Kernel (algebra)

For other uses, see kernel.

In algebra, the kernel of a homomorphism is the relation describing how elements in the domain of the homomorphism become related in the image.^[1] A homomorphism is a function that preserves the underlying algebraic structure to its image, which will have the same type of structure as its domain. 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. 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.^[2]

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

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.^[2] 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.^[1]

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.^[1][2]

Definition

Group homomorphisms

Let G and H be groups and let f be a group homomorphism from G to H. If 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.^[2][3][4]

The kernel is usually denoted ker f (or a variation)^[2]. In symbols:

${\displaystyle \ker f=\{g\in G:f(g)=e_{H}\}.}$

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}. If f were not injective, then the non-injective elements can form a distinct element of its kernel: there would exist a, b ∈ G such that a ≠ b and f(a) = f(b). Thus f(a)f(b)⁻¹ = e_H. f is a group homomorphism, so inverses and group operations are preserved, giving f(ab⁻¹) = e_H; in other words, ab⁻¹ ∈ ker f, and ker f would not be the singleton. Conversely, distinct elements of the kernel violate injectivity directly: if there would exist an element g ≠ e_G ∈ ker f, then f(g) = f(e_G) = e_H, thus f would not be injective.^[2]

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.^[2]

Ring homomorphisms

Let R and S be rings (assumed unital) and let f be a ring homomorphism from R to S. If 0_S is the zero element of S, then the kernel of f is its kernel as additive groups.^[3] 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:

${\displaystyle \operatorname {ker} f=\{r\in R:f(r)=0_{S}\}.}$

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.^[2]

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). (Note that rings need not be unital for the kernel definition).^[2]

Linear maps

Main article: Kernel (linear algebra)

Let V and W be vector spaces over a field (or more generally, modules over a ring) and let T be a linear map from V to W. If 0_W is the zero vector of W, then the kernel of T (or null space^[5][2]) 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 usually denoted as ker T, or some variation thereof:

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

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.^[5]

The kernel ker T is always a linear subspace of V.^[2] 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.

One can define kernels for homomorphisms between modules over a ring in an analogous manner. This includes kernels for homomorphisms between abelian groups as a special case.^[2]

Module homomorphisms

Let ${\displaystyle R}$ be a ring, and let ${\displaystyle M}$ and ${\displaystyle N}$ be ${\displaystyle R}$-modules. If ${\displaystyle \varphi :M\to N}$ is a module homomorphism, then the kernel is defined to be:^[2]

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

Every kernel is a submodule of the domain module.^[2]

Survey of 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_H. The quotient group G / (ker f) has two elements: {0, 2, 4} and {1, 3, 5}, and is isomorphic to H.^[2]

Given a isomorphism ${\displaystyle \varphi :G\to H}$, one has ${\displaystyle \ker \varphi =1}$.^[2] On the other hand, if this mapping is merely a homomorphism where H is the trivial group, then ${\displaystyle \varphi (g)=1}$ for all ${\displaystyle g\in G}$, so thus ${\displaystyle \ker \varphi =G}$.^[2]

Let ${\displaystyle \varphi$ :\mathbb {R} ^{2}\to \mathbb {R} } be the map defined as ${\displaystyle \varphi ((x,y))=x}$. Then this is a homomorphism with the kernel consisting precisely the points of the form ${\displaystyle (0,y)}$. This mapping is considered the "projection onto the x-axis." ^[2] A similar phenomenon occurs with the mapping ${\displaystyle f:(\mathbb {R} ^{\times })^{2}\to \mathbb {R} ^{\times }}$ defined as ${\displaystyle f(a,b)=b}$, where the kernel is the points of the form ${\displaystyle (a,1)}$^[4]

For a non-abelian example, let ${\displaystyle Q_{8}}$ denote the Quaternion group, and ${\displaystyle V_{4}}$ the Klein 4-group. Define a mapping ${\displaystyle \varphi :Q_{8}\to V_{4}}$ to be^[2]:

${\displaystyle \varphi (\pm 1)=1}$

${\displaystyle \varphi (\pm i)=a}$

${\displaystyle \varphi (\pm j)=b}$

${\displaystyle \varphi (\pm k)=c}$

Then this mapping is a homomorphism where ${\displaystyle \ker \varphi =\{\pm 1\}}$.^[2]

Ring homomorphisms

Consider the mapping ${\displaystyle \varphi$ :\mathbb {Z} \to \mathbb {Z} /2\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.^[2]

Let ${\displaystyle \varphi$ :\mathbb {Q} [x]\to \mathbb {Q} } be defined as ${\displaystyle \varphi (p(x))=p(0)}$. 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.^[2] If we instead work with polynomials with real coefficients, then we again receive a homomorphism with its kernel being the polynomials with constant term 0.^[4]

Linear maps

Let ${\displaystyle \varphi$ :\mathbb {C} ^{3}\to \mathbb {C} } be defined as ${\displaystyle \varphi (x,y,z)=x+2y+3z}$, then the kernel of ${\displaystyle \varphi }$ (that is, the null space) will be the set of points ${\displaystyle (x,y,z)\in \mathbb {C} ^{3}}$ such that ${\displaystyle x+2y+3z=0}$, and this set is a subspace of ${\displaystyle \mathbb {C} ^{3}}$ (the same is true for every kernel of a linear map).^[5]

If ${\displaystyle D}$ represents the derivative operator on real polynomials, then the kernel of ${\displaystyle D}$ will consist of the polynomials with deterivative equal to 0, that is the constant functions.^[5]

Consider the mapping ${\displaystyle (Tp)(x)=x^{2}p(x)}$, where ${\displaystyle p}$ is a polynomial with real coefficients. Then ${\displaystyle T}$ is a linear map whose kernel is precisely 0, since it is the only polynomial to satisfy ${\displaystyle x^{2}p(x)=0}$ for all ${\displaystyle x\in \mathbb {R} }$.^[5]

Quotient algebras

The kernel of a homomorphism can be used to define a quotient algebra. For instance, if ${\displaystyle \varphi :G\to H}$ denotes a group homomorphism, and we set ${\displaystyle K=\ker \varphi }$, we can consider ${\displaystyle G/K}$ to be the set of fibers of the homomorphism ${\displaystyle \varphi }$, where a fiber is merely the set of points of the domain mapping to a single chosen point in the range.^[2] If ${\displaystyle X_{a}\in G/K}$ denotes the fiber of the element ${\displaystyle a\in H}$, then we can give a group operation on the set of fibers by ${\displaystyle X_{a}X_{b}=X_{ab}}$, and we call ${\displaystyle G/K}$ the quotient group (or factor group), to be read as "G modulo K" or "G mod K".^[2] The terminology arises from the fact that the kernel represents the fiber of the identity element of the range, ${\displaystyle H}$, and that the remaining elements are simply "translates" of the kernel, so the quotient group is obtained by "dividing out" by the kernel.^[2]

The fibers can also be described by looking at the domain relative to the kernel; given ${\displaystyle X\in G/K}$ and any element ${\displaystyle u\in X}$, then ${\displaystyle X=uK=Ku}$ where:^[2]

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

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

these sets are called the left and right cosets respectively, and can be defined in general for any arbitrary subgroup aside from the kernel.^[2][4][3] The group operation can then be defined as ${\displaystyle uK\circ vK=(uk)K}$, which is well-defined regardless of the choice of representatives of the fibers.^[2][3]

According to the first isomorphism theorem, we have an isomorphism ${\displaystyle \mu :G/K\to \varphi (G)}$, where the later group is the image of the homomorphism ${\displaystyle \varphi }$, and the isomorphism is defined as ${\displaystyle \mu (uK)=\varphi (u)}$, and such map is also well-defined.^[2][3]

For rings, modules, and vector spaces, one can define the respective quotient algebras via the underlying additive group structure, with cosets represented as ${\displaystyle x+K}$. Ring multiplication can be defined on the quotient algebra the same way as in the group (and be well-defined). For a ring ${\displaystyle R}$ (possibly a field when describing vector spaces) and a module homomorphism ${\displaystyle \varphi :M\to N}$ with kernel ${\displaystyle K=\ker \varphi }$, one can define scalar multiplication on ${\displaystyle G/K}$ by ${\displaystyle r(x+K)=rx+K}$ for ${\displaystyle r\in R}$ and ${\displaystyle x\in M}$, which will also be well-defined.^[2]

Kernel structures

The structure of kernels allows for the building of quotient algebras from structures satisfying the properties of kernels. Any subgroup ${\displaystyle N}$ of a group ${\displaystyle G}$ can construct a quotient ${\displaystyle G/N}$ by the set of all cosets of ${\displaystyle N}$ in ${\displaystyle G}$.^[2] 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 ${\displaystyle uN\cdot vN=(uv)N}$, however this operation is well defined if and only if the subgroup ${\displaystyle N}$ is closed under conjugation under ${\displaystyle G}$, that is, if ${\displaystyle g\in G}$ and ${\displaystyle n\in N}$, then ${\displaystyle gng^{-1}\in N}$. Furthermore, the operation being well defined is sufficient for the quotient to be a group.^[2] Subgroups satisfying this property are called normal subgroups.^[2] Every kernel of a group is a normal subgroup, and for a given normal subgroup ${\displaystyle N}$ of a group ${\displaystyle G}$, the natural projection ${\displaystyle \pi (g)=gN}$ is a homomorphism with ${\displaystyle \ker \pi =N}$, so the normal subgroups are precisely the subgroups which are kernels.^[2] The closure under conjugation, however, gives an "internal"^[2] criterion for when a subgroup is a kernel for some homomorphism.

For a ring ${\displaystyle R}$, treating it as a group, one can take a quotient group via an arbitrary subgroup ${\displaystyle I}$ of the ring, which will be normal due to the ring's additive group being abelian. To define multiplication on ${\displaystyle R/I}$, the multiplication of cosets, defined as ${\displaystyle (r+I)(s+I)=rs+I}$ needs to be well-defined. Taking representative ${\displaystyle r+\alpha }$ and ${\displaystyle s+\beta }$ of ${\displaystyle r+I}$ and ${\displaystyle s+I}$ respectively, for ${\displaystyle r,s\in R}$ and ${\displaystyle \alpha ,\beta \in I}$, yields:^[2]

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

Setting ${\displaystyle r=s=0}$ implies that ${\displaystyle I}$ is closed under multiplication, while setting ${\displaystyle \alpha =s=0}$ shows that ${\displaystyle r\beta \in I}$, that is, ${\displaystyle I}$ is closed under arbitrary multiplication by elements on the left. Similarly, taking ${\displaystyle r=\beta =0}$ implies that ${\displaystyle I}$ is also closed under multiplication by arbitrary elements on the right.^[2] Any subgroup of ${\displaystyle R}$ that is closed under multiplication by any element of the ring is called an ideal.^[2] Analogously to normal subgroups, the ideals of a ring are precisely the kernels of homomorphisms.^[2]

Exact sequence

Main article: Exact sequence

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. If A, B, and C are modules, then a pair of homomorphisms ${\displaystyle \psi :A\to B,\varphi :B\to C}$ is said to be exact if ${\displaystyle {\text{image }}\psi =\ker \varphi }$. An exact sequence is then a sequence of modules and homomorphism ${\displaystyle \cdots \to X_{n-1}\to X_{n}\to X_{n+1}\to \cdots }$ where each adjacent pair of homomorphisms is exact.^[2]

Universal algebra

All the above cases may be unified and generalized in universal algebra. Let A and B be algebraic structures of a given type and let f be a homomorphism of that type from A to B. Then the kernel of f is the subset of the direct product A × A consisting of all those ordered pairs of elements of A whose components are both mapped by f to the same element in B.^[6][1] The kernel is usually denoted ker f (or a variation). In symbols:

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

The homomorphism f is injective if and only if its kernel is exactly the diagonal set {(a, a) : a ∈ A}, which is always at least contained inside the kernel.^[6][1]

It is easy to see that ker f is an equivalence relation on A, and in fact a congruence relation. Thus, it makes sense to speak of the quotient algebra A / (ker f). The first isomorphism theorem in general universal algebra states that this quotient algebra is naturally isomorphic to the image of f (which is a subalgebra of B).^[6][1]

See also

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