1.3 The characteristic of a field

One checks easily that the map

\mathbb{Z}\rightarrow F,\quad n\mapsto n\cdot 1_{F}\overset{\textup{{def}}}{=}% 1_{F}+1_{F}+\cdots+1_{F}\quad(n\text{ copies of }1_{F}),

is a homomorphism of rings. For example,

(\underbrace{1_{F}+\cdots+1_{F}}_{m})+(\underbrace{1_{F}+\cdots+1_{F}}_{n})=% \underbrace{1_{F}+\cdots+1_{F}}_{m+n}

because of the associativity of addition. Therefore its kernel is an ideal in $\mathbb{Z}{}$ .

Case 1: The kernel of the map is $(0)$ , so that

n\cdot 1_{F}=0\quad\text{(in }F\text{)}\implies n=0\quad\text{(in }\mathbb{Z}% \text{).}

Nonzero integers map to invertible elements of $F$ under $n\mapsto n\cdot 1_{F}\colon\mathbb{Z}{}\rightarrow F$ , and so this map extends to a homomorphism

\genfrac{}{}{}{}{\raisebox{-2.0pt}{$m$}}{\raisebox{2.0pt}{$n$}}\mapsto(m\cdot 1% _{F})(n\cdot 1_{F})^{-1}\colon\mathbb{Q}\hookrightarrow F.

In this case, $F$ contains a copy of $\mathbb{Q}$ , and we say that it has characteristic zero.

Case 2: The kernel of the map is $\neq(0)$ , so that $n\cdot 1_{F}=0$ for some $n\neq 0$ . The smallest positive such $n$ will be a prime $p$ (otherwise there will be two nonzero elements in $F$ whose product is zero), and $p$ generates the kernel. Thus, the map $n\mapsto n\cdot 1_{F}\colon\mathbb{Z}{}\rightarrow F$ defines an isomorphism from $\mathbb{Z}/p\mathbb{Z}{}$ onto the subring

\{m\cdot 1_{F}\mid m\in\mathbb{Z}\}

of $F$ . In this case, $F$ contains a copy of $\mathbb{F}_{p}$ , and we say that it has characteristic $p$ .

A field isomorphic to one of the fields $\mathbb{F}_{2},\mathbb{F}{}_{3},\mathbb{F}{}_{5},\ldots,\mathbb{Q}$ is called a prime field. Every field contains exactly one prime field (as a subfield).

1.4

More generally, a commutative ring $R$ is said to have characteristic $p$ (resp. $0$ ) if it contains a prime field (as a subring) of characteristic $p$ (resp. $0$ ).²² 2 A commutative ring has a characteristic if and only if it contains a field as a subring. For example, neither $\mathbb{Z}{}$ nor $\mathbb{F}{}_{2}\times\mathbb{F}{}_{3}$ has a characteristic. Then the prime field is unique and, by definition, contains $1_{R}$ . Thus, if $R$ has characteristic $p\neq 0$ , then $1_{R}+\cdots+1_{R}=0$ ( $p$ terms).

Let $R$ be a nonzero commutative ring. If $R$ has characteristic $p\neq 0$ , then

pa\overset{\smash{\lower 1.31395pt\hbox{{def}}}}{=}\underbrace{a+\cdots+a}_{p% \text{ terms}}=\underbrace{(1_{R}+\cdots+1_{R})}_{p\text{ terms}}a=0a=0

for all $a\in R$ . Conversely, if $pa=0$ for all $a\in R$ , then $R$ has characteristic $p$ .

Let $R$ be a nonzero commutative ring. The usual proof by induction shows that the binomial theorem

(a+b)^{m}=a^{m}+\tbinom{m}{1}a^{m-1}b+\tbinom{m}{2}a^{m-2}b^{2}+\cdots+b^{m}

holds in $R$ . If $p$ is prime, then it divides

\begin{pmatrix}p\\ r\end{pmatrix}\overset{\smash{\lower 1.31395pt\hbox{{def}}}}{=}\frac{p!}{r!(p-% r)!}

for all $r$ with $1\leq r\leq p-1$ because it divides the numerator but not the denominator. Therefore, when $R$ has characteristic $p$ ,

(a+b)^{p}=a^{p}+b^{p}\quad\text{for all }a,b\in R,

and so the map $a\mapsto a^{p}\colon R\rightarrow R$ is a homomorphism of rings (even of $\mathbb{F}{}_{p}$ -algebras). It is called the Frobenius endomorphism of $R$ . The map $a\mapsto a^{p^{n}}\colon R\rightarrow R$ , $n\geq 1$ , is the composite of $n$ copies of the Frobenius endomorphism, and so it also is a homomorphism. Therefore,

(a_{1}+\cdots+a_{m})^{p^{n}}=a_{1}^{p^{n}}+\cdots+a_{m}^{p^{n}}

for all $a_{i}\in R$ .

When $F$ is a field, the Frobenius endomorphism is injective, and hence an automorphism if $F$ is finite.

The characteristic exponent of a field $F$ is $1$ if $F$ has characteristic $0$ , and $p$ if $F$ has characteristic $p\neq 0$ . Thus, if $q$ is the characteristic exponent of $F$ and $n\geq 1$ , then $x\mapsto x^{q^{n}}$ is an isomorphism of $F$ onto a subfield of $F$ (denoted $F^{q^{n}}$ ).