Suppose $F$ is a finite field of characteristic $p$ ($p$ a prime). Prove $\exists u \in F$ such that $F = \mathbb{F}_{p}(u)$. Here, $\mathbb{F}_{p}$ denotes the field with $p$ elements.
Here is what I know:
$F^{\times}$, the group under multiplication consisting of $F - \{ 0 \}$, is cyclic since $F$ is a finite field. I also know since $F$ is of characteristic $p$ that $\mathbb{F}_{p}$ is a subgroup of $F$.
Now, if we let $u$ be the generator of $F^{\times}$, then every nonzero element can be written as $u^{m}$ for some $m \in \mathbb{N}$. This shows that $F \subseteq \mathbb{F}_{p}(u)$. Also, $F$ is field that contains $u$ and $\mathbb{F}_{p}$, so it must contain the smallest field containing both of these, which is $\mathbb{F}_{p}(u)$. Thus, we have $\mathbb{F}_{p}(u) = F$.
So, if I can prove the claim, then what is my question? Well, we proved the sets are equal. But we didn't prove they are isomorphic. Are they? What is the homomorphism? Do you just assume the operations on $\mathbb{F}_{p}(u)$ and $\mathbb{F}$ behave the same?
Proving two sets are equal is not enough to prove the underlying structure is the same, and this is what I am doubting. Is the field $\mathbb{F}_{p}(u)$ the same as the field $F$?
Think about $\mathbb F_p$ as being the field generated by $1$ in $F$. That means that $$\mathbb F_p = \{ 1_F, 1_F+1_F,..., 1_F+1_F+...+1_F \}$$
Then $\mathbb F_p \subset F$ and $u \in F$. Now when you construct $\mathbb F_p(u)$ you construct it as a subfield of $F$. If the underlying set is the same, it follows that the subfield contains all elements.