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''ring with domain of operators  $  \Sigma $''

A [[Ring|ring]] on which a an action ( "multiplication" ) of elements of the ring by elements from a fixed set  $  \Sigma $
is defined (an external law of composition), such that the following axioms are satisfied:

\begin{equation} \label{eq1}
( a + b) \alpha  =  a \alpha + b \alpha ,
\end{equation}

\begin{equation} \label{eq2}
( ab) \alpha  =  ( a \alpha ) b  =  a( b \alpha ),
\end{equation}

where  $  \alpha $
is an element of  $  \Sigma $
while  $  a $, 
$  b $, 
$  a \alpha $, 
$  b \alpha $
are elements of the ring. In this way, the operators act as endomorphisms of the additive group, commuting with multiplication by an element of the ring. A ring with domain of operators  $  \Sigma $, 
or, more succinctly, a  $  \Sigma $-
operator ring, can also be treated as a [[Universal algebra|universal algebra]] with two binary operations (addition and multiplication) and with a set  $  \Sigma $
of unary operations linked by the usual ring identities as well as by the identities \eqref{eq1} and \eqref{eq2}. The concepts of a  $  \Sigma $-
permissible subring, a  $  \Sigma $-
permissible ideal, a  $  \Sigma $-
operator isomorphism, and a  $  \Sigma $-
operator homomorphism can be defined in the same way as for groups with operators (cf. [[Operator group|Operator group]]). If a  $  \Sigma $-
operator ring  $  R $
possesses a unit element, then all ideals and all one-sided ideals of the ring  $  R $
are  $  \Sigma $-
permissible.

A ring  $  R $
is called a ring with a ring of operators  $  \Sigma $
if it is a  $  \Sigma $-
operator ring whose domain of operators  $  \Sigma $
is itself an associative ring, while for any  $  \alpha , \beta \in \Sigma $
and  $  a \in R $
the following equalities hold:

\begin{equation} \label{eq3}
a( \alpha + \beta )  =  a \alpha + a \beta ,
\end{equation}

\begin{equation} \label{eq4}
a( \alpha \beta )  =  ( a \alpha ) \beta .
\end{equation}

A ring with a ring of operators can also be defined as a ring which is simultaneously a  $  \Sigma $-
module and which satisfies axiom \eqref{eq2}. Every ring can naturally be considered as an operator ring over the ring of integers.

For all  $  a $
from  $  R $
and  $  \alpha , \beta $
from  $  \Sigma $, 
the element  $  a( \alpha \beta - \beta \alpha ) $
is an [[Annihilator|annihilator]] of  $  R $. 
Therefore, if  $  R $
is a ring with operators without annihilators, then its ring of operators  $  \Sigma $
must be commutative.

The most commonly studied rings with operators are those with an associative-commutative ring of operators possessing a unit element. This type of ring is usually called an algebra over a commutative ring, and also a linear algebra. The most commonly studied linear algebras are those over fields; the theory of these algebras is evolving in parallel with the general theory of rings (without operators).

====References====
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  A.G. Kurosh,   "Lectures on general algebra" , Chelsea  (1963)  (Translated from Russian)</TD></TR></table>

====Comments====
Thus, the bilinearity properties \eqref{eq1}, \eqref{eq2} and the module properties \eqref{eq3}, \eqref{eq4} are practically incompatible for rings $A$
with a non-commutative ring of operators $R$
in that  $  b \cdot a ( \alpha \beta - \beta \alpha ) = 0 $
for all  $  a, b \in A $, 
$  \alpha , \beta \in R $. 
This explains why algebras are usually only considered over commutative rings. Instead of algebra (over a ring) one also sometimes finds vector algebra. Both this phrase and the phrase linear algebra for an algebra over a ring are nowadays rarely used.

For algebras over non-commutative rings the bilinearity property \eqref{eq2} is weakened to  $  ( ab) \alpha = a( b \alpha ) $. 
Cf. also [[Algebra]] and [[Ring]].