By module over a ring, I mean always a right-module. All rings are supposed to be unital, and the module fulfills $m\cdot 1 = m$. If $R$ is commutative and $M$ a right-module, we can define $rx := xr$ and get also a left-module. In this situation we call $M$ a bimodule, and if not otherwise said this natural definition is applied if I speak of bimodules.

An algebra $A$ over a commutative ring $R$ is itself a ring with $1$, which is also a $R$-module, and such that for all $r \in R$ and $x,y \in A$ we have $$ (rx)y = r(xy) = x(ry). $$ As the ring is assumed to be commutative, we essentially have a bimodul over $R$ as written above.

A module $M$ over an algebra $A$ which is defined over some ring $R$ is itself an $R$-module, for which we have an operation $M \times A \to M$ such that for each $u, v \in M$ and $x,y \in A$ and $r \in R$ we have

(1) $(u+v)x = ux + vx$

(2) $v(x+y) = vx + vy$

(3) $(vx)y = v(xy)$

(4) $v1 = v$

(5) $(rv)x = r(vx) = v(rx)$.

This is the standard definition I see everywhere. But I guess I observed the following. For an algebra we can embed $R$ into $A$ by identifying it with $R\cdot 1$ for $1 \in A$. So the assertion that $A$ must be an $R$-module is implied by (1), (2), (3) and (4). So we can just define a module over an algebra as an ordinary module over $A$ seen as a ring with $1$. As $A$ is in general not commutative, this is just a right-module. But by the embedding, and the additional requirement for an algebra, every element of $R$ is in the center $Z(A) = \{ x : xy = yx \mbox{ for all y} \in A \}$, restricted to $R$ everything is fine and we again have a bimodule. So the only thing that makes modules over algebras special here is (5). But here we have \begin{align*} (rv)x & = (vr)x & \mbox{definition of $M$ as bimodul over $R$} \\ & = v(rx) & \mbox{by (3)} \\ & = v(xr) & \mbox{$R$ is central in $A$} \\ & = (vx)r & \mbox{by (3)} \\ & = r(vx) & \mbox{definition of $M$ as bimodul over $R$} \end{align*} so we see that in the third and last line we have recovered everything from (5).

So as I see it, a module over an algebra is just an ordinary module over the algebra seen as a ring, so why bother with these extra definition? I guess if we generalise further, for example look at non-associative algebras, then they are no longer rings and we cannot define modules over them as special cases of modules over rings. But most of the time (and in the textbooks I am reading right now) such generalizations are not considered, but nevertheless most of the time modules over algebras are defined separately to modules over rings.

So why that? Or have I overlooked something and my computations are wrong?

Remark: These modules over algebras come from the representation theory of (finite) groups.

  • always right module? That is unusual =O – Zelos Malum Jan 6 '16 at 15:18
  • How you mean that, I just said this to not have to write right-module everywhere, but can instead just write module. Does this contradicts any conventions? Then I can change it. – StefanH Jan 6 '16 at 15:19
  • Yes but most times people focus on left modules as things are virtually the same, to focus it on right modules is just unusual to see. – Zelos Malum Jan 6 '16 at 15:20
  • Ah okay, maybe because I come from a group theoretical background, and I am not that much involved into commutative algebra. In the part of group theory I am interested in most of the time groups are acting on the right, and also functions are written "from the right", for example $x^g$, the conjugation action is taken as an action from the right. Maybe because of that I am biased towards right-actions. EDIT: I looked it up in my group theory texts with representation theory part like Huppert Endliche Gruppen and Isaacs Character Theory of Finite groups they always consider right-modules. – StefanH Jan 6 '16 at 15:22
  • 1
    Right modules are not so unusual. While Lang's Algebra and Auslander, Reiten and Smalø's Representation Theory of Artin Algebras deal mostly with left modules, Jacobson's Basic Algebra II and Assem, Simson and Skowroński's Elements of the Representation Theory of Associative Algebras mostly deal with right modules. – Pierre-Guy Plamondon Jan 6 '16 at 15:58
up vote 2 down vote accepted

You need condition (5) because, in your definition of $A$-module, you asked for $M$ to be an $R$-module right at the start. Condition (5) is there to ensure that the structure of $A$-module which you define on $M$ is compatible with the already existing $R$-module structure.

If, in your definition, you had simply asked for $M$ to be an abelian group, then condition (5) would be redundant, for the reasons you noted. This is, for instance, what is done in Bourbaki's Algebra, chapter 3: if $A$ is an $R$-algebra, they define an $A$-module to be a module over the underlying (not necessarily commutative) ring of $A$.

  • But in this case, since condition (5) is needed only to ensurethat the induced structure of $R$-module and the already present structure of $R$-module on $M$ do coincide, couldn't one simply define a module over the algebra $A$ to be simply an $A$-module: since every $A$-module has naturally the induced $R$-module structure? – Giorgio Mossa Jan 6 '16 at 15:53
  • 1
    @GiorgioMossa Indeed, and it is done that way in some textbooks. In that case, the only starting assumption on $M$ is that it is an abelian group. – Pierre-Guy Plamondon Jan 6 '16 at 15:55
  • Okay, thanks, could you please give some references where it is done that way? I am curious as my books do not do it that way. – StefanH Jan 6 '16 at 15:57
  • @Stefan One in French is Assem's Algèbres et modules; I'm looking for one in English. – Pierre-Guy Plamondon Jan 6 '16 at 16:01
  • Would be nice if you can give some! Unfortunately I cannot speak French :( – StefanH Jan 6 '16 at 16:16

Indeed your proof seems to work fine assuming in the definition you require $M$ to be a right $R$-module. In that case one could simplify the definition by requiring that a module of an algebra $A$ is just an $A$-module (the $R$-module structure can be derived in the way you showed).

On the other hand if we reguard $M$ as a left $R$-module things change. If $M$ is just a left $R$-module the condition $(5)$ is required in order to ensure that the left action of $R$ and right action of $A$ (and so also the induced right action of $R$) are compatible: that is that $M$ is a $(R,A)$-module.

Note that in general even for $(R,R)$-bimodules (i.e. left and right $R$-modules where the two actions are compatible) it is not required that the left and right actions coincide.

Hope this helps.

  • $R$ is required to be a commutative ring, (for this is required in the definition of an algebra, but do not know if it could be dropped), so we have an induced right-action of $R$ on $M$ if $M$ is an left $R$-module. Yes, but in general for the left and right action we do not need to have $rv = vr$, so this induced action might be different from the given one (i.e. the one induced by $A$ acting on the right of $M$). So (5) guarantees them to be compatible that way, i.e. we have the relation $rv = vr$ between the left- and right action. Guess I got it now. – StefanH Jan 6 '16 at 16:11

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.