A Criterion for Zero Pure Tensors via Annihilators

Let $R$$R$ be a commutative ring and $M,N$$M,N$ be two $R$$R$-Module. This essay will discuss the sufficient and essential condition for

$$m\otimes n=0\in M{\otimes }_{R}N$$

Definition.

Let $L$$L$ be a $R$$R$-module, the Annihilator of $l\in L$$l\in L$ is

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

It is an ideal since it is the kernel of the $R$$R$-module homomorphism

$$\mu :R\to Rl,r⟼rl$$

Let $I\subseteq R$$I\subseteq R$ be an ideal and $N^{\prime }\subseteq N$$N'\subseteq N$ be a submodule, easy to see $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 ∈ M , n ∈ N$ . Then the following are equivalent

• A pure tensor $m\otimes n\in M{\otimes }_{R}N$$m\otimes n\in M\otimes_R N$ equal to $0$$0$.

• $n\in {\mathrm{Ann}}_R(m)N$$n\in\mathrm{Ann}_R(m)N$ or $m\in {\mathrm{Ann}}_R(n)M$$m\in\mathrm{Ann}_R(n)M$.

Proof.

Consider the following exact sequence:

$$0\stackrel{}{\to }{\mathrm{Ann}}_R(m)\stackrel{\iota }{\to }R\stackrel{\mu }{\to }Rm\stackrel{}{\to }0$$

Tensoring with $N$$N$ on the right hand side we have the following exact sequence:

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

This exact sequence is isomorphic to

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

Where ${\iota }^{\prime }(rn)=rn,{\mu }^{\prime }(n)=m\otimes n$$\iota'(rn)=rn,\mu'(n)=m\otimes n$.

Notice that ${\iota }^{\prime }$$\iota'$ is injective, hence we have the short exact sequence:

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

Hence $\ker ({\mu }^{\prime })={\mathrm{Ann}}_R(m)N$$\mathrm{ker}(\mu')=\mathrm{Ann}_R(m)N$. Thus for a fixed $m\in M,m\otimes n=0⟺n\in {\mathrm{Ann}}_R(m)N$$m\in M,m\otimes n=0\iff n\in\mathrm{Ann}_R(m)N$.

By symmetry we have for a fixed $n\in N,m\otimes n⟺m\in {\mathrm{Ann}}_R(n)M$$n\in N,m\otimes n\iff m\in\mathrm{Ann}_R(n)M$. $◻$$\square$

Corollary. Let $R$$R$ be an integral domain, and $M,N$$M,N$ be two torsion free module over $R$$R$.

Then $m\otimes n\in M{\otimes }_{R}N$$m\otimes n\in M\otimes_R N$ equal to $0⟺m=0$$0\iff m=0$ or $n=0$$n=0$.

Application to $S^1\otimes S^1$$S^1\otimes S^1$.

Let $[m]\otimes [n]\in S^1\otimes S^1$$[m]\otimes [n]\in S^1\otimes S^1$, then $[m]\otimes [n]=0⟺[m]\in {\mathrm{Ann}}_{\mathbb{Z}}([n])S^1$$[m]\otimes [n]=0\iff [m]\in\mathrm{Ann}_\mathbb Z([n])S^1$ or $[n]\in {\mathrm{Ann}}_{\mathbb{Z}}([m])S^1$$[n]\in\mathrm{Ann}_\mathbb Z([m])S^1$.

If $m$$m$ is irrational number, then ${\mathrm{Ann}}_{\mathbb{Z}}([m])=0$$\mathrm{Ann}_{\mathbb Z}([m])=0$, then $[m]\otimes [n]=0⟺[n]=0$$[m]\otimes [n]=0\iff [n]=0$.

If $m=\frac{p}{q}$$m=\frac{p}{q}$ is rational number, then ${\mathrm{Ann}}_{\mathbb{Z}}([m])=q\mathbb{Z}$$\mathrm{Ann_{\mathbb Z}}([m])=q\mathbb Z$, $q\mathbb{Z}{S}^1=\mathbb{Z}{S}^1=S^1$$q\mathbb ZS^1=\mathbb ZS^1=S^1$ since $S^1$$S^1$ is divisible.

Hence $[m]\otimes [n]=0$$[m]\otimes [n]=0$ for all $[n]\in S^1$$[n]\in S^1$ when $m\in \mathbb{Q}$$m\in\mathbb Q$.

In conclusion, $[m]\otimes [n]=0⟺m\in \mathbb{Q}$$[m]\otimes [n]=0\iff m\in\mathbb Q$ or $n\in \mathbb{Q}$$n\in\mathbb Q$.