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A kernel records the information lost by a homomorphism. For a group homomorphism it is the preimage of the identity element, giving the kernel of a group homomorphism; for a ring homomorphism or linear map it is the preimage of zero, giving the kernel of a ring homomorphism or kernel of a linear map. More generally, the relation $a\sim b⟺f(a)=f(b)$ records which inputs the homomorphism identifies. In the group case, its equivalence classes are cosets of the group kernel, as explained by fibres of a group homomorphism are cosets of its kernel.

The kernel of a group homomorphism is the group version of a kernel. For $\phi :G\to H$, it isIt is a normal subgroup of the domain: for $n$ in the group kernel, $\phi (gn{g}^{-1})=\phi (g)e_H\phi (g{)}^{-1}=e_H$. The group kernel describes inputs whose effect disappears after mapping. For reduction modulo three from $\mathbb{Z}/6\mathbb{Z}$, it is $\{0,3\}$; for the sign of a permutation on $S_3$, it is $A_3$. A group homomorphism is injective exactly when its group kernel contains only the identity element.

Map the integers into a field by $n\mapsto n1$. Its kernel is a prime ideal of $\mathbb{Z}$, since the image has no zero divisors. Thus the kernel is either zero or $(p)$ for a prime $p$. In the zero case, the map extends injectively to the rational numbers by taking quotients of nonzero integers. In the other case, its image is the field ${\mathbb{F}}_p$. This proves the prime-field assertion.

A finite field cannot contain the infinite rational field, so it has characteristic $p$. It is a finite-dimensional vector space over its prime field ${\mathbb{F}}_p$, say of dimension $n\ge 1$, and choosing a basis counts its elements asFor existence, put $q=p^n$ and take a splitting field of $x^q-x$ over ${\mathbb{F}}_p$. Its derivative is $-1$, so it has exactly $q$ distinct polynomial roots. The polynomial roots are closed under addition and multiplication because the Frobenius identity gives $(u+v{)}^q=u^q+v^q$ and $(uv{)}^q=u^q{v}^q$. They include $0,1$, are closed under additive inverses, and for a nonzero polynomial root $u$, $(u^{-1}{)}^q=(u^q{)}^{-1}=u^{-1}$. They therefore form a field of exactly $q$ elements. Indeed this field is the splitting field because it contains all polynomial roots and is generated by them.

In order 4, characteristic is 2. For $\omega \ne 0,1$, the multiplicative group has order 3, so ${\omega }^3=1$ and ${\omega }^2\ne 1,\omega$. Factorization of $x^3-1$ gives ${\omega }^2+\omega +1=0$. The quadratic $x^2+x+1$ has no polynomial root in ${\mathbb{F}}_2$ and is the minimal polynomial. Write $s={\omega }^2=\omega +1$. The complete tables areFor an element $\xi \ne 0,1$ in order 8, the tower law makes $[{\mathbb{F}}_2(\xi ):{\mathbb{F}}_2]$ divide3. It is not1, so its minimal polynomial is an irreducible monic cubic. Its constant term must be1. A cubic $x^3+a{x}^2+bx+1$ has no polynomial root at 1 exactly when $a+b=1$, and absence of a polynomial root is equivalent to cubic irreducibility. Consequently the two possibilities areBoth occur: together with $x$ and $x+1$ they factor $x^8-x$ over ${\mathbb{F}}_2$, whose distinct polynomial roots are exactly the eight elements of the field. Each cubic contributes its three polynomial roots.