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Dummit and Foote (in Chapter 14.3) construct the algebraic closure of the finite field $\mathbb{F}_p$ by the following union: $$\bar{\mathbb{F}}_p = \bigcup_{n \geq 1}\mathbb{\mathbb{F}}_{p^n}.$$I'm having trouble seeing how we can take this infinite union, because to do so we have to view all the fields $\mathbb{F}_{p^n}$ as subsets of some larger object. Finite unions of the form $$\bigcup_{k=1}^n \mathbb{F}_{p^k}$$ make sense because all the fields in this union can be seen as subfields of a finite field of order $p^{n!}$, but I'm having trouble with the infinite case.

The authors write:

...given any two finite fields, $\mathbb{F}_{p^{n_1}}$ and $\mathbb{F}_{p^{n_2}}$ there is a third finite field containing (an isomorphic copy of) them, namely $\mathbb{F}_{p^{n_1n_2}}$. This gives us a partial ordering on these fields and allows us to think of their union.

How does the partial ordering give us a well defined union in the infinite case? Does it involve something like Choice or the principle of recursive definition? Is it true in general that a partial ordering on a set gives us a well-defined notion of a union even when the elements aren't literally sets contained in a larger set?

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    $\begingroup$ The right way to think of this is as a direct limit.... $\endgroup$
    – Angina Seng
    Commented Apr 9, 2020 at 15:03
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    $\begingroup$ @AnginaSeng This seems like the right way to look at it, but to define a direct system we would need a homomorphism $\mathbb{F}_{p^i} \to \mathbb{F}_{p^j}$ for all $i \leq j$. But if $i$ doesn't divide $j$ then the only such homomorphism is the zero homorphism, so how would we make this work? (I might be understanding this wrong; I only just skimmed the wikipedia article on direct limits.) $\endgroup$
    – Nick A.
    Commented Apr 9, 2020 at 15:11
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    $\begingroup$ @NickA. You don't have to use that order (and shouldn't, for exactly the reason you give). The order you should use is $m\leq n$ when $m|n$. Although in some sense this just dodges the issue by pushing it into the construction of the direct limit. $\endgroup$
    – user208649
    Commented Apr 9, 2020 at 19:01
  • $\begingroup$ @TokenToucan Makes sense, I'll check out the details of that construction. Knowing that there are more details that D&F gloss over rather than something more obvious that I missed is good enough for me. $\endgroup$
    – Nick A.
    Commented Apr 9, 2020 at 20:31
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    $\begingroup$ Nick, you did well spotting the problem with forming the union (+1). I am actually a bit unhappy, if that's all D&F offered as an explanation (I have many other reasons to dislike D&F, this just adds to the list). True, the subfields are unique as sets, but at the level of individual elements there are problems arising from the several ways of embedding a subfield. For the purposes of showing what the end product looks like, I guess this suffices. But, I would want to go throught the process in Maxime's answer. Consistency of embeddings is not automatic. $\endgroup$
    – Jyrki Lahtonen
    Commented Apr 10, 2020 at 7:22

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As explained in the comments, the correct setup here is that of direct limit (or filtered colimit).

Given a preordered set $(I,\leq)$ and a direct system $(A_i)_i$ (so, with maps $A_i\to A_j$ for $i\leq j$) of algebraic gadgets (this is more general, but for algebraic things this always works), you can create their direct limit.

If all the $A_i\to A_j$ are inclusions, you can think of this direct limit as a union (because each $A_i\to \varinjlim_i A_i$ will be injective, and any element of the direct limit comes from at least one $A_i$; and because if you then regard $A_i$ as a subset of $\varinjlim_i A_i$, with this identification, then the limit is literally the union)

So here, what is meant, is to take the divisibility order on $\mathbb N$ : $n\preceq m$ if and only if $n\mid m$. Then you have to chose an embedding $\mathbb F_{p^n}\to \mathbb F_{p^m}$, and you have to make these choices sort of consistently (there are ways to ensure this here, e.g. using factorials)

Then this forms a direct system, and its direct limit is the algebraic closure. And then if you identify each $\mathbb F_{p^n}$ with its image in the direct limit, then it's just their union.

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    $\begingroup$ This is all well. I comment because I want to explain a trick avoiding some of the consistency problems with the inclusions. The trick is to build a nested tower $$\Bbb{F}_p\subset\Bbb{F}_{p^{2!}}\subset\Bbb{F}_{p^{3!}}\subset\cdots.$$ At each step we can construct the field of size $p^{(k+1)!}$ as an extension of the field of size $p^{k!}$. Then form the limit as in Maxime's answer. $\endgroup$
    – Jyrki Lahtonen
    Commented Apr 10, 2020 at 7:11
  • $\begingroup$ @JyrkiLahtonen : yes ! That's what I alluded to when I said "using factorials", but thanks for explicitly writing it ! $\endgroup$
    – Maxime Ramzi
    Commented Apr 10, 2020 at 7:13
  • $\begingroup$ (cont'd) Consistency is then automatic for all the maps are inclusions. We still get an algebraic closure because $\Bbb{F}_{p^n}$ is included in $\Bbb{F}_{p^{n!}}$ (if not sooner). Of course, we still have subtleties in that there is the freedom provided by the choice of the construction of the extension in each step. That cannot really be avoided, I think. $\endgroup$
    – Jyrki Lahtonen
    Commented Apr 10, 2020 at 7:14
  • $\begingroup$ I suspected that much, Maxime :-) $\endgroup$
    – Jyrki Lahtonen
    Commented Apr 10, 2020 at 7:14

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