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

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.

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

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• 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... – YCor Dec 7 '18 at 21:21
• Right you are. Perhaps the easiest way to remedy the author’s failing is to replace “any strictly increasing infinite sequence” with “the sequence $a_n=n!$”. – Lubin Dec 7 '18 at 22:15
• 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. – Martin Sleziak Dec 8 '18 at 7:06
• @MartinSleziak Thanks for your advice. As a new contributor, I'll learn from this. – Wembley Inter Dec 8 '18 at 15:08
• 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. – YCor Dec 9 '18 at 11:13

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 $$a_i$$.