Editing Kernel (algebra) (section)

== Definition ==

=== Group homomorphisms ===
{{Group theory sidebar}}
Let ''G'' and ''H'' be [[group (mathematics)|group]]s and let ''f'' be a [[group homomorphism]] from ''G'' to ''H''. If ''e''<sub>''H''</sub> is the [[identity element]] of ''H'', then the ''kernel'' of ''f'' is the preimage of the singleton set {''e''<sub>''H''</sub>}; that is, the subset of ''G'' consisting of all those elements of ''G'' that are mapped by ''f'' to the element ''e''<sub>''H''</sub>.<ref name=":0">{{Cite book |last1=Dummit |first1=David Steven |title=Abstract algebra |last2=Foote |first2=Richard M. |date=2004 |publisher=Wiley |isbn=978-0-471-43334-7 |edition=3rd |location=Hoboken, NJ}}</ref><ref name=":1">{{Cite book |last1=Fraleigh |first1=John B. |title=A first course in abstract algebra |last2=Katz |first2=Victor |date=2003 |publisher=Addison-Wesley |isbn=978-0-201-76390-4 |edition=7th |series=World student series |location=Boston}}</ref><ref name=":2">{{Cite book |last=Hungerford |first=Thomas W. |title=Abstract Algebra: an introduction |date=2014 |publisher=Brooks/Cole, Cengage Learning |isbn=978-1-111-56962-4 |edition=3rd |location=Boston, MA}}</ref>

The kernel is usually denoted {{nowrap|ker ''f''}} (or a variation). In symbols:
: <math> \ker f = \{g \in G : f(g) = e_{H}\} .</math>

Since a group homomorphism preserves identity elements, the identity element ''e''<sub>''G''</sub> of ''G'' must belong to the kernel.

The homomorphism ''f'' is injective if and only if its kernel is only the singleton set {''e''<sub>''G''</sub>}. If ''f'' were not injective, then the non-injective elements can form a distinct element of its kernel: there would exist {{nowrap|''a'', ''b'' &isin; ''G''}} such that {{nowrap|''a'' ≠ ''b''}} and {{nowrap|1=''f''(''a'') = ''f''(''b'')}}. Thus {{nowrap|1=''f''(''a'')''f''(''b'')<sup>−1</sup> = ''e''<sub>''H''</sub>}}. ''f'' is a group homomorphism, so inverses and group operations are preserved, giving {{nowrap|1=''f''(''ab''<sup>−1</sup>) = ''e''<sub>''H''</sub>}}; in other words, {{nowrap|''ab''<sup>−1</sup> &isin; 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 {{nowrap|''g'' ≠ ''e''<sub>''G''</sub> &isin; ker ''f''}}, then {{nowrap|1=''f''(''g'') = ''f''(''e''<sub>''G''</sub>) = ''e''<sub>''H''</sub>}}, thus ''f'' would not be injective.

{{nowrap|ker ''f''}} is a [[subgroup]] of ''G'' and further it is a [[normal subgroup]]. Thus, there is a corresponding [[quotient group]] {{nowrap|''G'' / (ker ''f'')}}. This is isomorphic to ''f''(''G''), the image of ''G'' under ''f'' (which is a subgroup of ''H'' also), by the [[isomorphism theorems|first isomorphism theorem]] for groups.

=== Ring homomorphisms ===
{{Ring theory sidebar}}
Let ''R'' and ''S'' be [[ring (mathematics)|ring]]s (assumed [[unital algebra|unital]]) and let ''f'' be a [[ring homomorphism]] from ''R'' to ''S''.
If 0<sub>''S''</sub> is the [[zero element]] of ''S'', then the ''kernel'' of ''f'' is its kernel as additive groups.<ref name=":1" /> It is the preimage of the [[zero ideal]] {{mset|0<sub>''S''</sub>}}, which is, the subset of ''R'' consisting of all those elements of ''R'' that are mapped by ''f'' to the element 0<sub>''S''</sub>.
The kernel is usually denoted {{nowrap|ker ''f''}} (or a variation).
In symbols:
: <math> \operatorname{ker} f = \{r \in R : f(r) = 0_{S}\} .</math>

Since a ring homomorphism preserves zero elements, the zero element 0<sub>''R''</sub> of ''R'' must belong to the kernel.
The homomorphism ''f'' is injective if and only if its kernel is only the singleton set {{mset|0<sub>''R''</sub>}}.
This is always the case if ''R'' is a [[field (mathematics)|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 sub[[rng (algebra)|rng]], and, more precisely, a two-sided [[ideal (ring theory)|ideal]] of ''R''.
Thus, it makes sense to speak of the [[quotient ring]] {{nowrap|''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).

=== Linear maps ===
{{Main|Kernel (linear algebra)}}
Let ''V'' and ''W'' be [[vector space]]s over a [[Field (mathematics)|field]] (or more generally, [[module (mathematics)|modules]] over a [[Ring (mathematics)|ring]]) and let ''T'' be a [[linear map]] from ''V'' to ''W''. If '''0'''<sub>''W''</sub> is the [[zero vector]] of ''W'', then the kernel of ''T'' (or null space<ref>{{Cite book |last=Axler |first=Sheldon |title=Linear Algebra Done Right |publisher=[[Springer Publishing|Springer]] |edition=4th}}</ref><ref name=":0" />) is the [[preimage]] of the [[zero space|zero subspace]] {'''0'''<sub>''W''</sub>}; that is, the [[subset]] of ''V'' consisting of all those elements of ''V'' that are mapped by ''T'' to the element '''0'''<sub>''W''</sub>. The kernel is usually denoted as {{nowrap|ker ''T''}}, or some variation thereof:
: <math> \ker T = \{\mathbf{v} \in V : T(\mathbf{v}) = \mathbf{0}_{W}\} . </math>

Since a linear map preserves zero vectors, the zero vector '''0'''<sub>''V''</sub> 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''.<ref name=":0" /> Thus, it makes sense to speak of the [[quotient space (linear algebra)|quotient space]] {{nowrap|''V'' / (ker ''T'')}}. The first isomorphism theorem for vector spaces states that this quotient space is [[natural isomorphism|naturally isomorphic]] to the [[image (function)|image]] of ''T'' (which is a subspace of ''W''). As a consequence, the [[dimension (linear algebra)|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 (mathematics)|ring]] in an analogous manner. This includes kernels for homomorphisms between [[abelian group]]s as a special case.<ref name=":0" /> This example captures the essence of kernels in general [[abelian categories]]; see [[Kernel (category theory)]].

=== Module homomorphisms ===
Let <math>R</math> be a [[Ring (mathematics)|ring]], and let <math>M</math> and <math>N</math> be <math>R</math>-[[Module (mathematics)|modules]]. If <math>\varphi: M \to N </math> is a module homomorphism, then the kernel is defined to be:<ref name=":0" />
: <math> \ker \varphi = \{m \in M \ | \ \varphi (m) = 0\} </math>
Every kernel is a [[submodule]] of the domain module.<ref name=":0" />

=== Monoid homomorphisms ===
{{unreferenced section|date=April 2025}}
Let ''M'' and ''N'' be [[monoid (algebra)|monoid]]s and let ''f'' be a [[monoid homomorphism]] from ''M'' to ''N''. Then the ''kernel'' of ''f'' is the subset of the [[direct product]] {{nowrap|''M'' × ''M''}} consisting of all those [[ordered pair]]s of elements of ''M'' whose components are both mapped by ''f'' to the same element in ''N''{{citation needed|date=April 2025}}. The kernel is usually denoted {{nowrap|ker ''f''}} (or a variation thereof). In symbols:
: <math>\operatorname{ker} f = \left\{\left(m, m'\right) \in M \times M : f(m) = f\left(m'\right)\right\}.</math>

Since ''f'' is a [[function (mathematics)|function]], the elements of the form {{nowrap|(''m'', ''m'')}} must belong to the kernel. The homomorphism ''f'' is injective if and only if its kernel is only the [[Equality (mathematics)|diagonal set]] {{nowrap|{{mset|(''m'', ''m'') : ''m'' in ''M''}}}}.

It turns out that {{nowrap|ker ''f''}} is an [[equivalence relation]] on ''M'', and in fact a [[congruence relation]]. Thus, it makes sense to speak of the [[quotient monoid]] {{nowrap|''M'' / (ker ''f'')}}. The first isomorphism theorem for monoids states that this quotient monoid is naturally isomorphic to the image of ''f'' (which is a [[submonoid]] of ''N''; for the congruence relation).

This is very different in flavor from the above examples. In particular, the preimage of the identity element of ''N'' is ''not'' enough to determine the kernel of ''f''.