# A question about finite field extension of a finite field

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. $$K\setminus \{0\} = \subset F(a)$$. 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 $$F(a)\subset K$$??

• $F(a)$ 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 $K$. – 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 $$F(a)=K$$.
• Thanks, a few questions. Why must $|F(a)|$ be a power of $p$? Also, why can $|K| - 1$ not be a power of $p$? – manifolded Apr 21 '19 at 18:52
• 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 $p$) – 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.
• Welcome to MSE. If you want to get $F\setminus E$, then type F\setminus E. – José Carlos Santos Dec 19 '20 at 14:20