Let $K$ be a finite extension field of a finite field $F$. Show that there is an element $a\in K$ s.t. $K = F(a)$.

My attempt:

$K$ is a finite field and $char(K) = char(F) := p$. I know that for a given prime $p$ and $n\in N$, there exists a unique finite field upto isomorphism of order $p^n$.

1) But how can I show that $K \simeq GF(p^n)$ where $p = char(K)$?

Once I have that, $K\setminus \{0\}$ is a cyclic group under multiplication and thus $\exists a\in K$, s.t. $K\setminus \{0\} = <a>\subset F(a)$. Thus $K\subset F(a)$.

2) Also, how can I show that $F(a)\subset K$ so that $F(a) = K$? $K$ is an finite field extension of $F$ but I can't understand why that would imply $F(a)\subset K$??

  • 1
    $\begingroup$ $F(a)$ is the smallest field containing $F,a$, so $F(a) \subset K$. That $a$ generates $K^\times$ means $K = F(a)$. That $K$ is the unique field (up to isomorphism/renaming) with $p^n = |K|$ elements is because $K$ is the unique splitting field of $x^{p^n}-x \in \Bbb{F}_p[x]$. Similarly, that $K^\times$ is cyclic is because for each $d | p^n-1$, $x^d-1$ has at most $d$ roots in the field $K$. $\endgroup$ – reuns Apr 21 '19 at 18:35

Let $a$ be a generator of the multiplicative group $K^\times$. The elements $a,a^2,...,a^{|K|-1}$ are all different elements in $F(a)$ and hence $|F(a)|\geq |K|-1$. On the other hand $F(a)\subseteq K$ and hence $|F(a)|\leq |K|$. Since the order of $F(a)$ must be a power of $p$ (and $|K|-1$ is not a power of $p$) we conclude that $|F(a)|=|K|$. Since $F(a)\subseteq K$ and both fields are finite this implies $F(a)=K$.

  • $\begingroup$ Thanks, a few questions. Why must $|F(a)|$ be a power of $p$? Also, why can $|K| - 1$ not be a power of $p$? $\endgroup$ – manifolded Apr 21 '19 at 18:52
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    $\begingroup$ The order of any finite field is a power of a prime. Since the characteristic of $F(a)$ is $p$ (which is the case because $F(a)$ is an extension of $F$ which has characteristic $p$) we know that here the prime is $p$. And $|K|-1$ is not a power of $p$ because $|K|$ is. (since again, $|K|$ is the order of a finite field with characteristic $p$) $\endgroup$ – Mark Apr 21 '19 at 18:53

Since the finite extension of a finite field is also a finite field, so there are at most finitely many middle fields $E_i$ such that $F\subset E_i\subset E$. so $E/F$ is a simplicial extension.

  • $\begingroup$ Welcome to MSE. If you want to get $F\setminus E$, then type F\setminus E. $\endgroup$ – José Carlos Santos Dec 19 '20 at 14:20
  • $\begingroup$ @JoséCarlosSantos my apologize. I just corrected the typo $\endgroup$ – Alexander Lau Dec 20 '20 at 9:43

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