PlanetMath (more info)
 Math for the people, by the people.
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
sections
Encyclopædia
Papers
Books
Expositions

meta
Requests (232)
Orphanage
Unclass'd
Unproven (519)
Corrections (22)
Classification

talkback
Polls
Forums
Feedback
Bug Reports

downloads
Snapshots
PM Book

information
News
Docs
Wiki
ChangeLog
TODO List
Legalese
About
Owner confidence rating: Very high Entry average rating: No information on entry rating
[parent] kernel of a homomorphism between algebraic systems (Definition)

Let $ f:(A,O)\to (B,O)$ be a homomorphism between two algebraic systems $ A$ and $ B$ (with $ O$ as the operator set). Each element $ b\in B$ corresponds to a subset $ K(b):=f^{-1}(b)$ in $ A$. Then $ \lbrace K(b)\mid b\in B\rbrace$ forms a partition of $ A$. The kernel $ \ker(f)$ of $ f$ is defined to be

$\displaystyle \ker(f):=\bigcup_{b\in B}K(b)\times K(b).$

It is easy to see that $ \ker(f)=\lbrace (x,y)\in A\times A\mid f(x)=f(y)\rbrace$. Since it is a subset of $ A\times A$, it is relation on $ A$. Furthermore, it is an equivalence relation on $ A$: 1

  1. $ \ker(f)$ is reflexive: for any $ a\in A$, $ a\in K(f(a))$, so that $ (a,a)\in K(f(a))^2\subseteq \ker(f)$
  2. $ \ker(f)$ is symmetric: if $ (a_1,a_2)\in \ker(f)$, then $ f(a_1)=f(a_2)$, so that $ (a_2,a_1)\in \ker(f)$
  3. $ \ker(f)$ is transitive: if $ (a_1,a_2),(a_2,a_3)\in \ker(f)$, then $ f(a_1)=f(a_2)=f(a_3)$, so $ (a_1,a_3)\in \ker(f)$.

We write $ a_1 \equiv a_2 \pmod {\ker(f)}$ to denote $ (a_1,a_2)\in \ker(f)$.

In fact, $ \ker(f)$ is a congruence relation: for any $ n$-ary operator symbol $ \omega\in O$, suppose $ c_1,\ldots,c_n$ and $ d_1,\ldots,d_n$ are two sets of elements in $ A$ with $ c_i\equiv d_i \mod \ker(f)$. Then

$\displaystyle f(\omega_A(c_1,\ldots,c_n) = \omega_B(f(c_1),\ldots,f(c_n))=\omega_B(f(d_1),\ldots,f(d_n)) = f(\omega_A(d_1,\ldots,d_n)),$
so $ \omega_A(c_1,\ldots,c_n)\equiv \omega_A(d_1,\ldots,d_n) \pmod {\ker(f)}$. For this reason, $ \ker(f)$ is also called the congruence induced by $ f$.

Example. If $ A,B$ are groups and $ f:A\to B$ is a group homomorphism. Then the kernel of $ f$, using the definition above is just the union of the square of the cosets of

$\displaystyle N=\lbrace x\mid f(x)=e\rbrace,$
the traditional definition of the kernel of a group homomorphism (where $ e$ is the identity of $ B$).

Remark. The above can be generalized. See the analog in model theory.


Footnotes

...1
In general, $ \lbrace N_i\rbrace$ is a partition of a set $ A$ iff $ \bigcup N_i^2$ is an equivalence relation on $ A$.


"kernel of a homomorphism between algebraic systems" is owned by CWoo.
(view preamble | get metadata)

View style:

See Also: kernel of a homomorphism is a congruence, kernel pair

Other names:  induced congruence
Also defines:  congruence induced by a homomorphism

This object's parent.
Log in to rate this entry.
(view current ratings)

Cross-references: model theory, identity, kernel of a group homomorphism, cosets, square, union, group homomorphism, groups, induced, congruence, operator symbol, congruence relation, transitive, symmetric, Reflexive, iff, equivalence relation, relation, easy to see, kernel, partition, subset, operator set, algebraic systems, homomorphism
There are 2 references to this entry.

This is version 8 of kernel of a homomorphism between algebraic systems, born on 2006-11-29, modified 2007-06-30.
Object id is 8593, canonical name is KernelOfAHomomorphismBetweenAlgebraicSystems.
Accessed 1470 times total.

Classification:
AMS MSC: 08A05 (General algebraic systems :: Algebraic structures :: Structure theory)
Pending Errata and Addenda
None.
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | add derivation | add example | add (any)