Lie algebroid

In mathematics, a quotient algebra, (where algebra is used in the sense of universal algebra), also called a factor algebra, is obtained by partitioning the elements of an algebra into equivalence classes given by a congruence relation, that is an equivalence relation that is additionally compatible with all the operations of the algebra, in the formal sense described below.

Compatible relation

Let A be a set (of the elements of an algebra ${\mathcal {A}}$), and let E be an equivalence relation on the set A. The relation E is said to be compatible with (or have the substitution property with respect to) an n-ary operation f if for all $a_{1},a_{2},\ldots ,a_{n},b_{1},b_{2},\ldots ,b_{n}\in A$ whenever $(a_{1},b_{1})\in E,(a_{2},b_{2})\in E,\ldots ,(a_{n},b_{n})\in E$ implies $(f(a_{1},a_{2},\ldots ,a_{n}),f(b_{1},b_{2},\ldots ,b_{n}))\in E$. An equivalence relation compatible with all the operations of an algebra is called a congruence.

Congruence lattice

For every algebra ${\mathcal {A}}$ on the set A, the identity relation on A, and $A\times A$ are trivial congruences. An algebra with no other congruences is called simple.

Let $\mathrm {Con} ({\mathcal {A}})$ be the set of congruences on the algebra ${\mathcal {A}}$. Because congruences are closed under intersection, we can define a meet operation: $\wedge :\mathrm {Con} ({\mathcal {A}})\times \mathrm {Con} ({\mathcal {A}})\to \mathrm {Con} ({\mathcal {A}})$ by simply taking the intersection of the congruences $E_{1}\wedge E_{2}=E_{1}\cap E_{2}$.

On the other hand, congruences are not closed under union. However, we can define the closure of any binary relation E, with respect to a fixed algebra ${\mathcal {A}}$, such that it is a congruence, in the following way: $\langle E\rangle _{\mathcal {A}}=\bigcap \{F\in \mathrm {Con} ({\mathcal {A}})|E\subseteq F\}$. Note that the (congruence) closure of a binary relation depends on the operations in ${\mathcal {A}}$, not just on the carrier set. Now define $\vee :\mathrm {Con} ({\mathcal {A}})\times \mathrm {Con} ({\mathcal {A}})\to \mathrm {Con} ({\mathcal {A}})$ as $E_{1}\vee E_{2}=\langle E_{1}\cup E_{2}\rangle _{\mathcal {A}}$.

For every algebra ${\mathcal {A}}$, $({\mathcal {A}},\wedge ,\vee )$ with the two operations defined above forms a lattice, called the congruence lattice of ${\mathcal {A}}$.

Quotient algebras and homomorphisms

A set A can be partitioned in equivalence classes given by an equivalence relation E, and usually called a quotient set, and denoted A/E. For an algebra ${\mathcal {A}}$, it is straightforward to define the operations induced on A/E if E is a congruence. Specifically, for any operation $f_{i}^{\mathcal {A}}$ of arity $n_{i}$ in ${\mathcal {A}}$ (where the superscript simply denotes that it's an operation in ${\mathcal {A}}$) define $f_{i}^{{\mathcal {A}}/E}:(A/E)^{n_{i}}\to A/E$ as $f_{i}^{{\mathcal {A}}/E}([a_{1}]_{E},\ldots ,[a_{n_{i}}]_{E})=[f_{i}^{\mathcal {A}}(a_{1},\ldots ,a_{n_{i}})]_{E}$, where $[a]_{E}$ denotes the equivalence class of a modulo E.

For an algebra ${\mathcal {A}}=(A,(f_{i}^{\mathcal {A}})_{i\in I})$, given a congruence E on ${\mathcal {A}}$, the algebra ${\mathcal {A}}/E=(A/E,(f_{i}^{{\mathcal {A}}/E})_{i\in I})$ is called the quotient algebra (or factor algebra) of ${\mathcal {A}}$ modulo E. There is a natural homomorphism from ${\mathcal {A}}$ to ${\mathcal {A}}/E$ mapping every element to its equivalence class. In fact, every homomorphism h determines a congruence relation; the kernel of the homomorphism, $\mathop {\mathrm {ker} } \,h=\{(a,a')\in A\times A|h(a)=h(a')\}$.

Given an algebra ${\mathcal {A}}$, a homomorphism h thus defines two algebras homomorphic to ${\mathcal {A}}$, the image h(${\mathcal {A}}$) and ${\mathcal {A}}/\mathop {\mathrm {ker} } \,h$ The two are isomorphic, a result known as the homomorphic image theorem. Formally, let $h:{\mathcal {A}}\to {\mathcal {B}}$ be a surjective homomorphism. Then, there exists a unique isomorphism g from ${\mathcal {A}}/\mathop {\mathrm {ker} } \,h$ onto ${\mathcal {B}}$ such that g composed with the natural homomorphism induced by $\mathop {\mathrm {ker} } \,h$ equals h.

See also

• quotient ring

References

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• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
  My blog: http://www.primaboinca.com/view_profile.php?userid=5889534