You are here
Homecongruence relation on an algebraic system
Primary tabs
congruence relation on an algebraic system
Let $(A,O)$ be an algebraic system. A congruence relation, or simply a congruence $\mathfrak{C}$ on $A$
1. is an equivalence relation on $A$; if $(a,b)\in\mathfrak{C}$ we write $a\equiv b\;\;(\mathop{{\rm mod}}\mathfrak{C})$, and
2. respects every $n$ary operator on $A$: if $\omega_{A}$ is an $n$ary operator on $A$ ($\omega\in O$), and for any $a_{i},b_{i}\in A$, $i=1,\ldots,n$, we have
$a_{i}\equiv b_{i}\;\;(\mathop{{\rm mod}}\mathfrak{C})\qquad\mbox{ implies }% \qquad\omega_{A}(a_{1},\ldots,a_{n})\equiv\omega_{A}(b_{1},\ldots,b_{n})\;\;(% \mathop{{\rm mod}}\mathfrak{C}).$
For example, $A^{2}$ and $\Delta_{A}:=\{(a,a)\mid a\in A\}$ are both congruence relations on $A$. $\Delta_{A}$ is called the trivial congruence (on $A$). A proper congruence relation is one not equal to $A^{2}$.
Remarks.

$\mathfrak{C}$ is a congruence relation on $A$ if and only if $\mathfrak{C}$ is an equivalence relation on $A$ and a subalgebra of the product $A\times A$.

The set of congruences of an algebraic system is a complete lattice. The meet is the usual set intersection. The join (of an arbitrary number of congruences) is the join of the underlying equivalence relations. This join corresponds to the subalgebra (of $A\times A$) generated by the union of the underlying sets of the congruences. The lattice of congruences on $A$ is denoted by $\operatorname{Con}(A)$.

(restriction) If $\mathfrak{C}$ is a congruence on $A$ and $B$ is a subalgebra of $A$, then $\mathfrak{C}_{B}$ defined by $\mathfrak{C}\cap(B\times B)$ is a congruence on $B$. The equivalence of $\mathfrak{C}_{B}$ is obvious. For any $n$ary operator $\omega_{B}$ inherited from $A$’s $\omega_{A}$, if $a_{i}\equiv b_{i}\;\;(\mathop{{\rm mod}}\mathfrak{C}_{B})$, then $\omega_{B}(a_{1},\ldots,a_{n})=\omega_{A}(a_{1},\ldots,a_{n})\equiv\omega_{A}(% b_{1},\ldots,b_{n})=\omega_{B}(b_{1},\ldots,b_{n})\;\;(\mathop{{\rm mod}}% \mathfrak{C})$. Since both $\omega_{B}(a_{1},\ldots,a_{n})$ and $\omega_{B}(b_{1},\ldots,b_{n})$ are in $B$, $\omega_{B}(a_{1},\ldots,a_{n})\equiv\omega_{B}(b_{1},\ldots,b_{n})\;\;(\mathop% {{\rm mod}}\mathfrak{C}_{B})$ as well. $\mathfrak{C}_{B}$ is the congruence restricted to $B$.

(extension) Again, let $\mathfrak{C}$ be a congruence on $A$ and $B$ a subalgebra of $A$. Define $B^{{\mathfrak{C}}}$ by $\{a\in A\mid(a,b)\in\mathfrak{C}\mbox{ and }b\in B\}$. In other words, $a\in B^{{\mathfrak{C}}}$ iff $a\equiv b\;\;(\mathop{{\rm mod}}\mathfrak{C})$ for some $b\in B$. We assert that $B^{{\mathfrak{C}}}$ is a subalgebra of $A$. If $\omega_{A}$ is an $n$ary operator on $A$ and $a_{1},\ldots,a_{n}\in B^{{\mathfrak{C}}}$, then $a_{i}\equiv b_{i}\;\;(\mathop{{\rm mod}}\mathfrak{C})$, so $\omega_{A}(a_{1},\ldots,a_{n})\equiv\omega_{A}(b_{1},\ldots,b_{n})\;\;(\mathop% {{\rm mod}}\mathfrak{C})$. Since $\omega_{A}(b_{1},\ldots,b_{n})\in B$, $\omega_{A}(a_{1},\ldots,a_{n})\in B^{{\mathfrak{C}}}$. Therefore, $B^{{\mathfrak{C}}}$ is a subalgebra. Because $B\subseteq B^{{\mathfrak{C}}}$, we call it the extension of $B$ by $\mathfrak{C}$.

Let $B$ be a subset of $A\times A$. The smallest congruence $\mathfrak{C}$ on $A$ such that $a\equiv b\;\;(\mathop{{\rm mod}}\mathfrak{C})$ for all $a,b\in B$ is called the congruence generated by $B$. $\mathfrak{C}$ is often written $\langle B\rangle$. When $B$ is a singleton $\{(a,b)\}$, then we call $\langle B\rangle$ a principal congruence, and denote it by $\langle(a,b)\rangle$.
Quotient algebra
Given an algebraic structure $(A,O)$ and a congruence relation $\mathfrak{C}$ on $A$, we can construct a new $O$algebra $(A/\mathfrak{C},O)$, as follows: elements of $A/\mathfrak{C}$ are of the form $[a]$, where $a\in A$. We set
$[a]=[b]\mbox{ iff }a\equiv b\;\;(\mathop{{\rm mod}}\mathfrak{C}).$ 
Furthermore, for each $n$ary operator $\omega_{A}$ on $A$, define $\omega_{{A/\mathfrak{C}}}$ by
$\omega_{{A/\mathfrak{C}}}\big([a_{1}],\ldots,[a_{n}]\big):=\big[\omega_{A}(a_{% 1},\ldots,a_{n})\big].$ 
It is easy to see that $\omega_{{A/\mathfrak{C}}}$ is a welldefined operator on $A/\mathfrak{C}$. The $O$algebra thus constructed is called the quotient algebra of $A$ over $\mathfrak{C}$.
Remark. The bracket $[\cdot]:A\to A/\mathfrak{C}$ is in fact an epimorphism, with kernel $\ker([\cdot])=\mathfrak{C}$. This means that every congruence of an algebraic system $A$ is the kernel of some homomorphism from $A$. $[\cdot]$ is usually written $[\cdot]_{{\mathfrak{C}}}$ to signify its association with $\mathfrak{C}$.
References
 1 G. Grätzer: Universal Algebra, 2nd Edition, Springer, New York (1978).
Mathematics Subject Classification
08A30 no label found Forums
 Planetary Bugs
 HS/Secondary
 University/Tertiary
 Graduate/Advanced
 Industry/Practice
 Research Topics
 LaTeX help
 Math Comptetitions
 Math History
 Math Humor
 PlanetMath Comments
 PlanetMath System Updates and News
 PlanetMath help
 PlanetMath.ORG
 Strategic Communications Development
 The Math Pub
 Testing messages (ignore)
 Other useful stuff
Recent Activity
new correction: Error in proof of Proposition 2 by alex2907
Jun 24
new question: A good question by Ron Castillo
Jun 23
new question: A trascendental number. by Ron Castillo
Jun 19
new question: Banach lattice valued Bochner integrals by math ias
Jun 13
new question: young tableau and young projectors by zmth
Jun 11
new question: binomial coefficients: is this a known relation? by pfb
Jun 6
new question: difference of a function and a finite sum by pfb