What is the algebraic closure of $\mathbb F_q$ with $q$ being some power of a prime $p$ ?
I wrote, ''the algebraic closure'' because, they're the same up to isomorphism right ?
It cannot be finite, otherwise it is not algebraically closed, so how does it look like ?
Given finite fields $\mathbb{F}_{p^m}$ and $\mathbb{F}_{p^n}$ with $\gcd(m, n) = 1$ then the compositum is the finite field $\mathbb{F}_{p^{mn}}$. This allows us to define the algebraic closure of $\mathbb{F}_{p}$ as the union $$ \overline{\mathbb{F}_{p}}=\bigcup_{k\ge 1} \mathbb{F}_{p^k}. $$ For prime powers $q=p^n$ the algebraic closure $\overline{\mathbb{F}_{q}}$ can be constructed by building and gluing $\ell$-adic towers $$ \mathbb{F}_{q}\subset \mathbb{F}_{q^{\ell}}\subset \mathbb{F}_{q^{\ell^2}}\subset \cdots $$ see here for an algorithm and an impressive picture on page $5$.