Modules which are both left and right modules over a noncommutative ring

Question

Given a noncommutative ring $R$, is there a name for those (left) $R$-modules $M$ for which $$ m.(rs) = m.(sr), ~~~ \textrm{ for all } r,s \in R, \textrm{ and } m \in M? $$ (Note that we have denoted the action of $R$ on $M$ by a dot.) In other words, what does one call modules which are both left and right modules.

Answer 1

I don't think there's a special word for an abelian group which is both a left module and a right module. Since the two actions are "unrelated" we don't think of them as functioning as a single structure.

But there is a special word for an abelian group which is a left and right $R$ module such that $r(ms)=(rm)s$ for all $r,s\in R$, and $m\in M$: this is called an $R, R$ bimodule.

You don't have to have the same ring on both sides: it could be two different rings, say $S$ on the right. And also, you can consider an $R,S$ bimodule structure as simply a left module structure over the ring $R\otimes_\mathbb Z S^{op}$.

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No, I think we're talking about different things. I am asking for a (let's say) left $$-module such that $()=()$, for all $∈$, and $,∈$

Well, that puts extra conditions on the right hand module action. I would call that "A right $R$ module whose annihilator contains the commutator ideal of $R$," because by what you wrote, $m(rs-sr)=0$ for all $r,s\in R$, $m\in M$.

Again, without some sort of relationship to the left hand action, I'm not aware of a word that encompasses both.

Answer 2

If $k \rightarrow A$ is a commutative $k$-algebra with $k$ a commutative unital ring, you may consider the tensor product $A(d):=A^{\otimes_k d}$ of $A$ with itself over $k$ d times. There are canonical maps

$$p_i: A \rightarrow A(d)$$

defined by

$$p_i(a):=1\otimes \cdots 1 \otimes a \otimes 1 \otimes \cdots \otimes 1$$

where the $a$ is "placed in position $i$". Here $i=1,..,d$. With this definition you get $d$ commuting $A$-module structures on $A(d)$: You define for any $a\in A$ the $i$'th left multiplication on $A(d)$ as follows. Let $b:=b_1\otimes \cdots \otimes b_d \in A(d)$. define

$$ a^ib:=b_1 \otimes \cdots \otimes ab_i \otimes \cdots \otimes b_d:=p_i(a)b.$$

You get for any elements $u,v\in A$

$$ u^i(v^jb):=p_i(u)(p_j(v)b)=p_j(v)(p_i(u)b)=v^j(u^ib)$$

hence these $d$ module structures commute. Hence you do not need $A$ to be non-commutative. I do not know if there is a word reserved for this structure. Maybe Bourbaki's "Commutative algebra" book can give some answers. They study bimodules and more generally "multi modules" and I believe the above construction is referred to as a multi module.

Answer 3

Basically, your condition amounts to the fact that the action of $R$ on $M$ factors through the "abelianization" of $R$ (not sure if this is standard terminology for rings).

Specifically, for any ring $R$, you can form the ideal $I$ generated by all elements of the form $xy-yx$, and $R^{ab} = R/I$ is then the minimal commutative quotient of $R$.

Then a left $R$-module $M$ is also a right module iff the action of $R$ factors through $R^{ab}$. So studying "left-and-right" modules over $R$ is the same thing as studying modules over the commutative ring $R^{ab}$.