3.6 Finite Field

Lemma

Let $F$ be a finite field. Then $\mathrm{char}F=p\ne 0$ and $|F|=p^n$ for some $n⩾1$.

Lemma

Let $F$ be any field, and let $G<F^{\times }$ be a finite subgroup. Then $G$ is a cyclic group. In particular, if $|F|<\infty$, then $F^{\times }$ is a cyclic group.

\begin{proof} See here.

Alternating proof. Since $G$ is abelian, $G\simeq \mathbb{Z}/m_1\mathbb{Z}\oplus \cdots \oplus \mathbb{Z}/m_k\mathbb{Z}$ with $m_1\mid \cdots \mid m_k$. For any $g\in G$, $g^{m_k}=1$ and so it is a root of $T^{m_k}-1\in F[T]$. It deduces that $|G|⩽m_k$ and so $k=1$, that is, $G$ is cyclic. \end{proof}

Remark. Consider $R=\mathbb{C}[x,y,z]/(x^2,y^2,z^2)$. Then $T^2=0$ has more than $2$ roots. Thus, $F$ is a field is necessary.

Corollary

If $F/{\mathbb{F}}_p$ is a finite extension, then $F$ is a simple extension.

\begin{proof} By ^91fdac, $F^{\times }$ is cyclic. \end{proof}

Theorem

Let $F$ be a finite field. Then $|F|=p^n$ iff $F$ is a splitting field of $T^{p^n}-T$ over ${\mathbb{F}}_p$.

\begin{proof} "→": Assume that $|F|=p^n$, then $F^{\times }=⟨u⟩$ with $u^{p^n-1}=1$. Thus all elements of $F$ are roots of $f(T)=T^{p^n}-T$ and $F$ is a splitting field of $f$.

"←": Assume that $F$ is a splitting field of $f(T)=T^{p^n}-T$. Since $f^{\prime }=-1$, it has $p^n$‘s distinct roots, denoted by $u_1,\cdots ,u_{p^n}$. Note that $\{u_1,\cdots ,u_{p^n}\}$ is already a field containing ${\mathbb{F}}_p$, because $(u_i\pm u_j{)}^{p^n}=u_i\pm u_j$. Now we finish the proof. \end{proof}

Remark. If $n⩾2$, ${\mathbb{F}}_{p^n}\not\simeq \mathbb{Z}/p^n\mathbb{Z}$ as $1\cdot p\ne 0$ for $1\in \mathbb{Z}/p^n\mathbb{Z}$.

Corollary

The followings hold.

• If $|F_1|=|F_2|=p^n$, then $F_1\simeq F_2$.
• ${\mathbb{F}}_{p^n}/{\mathbb{F}}_p$ is Galois.

\begin{proof} i) is obvious by ^4a9b03 and ^rkttex.

ii) By ^m1looz, it suffices to show $|{\mathrm{Aut}}_{{\mathbb{F}}_p}({\mathbb{F}}_{p^n})|=n$. Define the Frobenius map $\phi :u\mapsto u^p$, then $\phi \in {\mathrm{Aut}}_{{\mathbb{F}}_p}({\mathbb{F}}_{p^n})$ and $|\phi |=n$. \end{proof}

Proposition

Let $K={\mathbb{F}}_{p^r}$ with $r⩾1$. If $E/K$ is a finite extension of degree $s$, then $E\simeq {\mathbb{F}}_{p^{rs}}$, and $E/K$ is Galois with $\mathrm{Gal}(E/K)\simeq \mathbb{Z}/s\mathbb{Z}$ with a generator given by the ${\mathbb{F}}_{p^r}$-Frobenius ${\phi }_r:x\to x^{p^r}$.

\begin{proof} Since $E/K$ is of degree $s$, we have $|E|=p^{rs}$ and so by ^de0904 $E\simeq {\mathbb{F}}_{p^{rs}}$. Since $\mathrm{Aut}({\mathbb{F}}_{p^{rs}}/{\mathbb{F}}_{p^r})$ contains Frobenius map ${\phi }_r$, the extension ${\mathbb{F}}_{p^{rs}}/{\mathbb{F}}_{p^r}$ is Galois and so for $E/K$. \end{proof}

Remark. A natural corollary is

${\mathbb{F}}_{p^s}$ is a subfield of ${\mathbb{F}}_{p^r}$ if and only if $s\mid r$, and such subfield is unique.

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