Definition:Homomorphism (Abstract Algebra)

From ProofWiki
Jump to navigation Jump to search

This page is about Homomorphism in the context of Abstract Algebra. For other uses, see Homomorphism.

Definition

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

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 algebraic structures with more than one operation:

Let:

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

be algebraic structures.


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

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


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

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


Then $\phi: \struct {S, \circ} \to \struct {T, *}$ is a semigroup homomorphism.


Monoid Homomorphism

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

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

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

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

Suppose further that $\phi$ preserves identities, that is:

$\map \phi {e_S} = e_T$


Then $\phi: \struct {S, \circ} \to \struct {T, *}$ 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, $\forall a, b \in R$:

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


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

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


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

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


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

$(1): \quad \forall k \in \closedint 1 n: \forall x, y \in S: \map \phi {x \ast_k y} = \map \phi x \odot_k \map \phi y$
$(2): \quad \forall x \in S: \forall \lambda \in R: \map \phi {\lambda \circ x} = \lambda \otimes \map \phi x$

where $\closedint 1 n = \set {1, 2, \ldots, n}$ denotes an integer interval.


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


$G$-Module Homomorphism

Let $\struct {G, \cdot}$ be a group.

Let $\struct {V, \phi}$ and $\struct {W, \mu}$ be $G$-modules.


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

$\forall g \in G: \forall v \in V: \map f {\map \phi {g, v} } = \map \mu {g, \map f v}$


Homomorphism of Complexes

Let $\struct {R, +, \cdot}$ 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 $\phi = \set {\phi_i: i \in \Z}$ 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 $\phi: \struct {S, \circ} \to \struct {T, *}$ be a mapping from one algebraic structure $\struct {S, \circ}$ to another $\struct {T, *}$.

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

$\forall \tuple {x, y} \in S \times S: \map {\paren {\phi \times \phi} } {x, y} = \tuple {\map \phi x, \map \phi y}$


Hence we can state that $\phi$ is a homomorphism if and only if:

$\map \ast {\map {\paren {\phi \times \phi} } {x, y} } = \map \phi {\map \circ {x, y} }$

using the notation $\map \circ {x, y}$ 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:

$\begin{xy} \xymatrix@L+2mu@+1em{ S \times S \ar[r]^*{\circ} \ar[d]_*{\phi \times \phi} & S \ar[d]^*{\phi} \\ T \times T \ar[r]_*{\ast} & T }\end{xy}$

Thus we see that $\phi$ is a homomorphism if and only if 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


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=686856"