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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!

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closed as off-topic by YCor, Andreas Thom, Andreas Blass, Andrés E. Caicedo, Felipe Voloch Dec 8 '18 at 14:38

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    $\begingroup$ It's not a field, for instance is $a_n$ is the $n$-th prime. If $a_n=2^n$ it's a field, but it's not algebraically closed... $\endgroup$ – YCor Dec 7 '18 at 21:21
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    $\begingroup$ Right you are. Perhaps the easiest way to remedy the author’s failing is to replace “any strictly increasing infinite sequence” with “the sequence $a_n=n!$”. $\endgroup$ – Lubin Dec 7 '18 at 22:15
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    $\begingroup$ The same question on Mathematics: Write the algebra closure of $F_p$ as union of finite fields. This answer has some reasonable advice about cross-posting. Another things to keep in mind is that this site has different tags. For example, the tag (abstract-algebra) is deprecated and it is recommended to use at least one top-level tag. $\endgroup$ – Martin Sleziak Dec 8 '18 at 7:06
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    $\begingroup$ @MartinSleziak Thanks for your advice. As a new contributor, I'll learn from this. $\endgroup$ – Wembley Inter Dec 8 '18 at 15:08
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    $\begingroup$ By the way, the "I believe that" addendum was added by the OP after it was suggested as a comment (by @reuns) on the MathSE site. This is not very fair conduct. $\endgroup$ – YCor Dec 9 '18 at 11:13
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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 $a_i$.

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    $\begingroup$ ... it's about research level mathematics. $\endgroup$ – Andreas Thom Dec 8 '18 at 7:18

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