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 $a_1,a_2,\dots ,a_n,b_1,b_2,\dots ,b_n\in A$ whenever $(a_1,b_1)\in E,(a_2,b_2)\in E,\dots ,(a_n,b_n)\in E$ implies $(f(a_1,a_2,\dots ,a_n),f(b_1,b_2,\dots ,b_n))\in E$. 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 $\mathrm{C}\mathrm{o}\mathrm{n}(\mathcal{𝒜})$ be the set of congruences on the algebra $\mathcal{𝒜}$. Because congruences are closed under intersection, we can define a meet operation: $\wedge :\mathrm{C}\mathrm{o}\mathrm{n}(\mathcal{𝒜})\times \mathrm{C}\mathrm{o}\mathrm{n}(\mathcal{𝒜})\to \mathrm{C}\mathrm{o}\mathrm{n}(\mathcal{𝒜})$ 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: $⟨E{⟩}_{\mathcal{𝒜}}=\bigcap \{F\in \mathrm{C}\mathrm{o}\mathrm{n}(\mathcal{𝒜})|E\subseteq F\}$. Note that the (congruence) closure of a binary relation depends on the operations in $\mathcal{𝒜}$, not just on the carrier set. Now define $\vee :\mathrm{C}\mathrm{o}\mathrm{n}(\mathcal{𝒜})\times \mathrm{C}\mathrm{o}\mathrm{n}(\mathcal{𝒜})\to \mathrm{C}\mathrm{o}\mathrm{n}(\mathcal{𝒜})$ as $E_1\vee E_2=⟨E_1\cup E_2{⟩}_{\mathcal{𝒜}}$.

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 $f_i^{\mathcal{𝒜}/E}([a_1{]}_E,\dots ,[a_{n_i}{]}_E)=[f_i^{\mathcal{𝒜}}(a_1,\dots ,a_{n_i}){]}_E$, 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, $\mathrm{k}\mathrm{e}\mathrm{r}h=\{(a,a^{\prime })\in A\times A|h(a)=h(a^{\prime })\}$.

Given an algebra $\mathcal{𝒜}$, a homomorphism h thus defines two algebras homomorphic to $\mathcal{𝒜}$, the image h($\mathcal{𝒜}$) and $\mathcal{𝒜}/\mathrm{k}\mathrm{e}\mathrm{r}h$ 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 $\mathcal{𝒜}/\mathrm{k}\mathrm{e}\mathrm{r}h$ onto $\mathcal{ℬ}$ such that g composed with the natural homomorphism induced by $\mathrm{k}\mathrm{e}\mathrm{r}h$ 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