On modules in which tensor of each two elements is commutative

Question

Let $R$ be a commutative ring with unity. Let $M $ be a finitely generated $R$-module. Is it true that the following conditions are equivalent :

(1) In $M \otimes_RM$, we have $m\otimes n=n\otimes m, \forall m,n \in M$.

(2) $M_\mathfrak p$ is cyclic $R_\mathfrak p$-module for every $\mathfrak p \in \operatorname{Spec} R$.

(3) $M_\mathfrak m$ is cyclic $R_\mathfrak m$-module for every maximal ideal $\mathfrak m$ of $R$ .

?

I believe Nakayama lemma should be useful here, but I don't see how.

Please help.

Answer 1

There is always the natural antisymmetrization map $\iota_M:\bigwedge^2_R(M) \to M \otimes_R M$ given on elementary wedges by $m \wedge n \mapsto m \otimes n-n \otimes m$. Your statement (1) is then obviously equivalent to saying that $\iota_M$ is the $0$ map. Being $0$ is a local property, so it will suffice to show the claim when $(R,\mathfrak{m},k)$ is local. In particular, this reduces the claim to showing the following:

Suppose $(R,\mathfrak{m},k)$ is local. Then $\iota_M=0$ if and only if $M$ is cyclic.

Proof: The forward direction first: The map $\iota_{M \otimes_R k}$ is naturally identified with $\iota_M \otimes \operatorname{id}_k$. In particular, we have that $\iota_{M \otimes_R k}=0$, by hypothesis. But, $\iota_V$ is always injective when $V$ is a $k$-vector space; in fact, $\operatorname{id}_F$ is always injective when $F$ is a free $R$-module. I'll leave this as an exercise.

In particular, this means $\bigwedge^2_R(M \otimes_R k)=0$. But the vector space dimension of this module is always ${\dim_k(M \otimes_R k) \choose 2}={\mu_R(M) \choose 2}$, where $\mu_R(M)$ denotes the minimal number of generators of $M$. But this value can only be $0$ if $\mu_R(M) \le 1$, i.e., if $M$ is cyclic, and we have the forward direction.

For the backwards direction, if $M$ is cyclic, then $\bigwedge^2_R(M)=0$ so $\iota_M$ is trivially $0$.

Answer 2

(2) implies (3) is obvious, and (3) implies (1) follows from the fact that (1) can be tested by something of the form $f=0$ for some map $f: M\otimes M\to M\otimes M$ which can be tested locally on maximal ideals.

So the only interesting bit is (1) implies (2). Clearly $M_\mathfrak p$ satisfies the same condition over $R_\mathfrak p$, so one way to phrase the question would be over a local ring : if $(R,\mathfrak m)$ is local and $M$ f.g. over $R$ satisfying the condition, then is it cyclic ?

Now clearly the property is true over fields, so perhaps one could try to use Nakayama lemma to lift the statement for fields to local rings : let $x_1,...,x_n$ be a generating family for $M$. Then they generate $M/\mathfrak m M$ which is $\leq 1$-dimensional by the property for fields ($R/\mathfrak m$ is a field).

Therefore there is $i$ such that $\overline{x_i}$ generates $M/\mathfrak mM$. But then the Nakayama lemma implies that $x_i$ generates $M$ : $M$ is cyclic.