I have recently started studying algebraic field extensions and I got to know that algebraic closures $\overline{F}$ of finite fields $F$ are infinite. Therefore, I've asked myself the following question and couldn't come up with an answer.
Suppose $F$ was a finite field, i.e. $|F| < \infty$. Is it possible to construct an algebraic extension $F'$ of $F$ sucht that $|F'| = \infty$ and $F' \subsetneq \overline{F}$? Meaning, that $F'$ is not algebraically closed.
Best Answer
Fix an algebraic closure $\overline{\mathbb{F}}_3$ and take $F = \varinjlim_{n = 3^k} \mathbb{F}_{3^n}\subseteq\overline{\mathbb{F}}_3$ (i.e., $F = \bigcup_{n = 3^k}\mathbb{F}_{3^n}$). Now, the polynomial $x^2 + 1$ is irreducible over $\mathbb{F}_3$ (it's quadratic, so just test to see that it has no roots). Let $i\in\mathbb{F}_9 = \mathbb{F}[x]/(x^2+1)$ be an element such that $i^2 + 1 = 0$. I claim that $i\not\in F$. To see this, we note that any element $f\in F$ lies in some finite extension $\mathbb{F}_{3^{3^k}}$, and say that no smaller extension of $\mathbb{F}_3$ contains $f$. Then $f$ satisfies a minimal polynomial with $\mathbb{F}_3$ coefficients of degree $3^k$, and as such, $f\neq i$ (because $i$'s minimal polynomial over $\mathbb{F}_3$ has degree $2$).