Properties of Finite Field

Proposition

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

\begin{proof} Note that a finite field ${\mathbb{F}}_q$ of characteristic $p$ has cardinality $q=p^r$ for some number $r⩾1$. Furthermore, ${\mathbb{F}}_q$ can be realized as the set of the roots of the polynomial $X^q-X$ in some algebraic closure of ${\mathbb{F}}_p$.

If ${\mathbb{F}}_q\supseteq {\mathbb{F}}_{q^{\prime }}\supseteq {\mathbb{F}}_p$ is a sub-extension with $q=p^r$ and $q^{\prime }=p^s$. Since ${\mathbb{F}}_q$ is also a ${\mathbb{F}}_{q^{\prime }}$-vector space, we have that $s\mid r$. On the other hand, the condition is also sufficient, because the roots of the polynomial $X^{p^s}-X$ roots are also of $X^{p^r}-X$.

Therefore, ${\mathbb{F}}_{p^r}$ contains a field with $p^s$ elements if and only if $s\mid r$, and such subfield is unique.

\end{proof}

Proposition

The followings hold.

• For any prime $p$ and positive integer $n$, there exists a finite field $F$ with $|F|=p^n$. Furthermore, $F$ is unique up to isomorphism.
• In fact, $F$ is isomorphic to the splitting field of $x^{p^n}-x$ over ${\mathbb{F}}_p$.
• $F^{\times }$ is a cyclic group.