Doubt in algebraic closure of a finite fields

Question

In Dummit and Foote's Abstract Algebra, I came across the following: I was not able not understand what "allows us to think of their union" means. I know that any two finite fields of same cardinality are isomorphic. So I am able to interpret what $\bigcup_{n=1}^{k} \mathbb{F}_{p^n}$ means. But still I am not able to grasp what is $\bigcup_{n=1}^{\infty} \mathbb{F}_{p^n}$. I found it convenient to "construct" $\bigcup_{n=1}^{\infty} \mathbb{F}_{p^{n!}}$. I tried doing the following:
Given a finite field $F_k$ having cardinality $p^{k!}$, take an arbitrary finite field $G_{k+1}$ having cardinality $p^{(k+1)!}$. There is an embedding $f:F_k \to G_{k+1}$. A standard result from set theory is that: Given a set $X$ and a cardinal $\alpha$, there is a set $B$ with cardinality $\alpha$ and $A \cap B = \emptyset$. Using this I can find a set $E_k$ (disjoint from $F_k$) having cardinality $p^{(k+1)!} - p^{k!}$ ie. $|G_{k+1} \setminus f(F_k)|$, so there is a bijection $g:E_k \to G_{k+1} \setminus f(F_k)$, so that we can construct a bijection $h:F_k \cup E_k \to G_{k+1}$ such that $h\restriction_{F_k} = f$ and $h\restriction_{E_k} = g$. A field structure is induced on the set $F_k \cup E_k$ via the bijection $h$. Note that $F_k$ is a subfield of $F_k \cup E_k$ (under this new structure) and by our construction, this is exactly the same as our original structure on $F_k$. Define $F_{k+1} := F_k \cup E_k$.
Thus we constructed a field $F_{k+1}$ having $p^{(k+1)!}$ elements such that $F_{k}$ is a subfield of $F_{k+1}$.
Then I tried to justify that using induction that starting from $F_1 = \mathbb{F}_p$ we can define a sequence $F_1,F_2, F_3 \ldots$ and we just have to take the union $\bigcup_{k \ge 1} F_k$. However this reasoning is wrong. I think having an arbitrary choice of $E_k$ during construction of $F_{k+1}$ is a problem.
So to avoid this I took $F_1= \{0,1,2, \ldots p-1 \}$ and gave $\mathbb{F}_p$ structure in the standard way and defined the recursion $F_{k+1} = F_k \cup \{p^{k!},p^{k!}+1 \ldots p^{(k+1)!}-1 \}$ and I gave $F_{k+1}$ a $\mathbb{F}_{p^{k+1}}$ structure using the same process I have described above, so that $F_k$ is a subfield of $F_{k+1}$. Now, I think the sequence of fields $F_1,F_2, F_3 \ldots$ is well defined and moreover $F_n$ is a subfield of $F_{n+1}$ for $n \in \mathbb{N}$. Now we can take the union $\bigcup_{n \ge 1} F_k$. I am still not sure whether this works (although I am more confident in this method since now there are only finitely different ways to extend the field structure from $F_k$ to $F_{k+1}$).
I asked my professor about this and he said that the proper construction involves category theory. I am curious to see whether there is a nice set theoretic construction for $\overline{\mathbb{F}_p}$ (which was probably intended by Dummit and Foote). Is my construction valid?

Answer 1

You are right to be concerned, I think there's much more to say about this and Dummit and Foote are being sloppy here.

First, here is the construction that Dummit and Foote are alluding to with the use of the word "union." Suppose $S_1 \subseteq S_2 \subseteq \dots $is a sequence of sets each of which is literally a subset of the next one. Then we can consider the set

$$S = \bigcup_i S_i$$

given by the literal union of these sets; this is called the increasing union. A nice example here is that if $S$ is the set of polynomials then we can take $S_i$ to be the set of polynomials of degree $\le i$. You can think of $S$ as the disjoint union of the sets $S_i \setminus S_{i-1}$.

We might want a more abstract version of this construction, where instead of having literal subsets we just have a sequence $S_1 \to S_2 \to \dots$ of injections (or more generally monomorphisms in some category). In that case we can't just take the union, but we have an abstract replacement for it, which is what is called the directed colimit (the term "direct limit" is older terminology but it clashes with modern terminology and IMO should be deprecated) of this sequence. Explicitly, for sets and other familiar algebraic structures like groups, rings, and modules, the directed colimit is the quotient of the disjoint union $\bigsqcup_i S_i$ by the equivalence relation that $s_i \in S_i$ and $s_j \in S_j$ are equivalent if, after applying our sequence of injections, they become equal in $S_{\text{max}(i, j)}$.

It is not entirely obvious, but it is true, that the increasing union or more generally the directed colimit of a sequence of groups, rings, modules, or even fields (and injective group, ring, etc. homomorphisms) remains a group, ring, etc. respectively. The idea is that any operation you might want to perform on a finite set of elements takes place at some "finite stage" $S_i$ so you can verify all the group, ring, etc. axioms at that stage.

The directed colimit turns out to be a very flexible and general construction, and we can take it over a more complicated diagram than just a sequence. In fact we can take it over what is called a directed set. I don't want to spell out the definition but for this application it suffices to know that the poset $\mathbb{N}$ of natural numbers ordered by divisibility is a directed set.

Now the analogue of a sequence of injections in this case is more complicated. It is a collection $S_n, n \in \mathbb{N}$ of sets (or groups, rings, modules, etc.) such that whenever $n \mid m$ we have an injection $\varphi_{n, m} : S_n \to S_m$, and these injections need to satisfy the compatibility condition that if $n \mid m \mid k$ then

$$\varphi_{m, k} \circ \varphi_{n, m} = \varphi_{n, k}.$$

This is the reason I think Dummit and Foote are being sloppy. We know that for every $n \mid m$ we can write down an injection $\mathbb{F}_{p^n} \to \mathbb{F}_{p^m}$. But it is not obvious a priori, although it is true, that we can write down a collection of such injections which also satisfies the compatibility condition, because these injections are not unique! We actually can do this but it requires some work to establish.

Given a compatible collection of injections as above, the directed colimit $\text{colim}_i S_i$ is the quotient of the disjoint union $\bigsqcup_i S_i$ by the equivalence relation that $s_i \in S_i$ and $s_j \in S_j$ are equivalent if, after applying suitable injections, they become equal in $S_{\text{lcm}(i, j)}$. This construction is totally canonical given the injections and requires making no further choices, which saves us from having to make the arbitrary choices of the $E_k$ in your construction.

And this construction, applied to the finite fields $\mathbb{F}_{p^i}$ with a suitable compatible collection of injections, does in fact produce the algebraic closure. But I think it's misleading to just call this a "union" without any further elaboration since writing down that compatible collection of injections is exactly what is needed to actually think of the various finite fields as subsets of each other in a coherent way, and that hasn't been done in the text.

Your idea to work with the factorials $\mathbb{F}_{p^{n!}}$ is a good way to get around this because instead of working with the more complicated poset given by divisibility we just reduce to the case of a sequence. This is done, for example, on PlanetMath.

Answer 2

A simple way to view this is as follows: The underlying set is the union of the underlying sets (where the underlying set of $\mathbb F_{p^m} \subseteq \mathbb F_{p^n}$ when $m \mid n$) and when you want to do an operation to elements, just perform that operation in some $\mathbb F_{p^k}$ that both of them are contained in (e.g $k = \mathrm {lcm}(m, n)$.)

Answer 3

1. Existence/construction of algebraic closure.
2. Description of the elements of closure.

Existence/construction of algebraic closure. At this point of the exposition, existence of algebraic closure of any field should be already proved (as in Lang's "Algebra"). Any union of directed set of subfields of $\overline{\mathbb{F}_{p^n}}$ can be safely performed as union of sets. That is why authors semi-formally refer to union.

Proof of existence of generic algebraic closure is non-constructive (uses Zorn's lemma).

Description of the elements of closure. Still there is an issue with identification of abstractly given algebraic extension $\mathbb{F}_{p^m}$ of field $\mathbb{F}_{p^n}, n|m$ with specific subfields of the algebraic closure $\overline{\mathbb{F}_{p^n}}.$ If there were multiple candidates for $\mathbb{F}_{p^m}$ in $\overline{\mathbb{F}_{p^n}},$ authors would need to provide details re. the choice of subfield isomorphic to $\mathbb{F}_{p^m}$ and would not refer to "union". Authors omitted details about the choice on purpose: there is no ambiguity.

Side note. Uniqueness of field $\mathbb{F}_{p^m}$ upto isomorphism (as abstractly given field) should not be confused with the uniqueness of subfield isomorphic to $\mathbb{F}_{p^m}$ in $\overline{\mathbb{F}_{p^n}}.$ Consider, for example, $\mathbb{Q}[\sqrt[3]{2}\zeta]$ where $\zeta$ is the root of $x^3-1.$ All of these 3 fields are isomorphic to the quotient ring $\mathbb{Q}[X]/(x^3-2),$ but they are different subfields of $\overline{\mathbb{Q}}.$

To analyze uniqueness of subfields one can use normality of extensions. If extension $L$ of $K$ is normal then any embedding of $L$ into algebraic closure $\overline{K}$ has the same image. That is there is unique subfield of $\overline{K}$ isomorphic to $L.$

$\mathbb{F}_{p^m}$ is normal extension of $\mathbb{F}_{p^n},$ so when referring to $\mathbb{F}_{p^m}$ in $\overline{\mathbb{F}_{p^n}}$ there is no ambiguity. The only subfield $\mathbb{F}_{p^m}$ of $\overline{\mathbb{F}_{p^n}}$ consists of exactly those elements $x\in\overline{\mathbb{F}_{p^n}}$ which satisfy $x^{p^m}=x$.

Intuition related to "union" is justified, finally, as every $x\in\overline{\mathbb{F}_{p^n}}$ lies in some $\mathbb{F}_{p^m}, n|m.$