Modules over an algebra: confusion in understanding "left" and "right" business I came up with following definition of algebra $A$ over a (non necessarily commutative) ring $R$ with $1$ and module over $A$.

  
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*Let $R$ be a commutative ring with $1$, and $A$ be a left $R$-module. If $A$ is also a ring such that 
  $$(ra)b=r(ab)=a(rb)$$
  for $r\in R$ and $a,b\in A$ then we call $A$ an $R$-algebra.
  
*A left $R$-module $V$ is said to be an $A$-module if $V$ is a right $A$-module such that 
  $$(rv)a=r(va)=v(ra)$$
  for $r\in R$, $v\in V$ and $a\in A$.

My confusion is following: 
While defining an algebra over $R$, it was considered as left $R$ module. Then while defining module over algebra $A$, it was taken to be right $A$-module (and $A$ is left $R$-module); why this mix-up of left-right is done? I don't understand its necessity.
It may be possible to interpret this symbolic definitions (I mean equations) in terms of words which can make the definitions clear, but I was unable to write them. Can one clarify a little these definitions in some different words?
 A: You're right: there is no real significance to specifying the side of an $R$ module when $R$ is commutative. Also, any left $A$ module is automatically a left/right module over $R$ too, so that you have another category of modules.
There are, however, cases where this is an important convention. Given any ring $R$ and left $R$ module $V$, let $E$ be the ring of $R$ linear endomorphisms of $V$. Then $V$ has a natural right $E$ module structure given by $ve:=e(v)$. Moreover, it is a bimodule structure: $r(ve)=(rv)e$ for all choices of $r\in R$, $v\in V$, $e\in E$. 
Here it is impossible to drop the left/right specification. They have to live on opposite sides of the module so that they can commute with each other. However, you can do the same thing with a right $R$ module and the $R$ linear transformations written on the left, too.
This is done, for example, when proving the classical   double commutator theorem. You can't call this $R$ algebras though: the axioms require $R$ to be commutative for that.
