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Martin Sleziak
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This question follows Field theory by Steven Roman, ChpaterChapter 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!

This question follows Field theory by Steven Roman, Chpater 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!

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!

InThis question follows Field theory by Steven Roman, Chpater 9, Exercise 20, if we write.

Denote the algebraic closure of the finite field $F_q$ asby $\Gamma(q)$, and let $a_n$ be any strictly incresingincreasing infinite sequence of positive integers, the. 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 currentthe stated conditions offered.

In fact, I believe the conditionthat 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!

In Field theory by Steven Roman Chpater 9 Exercise 20, if we write the algebraic closure of finite field $F_q$ as $\Gamma(q)$ and $a_n$ be any strictly incresing 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.

In fact, I believe the condition 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!

This question follows Field theory by Steven Roman, Chpater 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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