finite field

A finite field (also called a Galois field) is a field that has finitely many elements.The number of elements in a finite field is sometimes called the order of the field.We will present some basic facts about finite fields.

1 Size of a finite field

Theorem 1.1.

A finite field $F$ has positive characteristic $p\mathrm{>}\mathrm{0}$ for some prime $p$. Thecardinality of $F$ is $p^n$ where $n\mathrm{:=}\mathrm{[}F\mathrm{:}{\mathrm{F}}_p\mathrm{]}$ and ${\mathrm{F}}_p$ denotesthe prime subfield of $F$.

Proof.

The characteristic of $F$ is positive because otherwise the additivesubgroup generated by $1$ would be an infinite subset of$F$. Accordingly, the prime subfield ${𝔽}_p$ of $F$ is isomorphic tothe field $\mathbb{Z}/p\mathbb{Z}$ of integers mod $p$. The integer $p$ is prime since otherwise $\mathbb{Z}/p\mathbb{Z}$ would have zero divisors. Since the field $F$ is an$n$–dimensional vector space over ${𝔽}_p$ for some finite $n$, it is set–isomorphic to${𝔽}_p^n$ and thus has cardinality $p^n$.∎

2 Existence of finite fields

Now that we know every finite field has $p^n$ elements, it is naturalto ask which of these actually arise as cardinalities of finitefields. It turns out that for each prime $p$ and each natural number$n$, there is essentially exactly one finite field of size $p^n$.

Lemma 2.1.

In any field $F$ with $m$ elements, the equation $x^m\mathrm{=}x$ is satisfied by all elements $x$ of $F$.

Proof.

The result is clearly true if $x=0$. We may therefore assume $x$ is not zero. By definition of field, the set $F^{\times }$ of nonzero elements of $F$ forms a group under multiplication. This set has $m-1$ elements, and by Lagrange’s theorem $x^{m-1}=1$ for any $x\in F^{\times }$, so $x^m=x$ follows.∎

Theorem 2.2.

For each prime $p\mathrm{>}\mathrm{0}$ and each natural number $n\mathrm{\in }\mathrm{N}$, thereexists a finite field of cardinality $p^n$, and any two such areisomorphic.

Proof.

For $n=1$, the finite field ${𝔽}_p:=\mathbb{Z}/p\mathbb{Z}$ has $p$ elements, and anytwo such are isomorphic by the map sending $1$ to $1$.

In general, the polynomial $f(X):=X^{p^n}-X\in {𝔽}_p[X]$ hasderivative $-1$ and thus is separable over ${𝔽}_p$. We claim that thesplitting field $F$ of this polynomial is a finite field of size$p^n$. The field $F$ certainly contains the set $S$ of roots of$f(X)$. However, the set $S$ is closed under the field operations, so$S$ is itself a field. Since splitting fields are minimal bydefinition, the containment $S\subset F$ means that $S=F$. Finally,$S$ has $p^n$ elements since $f(X)$ is separable, so $F$ is a field ofsize $p^n$.

For the uniqueness part, any other field $F^{\prime }$ of size $p^n$ contains asubfield isomorphic to ${𝔽}_p$. Moreover, $F^{\prime }$ equals the splitting field ofthe polynomial $X^{p^n}-X$ over ${𝔽}_p$, since by Lemma 2.1 every element of $F^{\prime }$ is a root of this polynomial, and all $p^n$ possible roots of the polynomial are accounted for in this way. By the uniqueness ofsplitting fields up to isomorphism, the two fields $F$ and $F^{\prime }$ areisomorphic.∎

Note: The proof of Theorem 2.2 given here, while standardbecause of its efficiency, relies on more abstract algebra than isstrictly necessary. The reader may find a more concrete presentationof this and many other results about finite fieldsin [1, Ch. 7].

Corollary 2.3.

Every finite field $F$ is a normal extension of its prime subfield${\mathrm{F}}_p$.

Proof.

This follows from the fact that field extensions obtained fromsplitting fields are normal extensions.∎

3 Units in a finite field

Henceforth, in light of Theorem 2.2, we will write ${𝔽}_q$for the unique (up to isomorphism) finite field of cardinality $q=p^n$. A fundamental step in the investigation of finite fields is theobservation that their multiplicative groups are cyclic:

Theorem 3.1.

The multiplicative group ${\mathrm{F}}_q^{\mathrm{*}}$ consisting of nonzero elements ofthe finite field ${\mathrm{F}}_q$ is a cyclic group.

Proof.

We begin with the formula

$$\sum _{d\mid k}\varphi (d)=k,$$(1)

where $\varphi$ denotes the Euler totient function. It is proved asfollows. For every divisor $d$ of $k$, the cyclic group $C_k$ of size$k$ has exactly one cyclic subgroup $C_d$ of size $d$. Let $G_d$ bethe subset of $C_d$ consisting of elements of $C_d$ which have themaximum possible order (http://planetmath.org/OrderGroup) of $d$. Since every element of $C_k$ hasmaximal order in the subgroup of $C_k$ that it generates, we see thatthe sets $G_d$ partition the set $C_k$, so that

$$\sum _{d\mid k}|G_d|=|C_k|=k.$$

The identity (1) then follows from the observation that thecyclic subgroup $C_d$ has exactly $\varphi (d)$ elements of maximal order$d$.

We now prove the theorem. Let $k=q-1$, and for each divisor $d$ of$k$, let $\psi (d)$ be the number of elements of ${𝔽}_q^{*}$ of order$d$. We claim that $\psi (d)$ is either zero or $\varphi (d)$. Indeed, ifit is nonzero, then let $x\in {𝔽}_q^{*}$ be an element of order $d$, andlet $G_x$ be the subgroup of ${𝔽}_q^{*}$ generated by $x$. Then $G_x$ hassize $d$ and every element of $G_x$ is a root of the polynomial $x^d-1$. But this polynomial cannot have more than $d$ roots in a field, soevery root of $x^d-1$ must be an element of $G_x$. In particular,every element of order $d$ must be in $G_x$ already, and we see that$G_x$ only has $\varphi (d)$ elements of order $d$.

We have proved that $\psi (d)\le \varphi (d)$ for all $d\mid q-1$. If$\psi (q-1)$ were 0, then we would have

which is impossible since the first sum must equal $q-1$ (becauseevery element of ${𝔽}_q^{*}$ has order equal to some divisor $d$ of$q-1$).∎

A more constructive proof of Theorem 3.1, which actuallyexhibits a generator for the cyclic group, may be foundin [2, Ch. 16].

Corollary 3.2.

Every extension of finite fields is a primitive extension.

Proof.

By Theorem 3.1, the multiplicative group of the extension field is cyclic. Any generator of the multiplicative group of the extension field also algebraically generates the extension field over the base field.∎

4 Automorphisms of a finite field

Observe that, since a splitting field for $X^{q^m}-X$ over ${𝔽}_p$contains all the roots of $X^q-X$, it follows that the field${𝔽}_{q^m}$ contains a subfield isomorphic to ${𝔽}_q$. We will showlater (Theorem 4.2) that this is the only way that extensions offinite fields can arise. For now we will construct the Galois group ofthe field extension ${𝔽}_{q^m}/{𝔽}_q$, which is normal byCorollary 2.3.

Theorem 4.1.

The Galois group of the field extension ${\mathrm{F}}_{q^m}\mathrm{/}{\mathrm{F}}_q$ is a cyclicgroup of size $m$ generated by the $q^{\mathrm{th}}$ power Frobenius map${\mathrm{Frob}}_q$.

Proof.

The fact that ${\mathrm{Frob}}_q$ is an element of $\mathrm{Gal}({𝔽}_{q^m}/{𝔽}_q)$, andthat ${({\mathrm{Frob}}_q)}^m={\mathrm{Frob}}_{q^m}$ is the identity on ${𝔽}_{q^m}$, isobvious. Since the extension ${𝔽}_{q^m}/{𝔽}_q$ is normal and of degree$m$, the group $\mathrm{Gal}({𝔽}_{q^m}/{𝔽}_q)$ must have size $m$, and we willbe done if we can show that ${({\mathrm{Frob}}_q)}^k$, for $k=0,1,\mathrm{\dots },m-1$, are distinct elements of $\mathrm{Gal}({𝔽}_{q^m}/{𝔽}_q)$.

It is enough to show that none of ${({\mathrm{Frob}}_q)}^k$, for $k=1,2,\mathrm{\dots },m-1$, is the identity map on ${𝔽}_{q^m}$, for then we will haveshown that ${\mathrm{Frob}}_q$ is of order exactly equal to $m$. But, if anysuch ${({\mathrm{Frob}}_q)}^k$ were the identity map, then the polynomial $X^{q^k}-X$ would have $q^m$ distinct roots in ${𝔽}_{q^m}$, which isimpossible in a field since .∎

We can now use the Galois correspondence between subgroups of theGalois group and intermediate fields of a field extension toimmediately classify all the intermediate fields in the extension${𝔽}_{q^m}/{𝔽}_q$.

Theorem 4.2.

The field extension ${\mathrm{F}}_{q^m}\mathrm{/}{\mathrm{F}}_q$ contains exactly one intermediatefield isomorphic to ${\mathrm{F}}_{q^d}$, for each divisor $d$ of $m$, and noothers. In particular, the subfields of ${\mathrm{F}}_{p^n}$ are precisely thefields ${\mathrm{F}}_{p^d}$ for $d\mathrm{\mid }n$.

Proof.

By the fundamental theorem of Galois theory, each intermediate fieldof ${𝔽}_{q^m}/{𝔽}_q$ corresponds to a subgroup of$\mathrm{Gal}({𝔽}_{q^m}/{𝔽}_q)$. The latter is a cyclic group of order $m$, soits subgroups are exactly the cyclic groups generated by${({\mathrm{Frob}}_q)}^d$, one for each $d\mid m$. The fixed field of${({\mathrm{Frob}}_q)}^d$ is the set of roots of $X^{q^d}-X$, which forms asubfield of ${𝔽}_{q^m}$ isomorphic to ${𝔽}_{q^d}$, so the resultfollows.

The subfields of ${𝔽}_{p^n}$ can be obtained by applying the aboveconsiderations to the extension ${𝔽}_{p^n}/{𝔽}_p$.∎