Kernel (algebra)

File:Group homomorphism ver.2.svg

A group homomorphism $h$ from the group $G$ to the group $H$ is illustrated, with the groups represented by a blue oval on the left and a yellow circle on the right respectively. The kernel of $h$ is the red circle on the left, as $h$ sends it to the identity element 1 of $H$.

File:Projection-on-diagonal.gif

An example for a kernel - the linear operator $L:(x,y)⟶(x,x)$ transforms all points on the $(x=0,y)$ line to the zero point $(0,0)$, thus they form the kernel for the linear operator

In algebra, the kernel of a homomorphism is the relation describing how elements in the domain of the homomorphism become related in the image. A homomorphism is a function that preserves the underlying algebraic structure in the domain to its image.

When the algebraic structures involved have an underlying group structure, the kernel is taken to be the preimage of the group's identity element in the image, that is, it consists of the elements of the domain mapping to the image's identity. For example, the map that sends every integer to its parity (that is, 0 if the number is even, 1 if the number is odd) would be a homomorphism to the integers modulo 2, and its respective kernel would be the even integers which all have 0 as its parity. The kernel of a homomorphism of group-like structures will only contain the identity if and only if the homomorphism is injective, that is if the inverse image of every element consists of a single element. This means that the kernel can be viewed as a measure of the degree to which the homomorphism fails to be injective.

For some types of structure, such as abelian groups and vector spaces, the possible kernels are exactly the substructures of the same type. This is not always the case, and some kernels have received a special name, such as normal subgroups for groups and two-sided ideals for rings. The concept of a kernel has been extended to structures such that the inverse image of a single element is not sufficient for deciding whether a homomorphism is injective. In these cases, the kernel is a congruence relation.

Kernels allow defining quotient objects (also called quotient algebras in universal algebra). For many types of algebraic structure, the fundamental theorem on homomorphisms (or first isomorphism theorem) states that image of a homomorphism is isomorphic to the quotient by the kernel.

Definition

Group homomorphisms

Template:Group theory sidebar A group is a set $G$ with a binary operation $\cdot$ satisfying the following three properties for all $a,b,c\in G$:

1. Associative: $(a\cdot b)\cdot c=a\cdot (b\cdot c)$
2. Identity: There is an $e\in G$ such that $e\cdot a=a\cdot e=a$
3. Inverses: There is an $a^{\prime }\in G$ for each $a\in G$ such that $a\cdot a^{\prime }=a^{\prime }\cdot a=e$

A group is also called abelian if it also satisfies $a\cdot b=b\cdot a$.

Let $G$ and $H$ be groups. A group homomorphism from $G$ to $H$ is a function $f:G\to H$ such that $f(ab)=f(a)f(b)$ for all $a,b\in G$. Letting $e_H$ is the identity element of $H$, then the kernel of $f$ is the preimage of the singleton set $\{e_H\}$; that is, the subset of $G$ consisting of all those elements of $G$ that are mapped by $f$ to the element $e_H$.

The kernel is usually denoted $\ker f$ (or a variation). In symbols:

$\ker f=\{g\in G:f(g)=e_H\}.$

Since a group homomorphism preserves identity elements, the identity element $e_G$ of $G$ must belong to the kernel. The homomorphism $f$ is injective if and only if its kernel is only the singleton set $\{e_G\}$.

$\ker f$ is a subgroup of $G$ and further it is a normal subgroup. Thus, there is a corresponding quotient group $G/\ker f$. This is isomorphic to $f(G)$, the image of $G$ under $f$ (which is a subgroup of $H$ also), by the first isomorphism theorem for groups.

Ring homomorphisms

Template:Ring theory sidebar A ring with identity (or unity) is a set $R$ with two binary operations $+$ and $\cdot$ satisfying:

1. $R$ with $+$ is an abelian group with identity $0$.
2. Multiplication $\cdot$ is associative.
3. Distributive: $a\cdot (b+c)=a\cdot b+a\cdot c$ and $(a+b)\cdot c=a\cdot c+b\cdot c$ for all $a,b,c\in R$
4. Multiplication $\cdot$ has an identity element $1$.Template:Efn

A ring is commutative if the multiplication is commutative, and such a ring is a field when every $0\ne a\in R$ has a multiplicative inverse, that is, some $b\in R$ where $ab=1$. Let $R$ and $S$ be rings. A ring homomorphism from $R$ to $S$ is a function $f:R\to S$ satisfying for all $a,b\in R$:

1. $f(a+b)=f(a)+f(b)$
2. $f(ab)=f(a)f(b)$

The kernel of $f$ is the kernel as additive groups. It is the preimage of the zero ideal $\{0_S\}$, which is, the subset of $R$ consisting of all those elements of $R$ that are mapped by $f$ to the element $0_S$. The kernel is usually denoted $\ker f$ (or a variation). In symbols:

$\ker f=\{r\in R:f(r)=0_S\}.$

Since a ring homomorphism preserves zero elements, the zero element $0_R$ of $R$ must belong to the kernel. The homomorphism $f$ is injective if and only if its kernel is only the singleton set $\{0_R\}$. This is always the case if $R$ is a field, and $S$ is not the zero ring.

Since $\ker f$ contains the multiplicative identity only when $S$ is the zero ring, it turns out that the kernel is generally not a subring of $R$. The kernel is a subrng, and, more precisely, a two-sided ideal of $R$. Thus, it makes sense to speak of the quotient ring $R/\ker f$. The first isomorphism theorem for rings states that this quotient ring is naturally isomorphic to the image of $f$ (which is a subring of $S$).

Linear maps

File:Kernel and image of linear map.svg

Kernel and image of a linear map Template:Mvar from Template:Mvar to Template:Mvar

Given a field $F$, a vector space (over $F$) is an abelian group $V$ (with binary operation $+$ and identity $0$) with scalar multiplication from $F$ satisfying for all $a,b\in F$ and $\alpha ,\beta \in V$:

1. $a(b\alpha )=(ab)\alpha$
2. $(a+b)\alpha =a\alpha +b\alpha$
3. $a(\alpha +\beta )=a\alpha +a\beta$
4. $1\alpha =\alpha$

Let $V$ and $W$ be vector spaces over the field $F$. A linear map (or linear transformation) from $V$ to $W$ is a function $T:V\to W$ satisfying for all $\alpha ,\beta \in V$ and $a\in F$:

1. $T(\alpha +\beta )=T(\alpha )+T(\beta )$
2. $T(a\alpha )=aT(\alpha )$

If $0_W$ is the zero vector of $W$, then the kernel of $T$ (or null space) is the preimage of the zero subspace $\{0_W\}$; that is, the subset of $V$ consisting of all those elements of $V$ that are mapped by $T$ to the element $0_W$. The kernel is denoted as $\ker T$, or some variation thereof, and is symbolically defined as:

$\ker T=\{\mathbf{𝐯}\in V:T(\mathbf{𝐯})={\mathbf{𝟎}}_W\}.$

Since a linear map preserves zero vectors, the zero vector $0_V$ of $V$ must belong to the kernel. The transformation $T$ is injective if and only if its kernel is reduced to the zero subspace.

The kernel $\ker T$ is always a linear subspace of $V$. Thus, it makes sense to speak of the quotient space $V/\ker T$. The first isomorphism theorem for vector spaces states that this quotient space is naturally isomorphic to the image of $T$ (which is a subspace of $W$). As a consequence, the dimension of $V$ equals the dimension of the kernel plus the dimension of the image.

Module homomorphisms

Let $R$ be a ring. A modules over $R$ is defined exactly like a vector space over a field, using the same axioms, expect the field is replaced with a ring. In fact, a module over a field is exactly the same as a vector space over a field. Let $M$ and $N$ be $R$-modules. A module homomorphism from $M$ to $N$ is also a function $\phi :M\to N$ satisfying the same analogous properties of a linear map. The kernel of $\phi$ is defined to be:

$\ker \phi =\{m\in M|\phi (m)=0\}$

Every kernel is a submodule of the domain module, which means they always contain 0, the additive identity of the module. Kernels of abelian groups can be considered a particular kind of module kernel when the underlying ring is the integers.

Examples

Group homomorphisms

Let $G$ be the cyclic group on 6 elements $\{0,1,2,3,4,5,\}$ with modular addition, $H$ be the cyclic on 2 elements $\{0,1\}$ with modular addition, and $f$ the homomorphism that maps each element $g\in G$ to the element $g$ modulo 2 in $H$. Then $\ker f=\{0,2,4\}$, since all these elements are mapped to $0\in H$. The quotient group $G/\ker f$ has two elements: $\{0,2,4\}$ and $\{1,3,5\}$, and is isomorphic to $H$.

Given a isomorphism $\phi :G\to H$, one has $\ker \phi =1$. On the other hand, if this mapping is merely a homomorphism where H is the trivial group, then $\phi (g)=1$ for all $g\in G$, so thus $\ker \phi =G$.

Let $\phi :{\mathbb{ℝ}}^2\to \mathbb{ℝ}$ be the map defined as $\phi ((x,y))=x$. Then this is a homomorphism with the kernel consisting precisely the points of the form $(0,y)$. This mapping is considered the "projection onto the x-axis." A similar phenomenon occurs with the mapping $f:({\mathbb{ℝ}}^{\times }{)}^2\to {\mathbb{ℝ}}^{\times }$ defined as $f(a,b)=b$, where the kernel is the points of the form $(a,1)$

For a non-abelian example, let $Q_8$ denote the Quaternion group, and $V_4$ the Klein 4-group. Define a mapping $\phi :Q_8\to V_4$ to be:

$\phi (\pm 1)=1$

$\phi (\pm i)=a$

$\phi (\pm j)=b$

$\phi (\pm k)=c$

Then this mapping is a homomorphism where $\ker \phi =\{\pm 1\}$.

Ring homomorphisms

Consider the mapping $\phi :\mathbb{\mathbb{Z}}\to \mathbb{\mathbb{Z}}/2\mathbb{\mathbb{Z}}$ where the later ring is the integers modulo 2 and the map sends each number to its parity; 0 for even numbers, and 1 for odd numbers. This mapping turns out to be a homomorphism, and since the additive identity of the later ring is 0, the kernel is precisely the even numbers.

Let $\phi :\mathbb{\mathbb{Q}}[x]\to \mathbb{\mathbb{Q}}$ be defined as $\phi (p(x))=p(0)$. This mapping, which happens to be a homomorphism, sends each polynomial to its constant term. It maps a polynomial to zero if and only if said polynomial's constant term is 0. Polynomials with real coefficients can receive a similar homomorphism, with its kernel being the polynomials with constant term 0.

Linear maps

Let $\phi :{\mathbb{ℂ}}^3\to \mathbb{ℂ}$ be defined as $\phi (x,y,z)=x+2y+3z$, then the kernel of $\phi$ (that is, the null space) will be the set of points $(x,y,z)\in {\mathbb{ℂ}}^3$ such that $x+2y+3z=0$, and this set is a subspace of ${\mathbb{ℂ}}^3$ (the same is true for every kernel of a linear map).

If $D$ represents the derivative operator on real polynomials, then the kernel of $D$ will consist of the polynomials with deterivative equal to 0, that is the constant functions.

Consider the mapping $(Tp)(x)=x^{2}p(x)$, where $p$ is a polynomial with real coefficients. Then $T$ is a linear map whose kernel is precisely 0, since 0 is the only polynomial to satisfy $x^{2}p(x)=0$ for all $x\in \mathbb{ℝ}$.

Quotient algebras

The kernel of a homomorphism can be used to define a quotient algebra. Let $G$ and $H$ be groups, $\phi :G\to H$ be a group homomorphism, and denote $K=\ker \phi$. Put $G/K$ to be the set of fibers of the homomorphism $\phi$, where a fiber is the set of points of the domain mapping to a single point in the range. Let $X_a\in G/K$ denotes the fiber of the element $a\in H$, then a group operation on the set of fibers can be endowed by $X_a{X}_b=X_{ab}$, and $G/K$ is called the quotient group (or factor group), to be read as "G modulo K" or "G mod K". The terminology arises from the fact that the kernel represents the fiber of the identity element of the range, $H$, and that the remaining elements are simply "translates" of the kernel, so the quotient group is obtained by "dividing out" the kernel.

The fibers can also be described by looking at the domain relative to the kernel; given $X\in G/K$ and any element $u\in X$, then $X=uK=Ku$ where:

$uK=\{uk|k\in K\}$

$Ku=\{ku|k\in K\}$

these sets are called the left and right cosets respectively, and can be defined in general for any arbitrary subgroup of $G$. The group operation can then be defined as $uK\circ vK=(uk)K$, which is well-defined regardless of the choice of representatives of the fibers.

According to the first isomorphism theorem, there is an isomorphism $\mu :G/K\to \phi (G)$, where the later group is the image of the homomorphism $\phi$, and the isomorphism is defined as $\mu (uK)=\phi (u)$, and such map is also well-defined.

For rings, modules, and vector spaces, one can define the respective quotient algebras via the underlying additive group structure, with cosets represented as $x+K$. Ring multiplication can be defined on the quotient algebra as $(x+K)(y+K)=xy+K$, and is well-defined. For a ring $R$ (possibly a field when describing vector spaces) and a module homomorphism $\phi :M\to N$ with kernel $K=\ker \phi$, one can define scalar multiplication on $G/K$ by $r(x+K)=rx+K$ for $r\in R$ and $x\in M$, which will also be well-defined.

Kernel structures

The structure of kernels allows for the building of quotient algebras from structures satisfying the properties of kernels. Any subgroup $N$ of a group $G$ can construct a quotient $G/N$ by the set of all cosets of $N$ in $G$. The natural way to turn this into a group, similar to the treatment for the quotient by a kernel, is to define an operation on (left) cosets by $uN\cdot vN=(uv)N$, however this operation is well defined if and only if the subgroup $N$ is closed under conjugation under $G$, that is, if $g\in G$ and $n\in N$, then $gn{g}^{-1}\in N$. Furthermore, the operation being well defined is sufficient for the quotient to be a group. Subgroups satisfying this property are called normal subgroups. Every kernel of a group is a normal subgroup, and for a given normal subgroup $N$ of a group $G$, the natural projection $\pi :G\to G/N$ defined as $\pi (g)=gN$ is a homomorphism with $\ker \pi =N$, so the normal subgroups are precisely the subgroups which are kernels. The closure under conjugation, however, gives a criterion for when a subgroup is a kernel for some homomorphism.

For a ring $R$, treating it as a group, one can take a quotient group via an arbitrary subgroup $I$ of the ring, which will be normal due to the ring's additive group being abelian. To define multiplication on $R/I$, the multiplication of cosets, defined as $(r+I)(s+I)=rs+I$ needs to be well-defined. Taking representatives $r+\alpha$ and $s+\beta$ of $r+I$ and $s+I$ respectively, for $r,s\in R$ and $\alpha ,\beta \in I$, yields:

$(r+\alpha )(s+\beta )+I=rs+I$

Setting $r=s=0$ implies that $I$ is closed under multiplication, while setting $\alpha =s=0$ shows that $r\beta \in I$, that is, $I$ is closed under arbitrary multiplication by elements on the left. Similarly, taking $r=\beta =0$ implies that $I$ is also closed under multiplication by arbitrary elements on the right. Any subgroup of $R$ that is closed under multiplication by any element of the ring is called an ideal. Analogously to normal subgroups, the ideals of a ring are precisely the kernels of homomorphisms.

Exact sequence

File:Illustration of an Exact Sequence of Groups.svg

An exact sequence of groups. At each pair of homomorphism, the image of the previous homomorphism becomes the kernel of the next homomorphism, that is they get sent to the identity element.

Kernels are used to define exact sequences of homomorphisms for groups and modules. Given modules $A$, $B$, and $C$, a pair of homomorphisms $\psi :A\to B,\phi :B\to C$ is said to be exact if $\text{image }\psi =\ker \phi$. An exact sequence is then a sequence of modules and homomorphisms $\cdots \to X_{n-1}\to X_n\to X_{n+1}\to \cdots$ where each adjacent pair of modules and homomorphisms is exact.

Universal algebra

Kernels can be generalized in universal algebra for homomorphisms between any two algebraic structures. An operation on a set $A$ is a function of the form $Q:A^n\to A$, where $n$ is the arity (or rank) of the operation. An $n$-ary operation takes an ordered list of $n$ elements from $A$ and maps them to a single element in $A$. An algebraic structure is a tuple $⟨A,F⟩$ where $A$ is the underlying set of the algebra, and $F$ is an indexed set of operations $Q\in F$ on $A$, with their interpretation denoted $Q^A$. The set indexing $F$ is the language, which also maps each operation symbol to their fixed arity (called the rank function). Two algebraic structures are similar when they share the same language, including their rank function.

Let $A$ and $B$ be algebraic structures of a similar type $F$. A homomorphism is a function $f:A\to B$ that respects the interpretation of each $Q\in F$, that is, taking $Q$ to be an $n$-ary operation, and $a_i\in A$ for $1\le i\le n$:

$f(Q^A(a_1,\dots a_n))=Q^B(f(a_1),\dots f(a_n))$

The kernel of $f$, denoted $\ker f$, is the subset of the direct product $A\times A$ consisting of all ordered pairs of elements of $A$ whose components are both mapped by $f$ to the same element in $B$. In symbols:

$\ker f=\{(a,b)\in A\times A:f(a)=f\left(b\right)\}\text{.}$

The homomorphism $f$ is injective if and only if its kernel is the diagonal set $\{(a,a)|a\in A\}$, which is always contained inside the kernel. $\ker f$ is an equivalence relation on $A$, and in fact a congruence relation, meaning that for an n-ary operation $Q\in F$, the relation $a_i\ker f{b}_i$ for $1\le i\le n$ implies $Q^A(a_1,\dots a_n)\ker f{Q}^A(b_1,\dots b_n)$. It makes sense to speak of the quotient algebra $A/\ker f$, with the set consisting of the equivalence classes of the kernel, and the well-defined operations defined for an $n$-ary operation $Q\in F$ as:

$Q^{A/\ker f}(a_1/\ker f,\dots a_n/\ker f)=Q^A(a_1,\dots a_n)/\ker f$

The first isomorphism theorem in universal algebra states that this quotient algebra is naturally isomorphic to the image of $f$ (which is a subalgebra of $B$).

See also

• Kernel (linear algebra)
• Kernel (category theory)
• Kernel of a function
• Equalizer (mathematics)
• Zero set