Definition:Homomorphism (Abstract Algebra)

From ProofWiki
Jump to: navigation, search

Definition

Let $\struct {S, \circ}$ and $\struct {T, *}$ be magmas.

Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a mapping from $\struct {S, \circ}$ to $\struct {T, *}$.

Let $\circ$ have the morphism property under $\phi$, that is:

$\forall x, y \in S: \map \phi {x \circ y} = \map \phi x * \map \phi y$


Then $\phi$ is a homomorphism.


This can be generalised to magmas with more than one operation:

Let:

$\struct {S_1, \circ_1, \circ_2, \ldots, \circ_n}$
$\struct {T, *_1, *_2, \ldots, *_n}$

be magmas.


Let $\phi: \struct {S_1, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *_2, \ldots, *_n}$ be a mapping from $\struct {S_1, \circ_1, \circ_2, \ldots, \circ_n}$ to $\struct {T, *_1, *_2, \ldots, *_n}$.

Let, $\forall k \in \closedint 1 n$, $\circ_k$ have the morphism property under $\phi$, that is:

$\forall x, y \in S: \map \phi {x \circ_k y} = \map \phi x *_k \map \phi y$


Then $\phi$ is a homomorphism.


Semigroup Homomorphism

Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be semigroups.

Let $\phi: S \to T$ be a mapping such that $\circ$ has the morphism property under $\phi$.


That is, $\forall a, b \in S$:

$\phi \left({a \circ b}\right) = \phi \left({a}\right) * \phi \left({b}\right)$


Then $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ is a semigroup homomorphism.


Monoid Homomorphism

Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be monoids.

Let $\phi: S \to T$ be a mapping such that $\circ$ has the morphism property under $\phi$.

That is, $\forall a, b \in S$:

$\phi \left({a \circ b}\right) = \phi \left({a}\right) * \phi \left({b}\right)$

Suppose further that $\phi$ preserves identities, i.e.:

$\phi \left({e_S}\right) = e_T$


Then $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ is a monoid homomorphism.


Group Homomorphism

Let $\struct {G, \circ}$ and $\struct{H, *}$ be groups.

Let $\phi: G \to H$ be a mapping such that $\circ$ has the morphism property under $\phi$.


That is, $\forall a, b \in G$:

$\map \phi {a \circ b} = \map \phi a * \map \phi b$


Then $\phi: \struct {G, \circ} \to \struct {H, *}$ is a group homomorphism.


Ring Homomorphism

Let $\struct {R, +, \circ}$ and $\struct{S, \oplus, *}$ be rings.

Let $\phi: R \to S$ be a mapping such that both $+$ and $\circ$ have the morphism property under $\phi$.


That is, $\forall a, b \in R$:

\((1):\quad\) \(\displaystyle \map \phi {a + b}\) \(=\) \(\displaystyle \map \phi a \oplus \map \phi b\) $\quad$ $\quad$
\((2):\quad\) \(\displaystyle \map \phi {a \circ b}\) \(=\) \(\displaystyle \map \phi a * \map \phi b\) $\quad$ $\quad$


Then $\phi: \struct {R, +, \circ} \to \struct {S, \oplus, *}$ is a ring homomorphism.


Field Homomorphism

Let $\struct {F, +, \times}$ and $\struct {K, \oplus, \otimes}$ be fields.

Let $\phi: F \to K$ be a mapping such that both $+$ and $\times$ have the morphism property under $\phi$.


That is, $\forall a, b \in F$:

\((1):\quad\) \(\displaystyle \map \phi {a + b}\) \(=\) \(\displaystyle \map \phi a \oplus \map \phi b\) $\quad$ $\quad$
\((2):\quad\) \(\displaystyle \map \phi {a \times b}\) \(=\) \(\displaystyle \map \phi a \otimes \map \phi b\) $\quad$ $\quad$


Then $\phi: \struct {F, +, \times} \to \struct {K, \oplus, \otimes}$ is a field homomorphism.


F-Homomorphism

Let $R, S$ be rings with unity.

Let $F$ be a subfield of both $R$ and $S$.


Then a ring homomorphism $\varphi: R \to S$ is called an $F$-homomorphism if:

$\forall a \in F: \map \phi a = a$


That is, $\phi \restriction_F = I_F$ where:

$\phi \restriction_F$ is the restriction of $\phi$ to $F$
$I_F$ is the identity mapping on $F$.


R-Algebraic Structure Homomorphism

Let $R$ be a ring.

Let $\left({S, \ast_1, \ast_2, \ldots, \ast_n, \circ}\right)_R$ and $\left({T, \odot_1, \odot_2, \ldots, \odot_n, \otimes}\right)_R$ be $R$-algebraic structures.

Let $\phi: S \to T$ be a mapping.


Then $\phi$ is an $R$-algebraic structure homomorphism if and only if:

$(1): \quad \forall k \in \left[{1 \,.\,.\, n}\right]: \forall x, y \in S: \phi \left({x \ast_k y}\right) = \phi \left({x}\right) \odot_k \phi \left({y}\right)$
$(2): \quad \forall x \in S: \forall \lambda \in R: \phi \left({\lambda \circ x}\right) = \lambda \otimes \phi \left({x}\right)$

where $\left[{1 \,.\,.\, n}\right] = \left\{{1, 2, \ldots, n}\right\}$ denotes an integer interval.


Note that this definition also applies to modules and vector spaces.


G-Module Homomorphism

Let $\left({G, \cdot}\right)$ be a group.

Let $\left({V, \phi}\right)$ and $\left({W, \mu}\right)$ be $G$-modules.


Then a linear mapping $f: V \to W$ is called a $G$-module homomorphism if and only if:

$\forall g \in G: \forall v \in V: f \left({\phi \left({g, v}\right)}\right) = \mu \left({g, f \left({v}\right)}\right)$


Homomorphism of Complexes

Let $\left({R, +, \cdot}\right)$ be a ring.

Let:

$M: \quad \cdots \longrightarrow M_i \stackrel {d_i} {\longrightarrow} M_{i + 1} \stackrel {d_{i + 1} } {\longrightarrow} M_{i + 2} \stackrel {d_{i + 2} } {\longrightarrow} \cdots$

and

$N: \quad \cdots \longrightarrow N_i \stackrel {d'_i} {\longrightarrow} N_{i + 1} \stackrel {d'_{i + 1} } {\longrightarrow} N_{i + 2} \stackrel {d'_{i + 2} } {\longrightarrow} \cdots$

be two differential complexes of $R$-modules.

Let $\phi = \left\{ {\phi_i: i \in \Z}\right\}$ be a family of module homomorphisms $\phi_i: M_i \to N_i$.


Then $\phi$ is a homomorphism of complexes if and only if for each $i \in \Z$:

$\phi_{i+1} \circ d_i = \phi_i \circ d'_i$


Homomorphic Image

As a homomorphism is a mapping, the homomorphic image of $\phi$ is defined in the same way as the image of a mapping:

$\Img \phi = \set {t \in T: \exists s \in S: t = \map \phi s}$


Homomorphism as Cartesian Product

Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be a mapping from one algebraic structure $\left({S, \circ}\right)$ to another $\left({T, *}\right)$.

We define the cartesian product $\phi \times \phi: S \times S \to T \times T$ as:

$\forall \left({x, y}\right) \in S \times S: \left({\phi \times \phi}\right) \left({x, y}\right) = \left({\phi \left({x}\right), \phi \left({y}\right)}\right)$


Hence we can state that $\phi$ is a homomorphism iff:

$\ast \left({\left({\phi \times \phi}\right) \left({x, y}\right)}\right) = \phi \left({\circ \left({x, y}\right)}\right)$

using the notation $\circ \left({x, y}\right)$ to denote the operation $x \circ y$.


The point of doing this is so we can illustrate what is going on in a commutative diagram:

Homomorphism.png

Thus we see that $\phi$ is a homomorphism iff both of the composite mappings from $S \times S$ to $T$ have the same effect on all elements of $S \times S$.


Also known as

Some sources refer to a homomorphism as a morphism, but this term is best reserved for its use in category theory.


Also see

  • Results about homomorphisms can be found here.


Linguistic Note

The word homomorphism comes from the Greek morphe (μορφή) meaning form or structure, with the prefix homo- meaning similar.

Thus homomorphism means similar structure.


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Definition:Homomorphism_(Abstract_Algebra)&oldid=372934"