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Let $\mathbb{F}^{\operatorname{alg}}_p$ be the algebraic closure of the finite field with $p$ elements.

I know that any finitely generated subfield of $\mathbb{F}^{\operatorname{alg}}_p $ is contained in some finite sub-union $\displaystyle{\bigcup_{n=1,2,\dots,N} \mathbb{F}_{p^n}}$.

However, I am lacking intuition for why this is so, and I'm having trouble conceiving what a finitely generated subfield of $\mathbb{F}^{\operatorname{alg}}_p $ would look look like.

Though I will continue reading on the matter, any examples of such a finitely generated, and in turn finite, subfield would be greatly appreciated.

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    $\begingroup$ A finite union of finite fields isn't a field... unless one of the fields contains all of the others. $\endgroup$ – user14972 May 27 '13 at 3:21
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What do you mean by $(\mathbf{F}_p^{\mathrm{alg}})^n$? A direct product of $n$ copies of $\mathbf{F}_p^{\mathrm{alg}}$? If you actually just mean a finitely generated subfield of $\mathbf{F}_p^{\mathrm{alg}}$, then this is a general fact about adjoining algebraic elements to a field.

If $F$ is a field and $\alpha_1,\ldots,\alpha_n$ are elements of some extension $K$ of $F$, each of which is algebraic over $F$, then $F(\alpha_1,\ldots,\alpha_n)$ is of finite degree over $F$. This is because you have a tower $F\subseteq F(\alpha_1)\subseteq F(\alpha_1,\alpha_2)\subseteq\cdots\subseteq F(\alpha_1,\ldots,\alpha_n)$, and in each step, you're adjoining a single algebraic element, so you get a finite extension.

In the case of $\mathbf{F}_p^{\mathrm{alg}}$, each of the $\alpha_i$ must lie in some finite extension $F_i$ of $\mathbf{F}_p$ in $\mathbf{F}_p^{\mathrm{alg}}$, and $F_i$ must be of the form $\mathbf{F}_{p^{n_i}}$ for some $n_i\geq 1$ (meaning the unique subfield of $\mathbf{F}_p^{\mathrm{alg}}$ of degree $n_i$ over $\mathbf{F}_p$). But if $n=\max_i n_i$, then each of these $F_i$ is contained in $\mathbf{F}_{p^n}$, so all the $\alpha_i$ lie in $\mathbf{F}_{p^n}$.

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  • $\begingroup$ I have edited the question, I did not intend to include the n copies of the field in the question $\endgroup$ – Nik Kumar May 27 '13 at 3:33
  • $\begingroup$ also, this clears things up greatly. Thank you. $\endgroup$ – Nik Kumar May 27 '13 at 3:33
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For typographical simplicity, let’s write $\mathscr F\!_n$ for $\mathbb F_{p^n}$. Then one sees that $\mathscr F\!_m\subset \mathscr F\!_n$ if and only if $m|n$. It follows that $\mathscr F\!_m \mathscr F\!_n=\mathscr F\!_\ell$, where $\ell$ is the least common multiple of $m$ and $n$. Now, since every simply generated subfield of your algebraically closed field is a finite field $\mathscr F\!_m$ for some $m$, it follows that every finitely generated subfield is also simply generated.

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