Definition:Kernel (Abstract Algebra)

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

Definition

Kernel of Magma Homomorphism

Let $\struct {S, \circ}$ be a magma.

Let $\struct {T, *}$ be an algebraic structure with an identity element $e$.

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

The kernel of $\phi$ is the subset of the domain of $\phi$ defined as:

$\map \ker \phi = \set {x \in S: \map \phi x = e}$

That is, $\map \ker \phi$ is the subset of $S$ that maps to the identity of $T$.

Kernel of Group Homomorphism

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

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

The kernel of $\phi$ is the subset of the domain of $\phi$ defined as:

$\map \ker \phi := \phi^{-1} \sqbrk {e_H} = \set {x \in G: \map \phi x = e_H}$

where $e_H$ is the identity of $H$.

That is, $\map \ker \phi$ is the subset of $G$ that maps to the identity of $H$.

Kernel of Ring Homomorphism

Let $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ be rings.

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.

The kernel of $\phi$ is the subset of the domain of $\phi$ defined as:

$\map \ker \phi = \set {x \in R_1: \map \phi x = 0_{R_2} }$

where $0_{R_2}$ is the zero of $R_2$.

That is, $\map \ker \phi$ is the subset of $R_1$ that maps to the zero of $R_2$.

From Ring Homomorphism Preserves Zero it follows that $0_{R_1} \in \map \ker \phi$ where $0_{R_1}$ is the zero of $R_1$.

Kernel of Linear Transformation

Let $\phi: G \to H$ be a linear transformation where $G$ and $H$ are $R$-modules.

Let $e_H$ be the identity of $H$.

The kernel of $\phi$ is defined as:

$\map \ker \phi := \phi^{-1} \sqbrk {\set {e_H} }$

where $\phi^{-1} \sqbrk S$ denotes the preimage of $S$ under $\phi$.

In Vector Space

Let $\struct {\mathbf V, +, \times}$ be a vector space.

Let $\struct {\mathbf V', +, \times}$ be a vector space whose zero vector is $\mathbf 0'$.

Let $T: \mathbf V \to \mathbf V'$ be a linear transformation.

Then the kernel of $T$ is defined as:

$\map \ker T := T^{-1} \sqbrk {\set {\mathbf 0'} } = \set {\mathbf x \in \mathbf V: \map T {\mathbf x} = \mathbf 0'}$

Kernel of Homomorphism of Differential 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 homomorphism $M \to N$.

For each $i \in \Z$ let $K_i$ be the kernel of $\phi_i$.

For each $i \in \Z$ let $f_i$ be the restriction of $d_i$ to $K_i$.

Then the kernel of $\phi$ is:

$\ker \phi : \quad \cdots \longrightarrow K_i \stackrel {f_i} \longrightarrow K_{i + 1} \stackrel {f_{i + 1} } \longrightarrow K_{i + 2} \stackrel {f_{i + 2} } \longrightarrow \cdots$