Assertion: ring | /r/HasProperty | commutative
Facets:
value: in the remainder of this article
type: OTHER
Source sentences:
text: Whether a ring is commutative or not (that is, whether the order in which two elements are multiplied changes the result or not) has profound implications on its behavior as an abstract object.
text: then A B = ( 0 0 0 1 ) {\displaystyle AB={\begin{pmatrix}0&0\\0&1\end{pmatrix}}} while B A = ( 1 0 0 0 ) {\displaystyle BA={\begin{pmatrix}1&0\\0&0\end{pmatrix}}} ; this example shows that the ring is noncommutative.
text: A ring is called commutative if its multiplication is commutative.
text: As an example, the nilradical of a ring, the set of all nilpotent elements, need not be an ideal unless the ring is commutative.
text: In the remainder of this article, all rings will be commutative, unless explicitly stated otherwise.
text: Many rings that appear in analysis are noncommutative.
text: When the ring is noncommutative these axioms define left modules; right modules are defined similarly by writing xa instead of ax.