Write the algebra closure of $F_p$ as union of finite fields [closed]

Question

This question follows Field theory by Steven Roman, Chapter 9, Exercise 20.

Denote the algebraic closure of the finite field $F_q$ by $\Gamma(q)$, and let $a_n$ be any strictly increasing infinite sequence of positive integers. The exercise wants us to prove that $\Gamma(q)=\bigcup_{n=0}^{\infty}GF(q^{a_n})$.

However, if $a_n$ is an arbitrary sequence, we are even unable to prove $\bigcup_{n=0}^{\infty}GF(q^{a_n})$ is a field. I wonder whether the exercise has omitted some condition since the equality doesn't hold under the stated conditions.

In fact, I believe that to demand that $a_n$ is any sequence of positive integers such that any positive integer $k$ divides some $a_n$ is both sufficient and necessary, though I'm not sure.

Hope for answers!

Answer 1

What you say sounds fine. The absolute Galois group of $F_q$ is $\widehat{\mathbb{Z}}$ and if you take some infinite quotient of this like $\mathbb{Z}_2$ (corresponding to a closed subgroup) then that corresponds to an infinite extension of $F_q$ with Galois group $\mathbb{Z}_2$, which you can write as the union of $GF(q^2)$, $GF(q^4)$, $GF(q^8)$, $GF(q^{16})$ and so on. If now you also throw in $GF(q^3)$ then you have something which is not a field. In general, as subfields of $\Gamma(q)$, $GF(q^a)$ is a subfield of $GF(q^b)$ if and only if $a$ divides $b$, so if you want the union to be all of $\Gamma(q)$ then you'd better have $GF(q^k)$ for all $k$ so you'd better have a multiple of $k$ in your sequence of $a_i$.