How do I show the existence of an intermediate field between $\mathbb{F}_p$ and its algebraic closure? I am currently struggling with field theory and I don't know how to solve the following problem:
Let $\mathbb{F}_q$ be a finite field and $\overline{\mathbb{F}}_q$ one of its algebraic closures. I want to find out that there is a unique intermediate field $\mathbb{F}_q\subset K \subset \overline{\mathbb{F}}_q$ with $q^n$ elements. I have to do it by showing that every such intermediate field $K$ has to be a splitting field of the same polynomial $f\in \mathbb{F}_q[t]$.
What I already know is that
$$\overline{\mathbb{F}}_q=\bigcup_{n\in\mathbb{N}}\mathbb{F}_{q^{n!}}$$
up to isomorphism (since all algebraic closures are isomorphic to each other). What is irritating me here is that the problem is about the existence of "a unique" field but I already know that there is only one field $K$ with $q^n$ elements (up to  isomorphism), which we call $\mathbb{F}_{q^n}$.
Well, before I get more confused about this problem, I hope you guys can help me. What $f$ should I take and what is the connection of $f$ to this intermediate field $K$?
Regards, Kira
 A: Fix an algebraic closure $\overline{\Bbb F}_q$ of $\Bbb F_q$. Note that any finite field extension $K/\Bbb F_q$ must have $q^n$ elements for some $n\in\Bbb N$, as $K$ is an $\Bbb F_q$-vector space of some finite dimension $n$.
First, uniqueness. Suppose we have a field $K\subseteq\overline{\Bbb F}_q$ with $q^n$ elements. Then by Lagrange's theorem, $\alpha^{q^n - 1} = 1$ for any $\alpha\in K^\times$, and it follows easily that $\alpha^{q^n} = \alpha$ for any $\alpha\in K$. Hence, every element $\alpha\in K$ is a root of $t^{q^n} - t\in\Bbb F_q[t]$, and there are exactly $q^n$ roots of $t^{q^n} - t$ (counting multiplicity) inside $\overline{\Bbb F}_q$. We can also see that this polynomial has no repeated roots, as its derivative is $-1$, which is clearly coprime to $t^{q^n} - t$. Thus, a field $K\subseteq\overline{\Bbb F}_q$ with $q^n$ elements must contain exactly the roots of $t^{q^n} - t$, and is hence unique within the fixed algebraic closure $\overline{\Bbb F}_q$.
Next, existence. The above argument suggests that $K$ should be the splitting field of $t^{q^n} - t$, so we must show that the splitting field of this polynomial indeed contains exactly $q^n$ elements (note that this is not immediate: the splitting field of $t^{q^n} - t$ exists over any given base field, and if the base field does not have characteristic $p\mid q$, it will be impossible for the splitting field to have $q^n$ elements). To that end, let $\alpha,\beta\in\overline{\Bbb F}_q$ be roots of $f(t) = t^{q^n} - t$. We must show that $-\alpha$, $\alpha + \beta$, $\alpha\beta$, and $\alpha^{-1}$ (for $\alpha\neq 0$) are also roots of this polynomial (so that the roots of $f$ within $\overline{\Bbb F}_q$ actually form a subfield and not just a subset). We compute:
\begin{align*}
(-\alpha)^{q^n} - (-\alpha) &= -(\alpha^{q^n} - \alpha)\\
&= 0,\\
(\alpha + \beta)^{q^n} - (\alpha + \beta) &= \alpha^{q^n} + \beta^{q^n} - \alpha - \beta\qquad\textrm{(because }\operatorname{char}\Bbb F_q\mid q^n)\\
&= \alpha^{q^n} - \alpha + \beta^{q^n} - \beta\\
&= f(\alpha) + f(\beta)\\
&= 0,\\
(\alpha\beta)^{q^n} - \alpha\beta &= \alpha^{q^n}\beta^{q^n} - \alpha\beta\\
&= \alpha\beta - \alpha\beta\qquad\textrm{(as }\alpha^{q^n} = \alpha, \beta^{q^n} = \beta)\\
&= 0.
\end{align*}
Finally, suppose $\alpha\neq 0$. Then $\alpha$ has a (nonzero) inverse $\alpha^{-1}\in\overline{\Bbb F}_q$, and because $\alpha\beta = 0$ in a field implies $\alpha = 0$ or $\beta = 0$ it is enough to show that $f(\alpha^{-1})\cdot\alpha^{q^n + 1} = 0$.
\begin{align*}
\left((\alpha^{-1})^{q^n} - \alpha^{-1}\right)\cdot\alpha^{q^n + 1} &= \left(\alpha^{-q^n} - \alpha^{-1}\right)\cdot\alpha^{q^n + 1}\\
&= \alpha^{-q^n}\alpha^{q^n + 1} - \alpha^{-1}\alpha^{q^n + 1}\\
&= \alpha^{q^n + 1 - q^n} - \alpha^{q^n + 1 - 1}\\
&= \alpha - \alpha^{q^n}\\
&= 0.
\end{align*}
Hence, the roots of $f$ within $\overline{\Bbb F}_q$ form a subfield, which has $q^n$ elements as previously shown.
