $M$ be a non-free and torsion-free $R$-module. Then, every element of $\mathfrak m$ is a zero-divisor on $M\otimes_R \Omega^n_R k$ for every $n\ge 1$.

proof: Let $M$ be as in hypothesis. As $R$ has depth $1$, so $\mathfrak m$ contains a non-zerodivisor on $R$, which then is a non-zerodivisor on $M$ as $M$ is torsion-free. Thus, $\text{depth } M\geq 1$. If $M$ had finite projective dimension, then $\text{pd } M\ge 1$ as $M$ is non-free, but then Auslander-Buchsbaum formula would give $1=\text{depth } R=\text{pd } M+\text{depth } M\geq 2$, contradiction! Thus, $M$ has infinite projective dimension, and so now the claim follows from the "Observation".

Corollary 2: Let $(R,\mathfrak m, k)$ be a non-regular local ring. Then, for every positive integers $a,b$, every element of $\mathfrak m$ is a zero-divisor on $\left(\Omega^a_R k \right)\otimes_R \left(\Omega^b_R k \right)$ .

Proof: this is immediate from the "Observation" as $R$ not being regular implies $\Omega^a_R k$ has infinite projective dimension for every $a\geq 1$.

Remark: In the "Observation", one could easily replace $\Omega^n_R k$ by $\Omega^n_R N$ where $N$ is a module locally free at non-maximal prime ideals, and moreover $N$ is strongly-rigid in the sense of Definition 2.1 of https://doi.org/10.2140/ANT.2010.4.1039 . Indeed, $N$ being locally free at non-maximal prime ideals would imply $\text{Tor}^R_1(\Omega^{n-1}_R N, M)\cong \text{Tor}^R_n( N, M)$ has finite length, and $N$ being strongly-rigid and $\text{pd } M=\infty$ implies that this is non-zero. Hence, $\text{Tor}^R_1(\Omega^{n-1}_R N, M)$ has depth $0$, so $k$ embeds in $\text{Tor}^R_1(\Omega^{n-1}_R N, M)$ which in turn embeds in $M\otimes_R \Omega^n_R N$ showing $M\otimes_R \Omega^n_R N$ has depth $0$ , i.e. every element of $\mathfrak m$ is a zero-divisor on $M\otimes_R \Omega^n_R N$.