modules over algebars and homomorphisms between them

Let $R$ be a ring and let $A$ be an associative algebra (not necessarily unital).

Definition. A (left) $A$-module over $R$ is a pair $(M,\circ )$ where $M$ is a (left) $R$-module and

$$\circ :A\times M\to M$$

is a $R$-bilinear map such that the following conditions hold:

1. $(a\circ b)\circ x=a\circ (b\circ x)$

2. $r(a\circ x)=(ra)\circ x=a\circ (rx)$

for any $a,b\in A$, $x\in M$ and $r\in R$. We will simply use capital letters to denote modules.

Let $M$ be an $A$-module over $R$. If $M^{\prime }\subseteq M$ and $A^{\prime }\subseteq A$ then by $A^{\prime }{M}^{\prime }$ we denote $R$-submodule of $M$ generated by elements of the form $am$ for $a\in A^{\prime }$ and $m\in M^{\prime }$. We will call $M$ unitary if $AM=M$. Note, that if $A$ has multiplicative identity $1$, then $M$ is unitary if and only if $1m=m$ for any $m\in M$.

The reason we use name ,,$A$-module over $R$” instead of ,,$A$-module” is that these to concepts may differ. The latter means that we treat $A$ simply as a ring and take modules over it. But such module need not be equiped with a ,,good” $R$-module structure. On the other hand this is always the case, when $M$ is unitary over unital algebra.

If $M$ and $N$ are two $A$-modules over $R$, then a function $f:M\to N$ is called an $A$-homomorphism iff $f$ is an $R$-homomorphism and additionaly $f(am)=af(m)$ for any $a\in A$ and $m\in M$.

It can be easily checked that $A$-modules over $R$ together with $A$-homomorphisms form a category which is abelian. Furthermore, if $A$ is unital, then its full subcategory consisting unitary $R$-modules over $A$ is equivalent to category of unitary $A$-modules.

In most cases it is important to assume that the base ring $R$ is a field, even algebraically closed.