# algebra without order

An algebra^{} (http://planetmath.org/Algebra) $A$ is said to be \PMlinkescapephraseorder without order if it is commutative^{} and

for each $a\in A$, there exists $b\in A$ such that $ab\ne 0$.

The phrase algebra without order seems first in the book “Multipliers of Banach algebras” by Ronald Larsen. In noncommutative case, the concept is divied into two parts – without left/right order. However, in the noncommutative case, it is defined in terms of the injectivity of the left (right) regular representation given by $x\in A\mapsto {L}_{x}\in L(A)$.

Note that for an algebra $A$ and an element $x\in A$, ${L}_{x}:A\to A$ is the map defined by ${L}_{x}(y)=xy$. Then ${L}_{x}$ is a linear operator on $A$. It is easy to see that $A$ is without left order if and only if the map $x\in A\mapsto {L}_{x}\in L(A)$ is one-one; equivalently, the left ideal^{} $\{x\in A:x\in A\}=\{0\}$. This ideal is is called the left annihilator of $A$.

Every commutative algebra with identity^{} is without order.

Example: ${\mathbb{R}}^{2}$ with multiplication defined by $({x}_{1},{x}_{2})*({y}_{1},{y}_{2})=({x}_{1}{y}_{1},0)$, ($({x}_{1},{x}_{2}),({y}_{1},{y}_{2})\in {\mathbb{R}}^{2}$) is not an algebra without order as multiplication of (0,1) with any other element gives (0,0).

Title | algebra without order |
---|---|

Canonical name | AlgebraWithoutOrder |

Date of creation | 2013-03-22 14:46:01 |

Last modified on | 2013-03-22 14:46:01 |

Owner | mathcam (2727) |

Last modified by | mathcam (2727) |

Numerical id | 11 |

Author | mathcam (2727) |

Entry type | Definition |

Classification | msc 13A99 |