Let $F$ be a finite field, and $f(x) \in F[x]$ is some non constant polynomial, Let $E$ be the splitting field of $f(x)$ over $F$ , is $E$ always a finite field?
My attempt let $\deg(f) = n$ then there exist some $a_1,...,a_n$ such that $f(x) = a(x-a_1)...(x-a_n)$ in $E$ such that $E = F(a_1,...,a_n)$.
I try to prove that $E$ is finite field by induction on $i$.
First the base case $F(a_1)$ is finite field, consider the evaluation map $$\Phi_{a_1} : F[x] \to E\\ g(x) \mapsto g(a_1)$$ since $a_1$ is algebraic (as root of $f$ ) ,there exist a minimal polynomial $m_{a_1}$ with $m_{a_1} \mid f$ therefore $k = \deg(m_{a_1}) \le \deg(f)$ and all elements in $$F(a_1) = \{c_{k-1} a_1^{k-1}+ c_{k-2}a_1^{k-2} +...+ c_0\mid c_i \in F\}$$ (which is the case $a_1 \notin F$ otherwise $F(a_1) = F$),therefore $$|F(a_1)| \le |F|^{k}$$ which is finite.
Consider the induction case $F(a_1,..,a_k)$ is finite field, needs to prove $F(a_1,...,a_{k+1})$ is finite field, just replace $F$ above by $F(a_1,..,a_k)$ gets the result, for the case $k$.
Therefore $E$ is always a finite field, is my proof correct?
Your proof looks fine. Also consider the following. Any finite extension is a finite dimensional vector space over the base field. If your base field is finite, then any finite dimensional vector space over it is also finite. Splitting fields are finite extension.
Another comment is that you seem to be a bit confused about the relationship between being algebraic and having a minimal polynomial. In both cases, it is important to keep in mind the context in which you are working. For example $a_1$ may or may not be algebraic over $F$ but it is by definition algebraic over $E$.