A Criterion for Zero Pure Tensors via Annihilators

Let $R$$R$ be a commutative ring and $M,N$$M, N$ be two $R$$R$-modules.
We give a correct characterization of when $m\otimes n=0$$m \otimes n = 0$ in $M{\otimes }_{R}N$$M \otimes_R N$, under appropriate flatness hypotheses. The original argument is repaired by ensuring that the functor $-{\otimes }_{R}N$$- \otimes_R N$ (or $M{\otimes }_R-$$M \otimes_R -$) preserves the relevant monomorphisms.

Definition

Let $L$$L$ be an $R$$R$-module. For $l\in L$$l \in L$, the annihilator of $l$$l$ is

$${\mathrm{Ann}}_R(l):=\{r\in R:rl=0\}.$$

It is an ideal of $R$$R$, being the kernel of the homomorphism $\mu :R\to Rl,r\mapsto rl$$\mu : R \to Rl,\; r \mapsto rl$.
If $I\subseteq R$$I \subseteq R$ is an ideal and $N^{\prime }\subseteq N$$N' \subseteq N$ a submodule, then $I{N}^{\prime }:=\{a{n}^{\prime }:a\in I,n^{\prime }\in N^{\prime }\}$$IN' := \{ an' : a \in I,\; n' \in N' \}$ is a submodule of $N$$N$.

Proposition

Let $M,N$$M, N$ be $R$$R$-modules and let $m\in M,n\in N$$m \in M, \, n \in N$.

1. If $N$$N$ is a flat $R$$R$-module, then

   $$m\otimes n=0\text{in}M{\otimes }_{R}N⟺n\in {\mathrm{Ann}}_R(m)N.$$

2. If $M$$M$ is a flat $R$$R$-module, then

   $$m\otimes n=0⟺m\in {\mathrm{Ann}}_R(n)M.$$

In particular, if both $M$$M$ and $N$$N$ are flat, then $m\otimes n=0$$m \otimes n = 0$ iff either $n\in {\mathrm{Ann}}_R(m)N$$n \in \operatorname{Ann}_R(m)N$ or $m\in {\mathrm{Ann}}_R(n)M$$m \in \operatorname{Ann}_R(n)M$.

Proof

Assume that $N$$N$ is flat over $R$$R$. Fix $m\in M$$m \in M$ and consider the exact sequence

$$0⟶{\mathrm{Ann}}_R(m)\stackrel{{\iota }^{\prime }}{⟶}R\stackrel{\mu }{⟶}Rm⟶0.$$

Because $N$$N$ is flat, tensoring with $N$$N$ preserves exactness; we obtain the short exact sequence

$$0⟶{\mathrm{Ann}}_R(m){\otimes }_{R}N\stackrel{{\iota }^{\prime }\otimes 1}{\to }R{\otimes }_{R}N\stackrel{\mu \otimes 1}{\to }Rm{\otimes }_{R}N⟶0.$$

Using the natural isomorphism $R{\otimes }_{R}N\cong N$$R \otimes_R N \cong N$ given by $r\otimes n\mapsto rn$$r \otimes n \mapsto rn$, the above sequence becomes

$$0⟶{\mathrm{Ann}}_R(m)N\stackrel{{\iota }^{″}}{⟶}N\stackrel{{\mu }^{\prime }}{⟶}Rm{\otimes }_{R}N⟶0,$$

where ${\iota }^{″}(\sum _i{a}_i{n}_i)=\sum _i{a}_i{n}_i$$\iota''( \sum_i a_i n_i ) = \sum_i a_i n_i$ and ${\mu }^{\prime }(n)=m\otimes n$$\mu'(n) = m \otimes n$.
In particular, ${\iota }^{″}$$\iota''$ is injective, so

$$\ker ({\mu }^{\prime })={\mathrm{Ann}}_R(m)N.$$

Hence, in $Rm{\otimes }_{R}N$$Rm \otimes_R N$, we have $m\otimes n=0$$m \otimes n = 0$ if and only if $n\in {\mathrm{Ann}}_R(m)N$$n \in \operatorname{Ann}_R(m)N$.

Now, the inclusion $Rm\hookrightarrow M$$Rm \hookrightarrow M$ induces a map $Rm{\otimes }_{R}N\to M{\otimes }_{R}N$$Rm \otimes_R N \to M \otimes_R N$.
Since $N$$N$ is flat, this map is injective. Therefore, $m\otimes n=0$$m \otimes n = 0$ in $M{\otimes }_{R}N$$M \otimes_R N$ exactly when it vanishes in $Rm{\otimes }_{R}N$$Rm \otimes_R N$. This establishes the first equivalence.

The second statement follows by symmetry, assuming that $M$$M$ is flat and fixing $n\in N$$n \in N$. $◻$$\square$

Remark. Without the flatness assumption, the claim is false in general. For example, take $R=\mathbb{Z}$$R = \mathbb{Z}$, $M=\mathbb{Z}$$M = \mathbb{Z}$, $N=\mathbb{Q}/\mathbb{Z}$$N = \mathbb{Q}/\mathbb{Z}$.
The element $2\otimes (\frac{1}{2}+\mathbb{Z})$$2 \otimes (\frac{1}{2}+\mathbb{Z})$ vanishes in $\mathbb{Z}{\otimes }_{\mathbb{Z}}\mathbb{Q}/\mathbb{Z}\cong \mathbb{Q}/\mathbb{Z}$$\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Q}/\mathbb{Z} \cong \mathbb{Q}/\mathbb{Z}$, but ${\mathrm{Ann}}_{\mathbb{Z}}(2)=0$$\operatorname{Ann}_{\mathbb{Z}}(2)=0$, while $\frac{1}{2}+\mathbb{Z}\ne 0$$\frac{1}{2}+\mathbb{Z} \neq 0$. Here $N$$N$ is not flat.

Corollary

Let $R$$R$ be an integral domain, and let $M,N$$M,N$ be $R$$R$-modules.
If $R$$R$ is a Prüfer domain (e.g., a Dedekind domain or a PID) and $M,N$$M,N$ are torsion-free, then $M$$M$ and $N$$N$ are flat. In this case

$$m\otimes n=0⟺m=0\text{or}n=0.$$

More generally, the equivalence holds whenever $M$$M$ and $N$$N$ are flat and torsion‑free over an integral domain.

Application to $S^1{\otimes }_{\mathbb{Z}}{S}^1$$S^1 \otimes_{\mathbb{Z}} S^1$

The abelian group $S^1=\mathbb{R}/\mathbb{Z}$$S^1 = \mathbb{R}/\mathbb{Z}$ is not flat as a $\mathbb{Z}$$\mathbb{Z}$-module, so the flatness‑based criterion above does not apply directly. Nevertheless, we can still describe the vanishing of pure tensors by a structural argument.

Recall that $S^1$$S^1$ is divisible and decomposes as

$$S^1\cong \mathbb{Q}/\mathbb{Z}\oplus V,$$

where $V$$V$ is a $\mathbb{Q}$$\mathbb{Q}$-vector space (isomorphic to $\mathbb{R}/\mathbb{Q}$$\mathbb{R}/\mathbb{Q}$). For any $[m],[n]\in S^1$$[m],[n] \in S^1$:

• If $m\in \mathbb{Q}$$m \in \mathbb{Q}$ (i.e. $[m]\in \mathbb{Q}/\mathbb{Z}$$[m] \in \mathbb{Q}/\mathbb{Z}$), then $[m]$$[m]$ is a torsion element. Because $V$$V$ is divisible and torsion‑free, one checks that $\mathbb{Q}/\mathbb{Z}{\otimes }_{\mathbb{Z}}{S}^1=0$$\mathbb{Q}/\mathbb{Z} \otimes_{\mathbb{Z}} S^1 = 0$. Hence $[m]\otimes [n]=0$$[m] \otimes [n] = 0$.

• The same holds if $n\in \mathbb{Q}$$n \in \mathbb{Q}$.

• If both $m,n\notin \mathbb{Q}$$m,n \notin \mathbb{Q}$, then their images in $V$$V$ are non‑zero. In the $\mathbb{Q}$$\mathbb{Q}$-vector space tensor product $V{\otimes }_{\mathbb{Q}}V$$V \otimes_{\mathbb{Q}} V$, we have $v\otimes w=0$$v \otimes w = 0$ iff $v=0$$v = 0$ or $w=0$$w = 0$. Therefore $[m]\otimes [n]\ne 0$$[m] \otimes [n] \neq 0$ in $V{\otimes }_{\mathbb{Q}}V\hookrightarrow S^1{\otimes }_{\mathbb{Z}}{S}^1$$V \otimes_{\mathbb{Q}} V \hookrightarrow S^1 \otimes_{\mathbb{Z}} S^1$.

Consequently, the original conclusion remains valid despite the flawed proof:

$$[m]\otimes [n]=0\text{in}{S}^1{\otimes }_{\mathbb{Z}}{S}^1⟺m\in \mathbb{Q}\text{or}n\in \mathbb{Q}.$$