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In Field theory by Steven Roman Chapter 9 Exercise 20, if we write the algebraic closure of finite field $F_q$ as $\Gamma(q)$ and $a_n$ be any strictly increasing infinite sequence of positive integers, the exercise wants 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 current conditions offered.

Hope for answers!

Wembley Inter
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    any sequence of positive integers such that any $k$ divides some $a_n$ – reuns Dec 07 '18 at 20:29
  • @reuns That condition is both necessary and sufficient, right? – Wembley Inter Dec 07 '18 at 20:47
  • If a<b isn’t $GF(q^a)$ a subfield of $GF(q^b)$, since the latter is the splitting field of $x^{q^b}-1$, which has $x^{q^a}-1$ a as factor? So then the union is a field if we are implicitly assuming the union incorporates these inclusions – usr0192 Dec 08 '18 at 02:49
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    No, @usr0192, for instance $\Bbb F_4\not\subset\Bbb F_8$. – Lubin Dec 08 '18 at 05:11
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    My earlier comment seems to have gotten lost: this author has been unacceptably sloppy in asking you to prove something that just isn’t so. The suggestion of @reuns will work, as will the specification of ${a_n}$ to be the sequence of factorials: $a_n=n!$. I think the best way of looking at this problem is to consider $\Gamma(q)$ to be the direct limit of the finite fields $\Bbb F_{q^n}$, with respect to the directed set of the positive integers, ordered by divisibility. – Lubin Dec 08 '18 at 05:18
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    The same question on MathOverflow: Write the algebra closure of $F_p$ as union of finite fields. This answer has a very reasonable advice about cross-posting - and you can also looks at other discussions on [meta-tag:cross-posting]. – Martin Sleziak Dec 08 '18 at 07:04
  • @Lubin Thank you! I now see the Galois group of $GF(q^b)$ is cyclic of order $b$, so the only subgroups are those of order dividing $b$. Or another way, the degree of $GF(q^b):GF(q)$ is b so the degree of any intermediate field has to divide b. – usr0192 Dec 08 '18 at 14:06

1 Answers1

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Since the question is now closed on MathOF (and rather belongs on here), here's an answer in the form of an exercise ((a),(b) below). Yes, the exercise in Roman's book is false and it is quite hardly reassuring that such an exercise appears as published in a GTM book.

First, to be rigorous, we should work within one given algebraically closed field $K$ with characteristic $p$ (where $p$ is the unique prime divisor of $q$), so that $\mathrm{GF}(q^k)$ is a well-defined subfield of $K$, and so that the union makes sense as a subset of $K$.

(a) As suggested by @reuns (and copied by the OP to the MO post without reference), for a sequence $a=(a_n)$, the union $L_a=\bigcup_n\mathrm{GF}(q^{a_n})$ is an algebraic closed subfield of $K$ iff for every $k\ge 1$ there exists $n$ such that $k$ divides $a_n$;

(b) this union $L_a$ is a subfield of $K$ iff for any $m,m'$, there exists $n$ such that $\mathrm{lcm}(a_m,a_{m'})$ divides $a_n$.

(c) for every subfield $L$ of $K$ algebraic over $\mathrm{GF}(q)$, there exists a sequence $a=(a_n)$, such that $a_n$ divides $a_{n+1}$ for every $n$, such that $L=L_a$.

For instance, if $a_n=2^n$ for all $n$, then the union is a field, but not algebraically closed. If $a_n$ is the $n$-th prime, the union is not a field.

YCor
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