# Ring with operators

*ring with domain of operators $ \Sigma $*

A 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:

$$ \tag{1 } ( a + b) \alpha = a \alpha + b \alpha , $$

$$ \tag{2 } ( ab) \alpha = ( a \alpha ) b = a( b \alpha ), $$

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 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 (1) and (2). 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). 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:

$$ \tag{3 } a( \alpha + \beta ) = a \alpha + a \beta , $$

$$ \tag{4 } a( \alpha \beta ) = ( a \alpha ) \beta . $$

A ring with a ring of operators can also be defined as a ring which is simultaneously a $ \Sigma $- module and which satisfies axiom (2). 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 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

[1] | A.G. Kurosh, "Lectures on general algebra" , Chelsea (1963) (Translated from Russian) |

#### Comments

Thus, the bilinearity properties (1), (2) and the module properties (3), (4) 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 (2) is weakened to $ ( ab) \alpha = a( b \alpha ) $. Cf. also Algebra and Ring.

**How to Cite This Entry:**

Ring with operators.

*Encyclopedia of Mathematics.*URL: http://encyclopediaofmath.org/index.php?title=Ring_with_operators&oldid=48574