I am studying from Patrick Morandi's Field and Galois Theory, and on page 32 he proves the existence of an algebraic closure of an arbitrary field, as follows:
Lemma 3.13. If $K/F$ is algebraic, then $|K| \leq \max\{ |F|,|\mathbb{N}| \}$.
Theorem 3.14. Let $F$ be a field. Then $F$ has an algebraic closure.
Proof. Let $S$ be a set containing $F$ with $|S| > \max\{ |F|, |\mathbb{N}| \}$. Let $\mathcal{A}$ be the set of all algebraic extension fields of $F$ inside $S$. Then $\mathcal{A}$ is ordered by defining $K \leq L$ if $L$ is an extension field of $K$. By Zorn's lemma, there is a maximal element $M$ of $\mathcal{A}$. We claim that $M$ is an algebraic closure of $F$. To show that $M$ is algebraically closed, let $L$ be an algebraic extension of $M$. By Lemma 3.13,
$$
|L| \leq \max\{ |M|, |\mathbb{N}| \} \leq \max\{ |F|,|\mathbb{N}| \} < |S|.
$$
Thus, there is a function $f:L \to S$ with $f|_M = \operatorname{id}$. By defining $+$ and $\cdot$ on $f(L)$ by $f(a) + f(b) = f(a+b)$ and $f(a) \cdot f(b) = f(ab)$, we see that $f(L)$ is a field extension of $M$ and $f$ is a field homomorphism. Maximality of $M$ shows that $f(L) = M$. Thus, $M$ is algebraically closed. Since $M$ is algebraic over $F$, we see that $M$ is an algebraic closure of $F$. $\tag*{$\blacksquare$}$
I do not understand the second line in the proof, where he says
Let $\mathcal{A}$ be the set of all algebraic extension fields of $F$ inside $S$.
Since we are assuming $S$ is just a set containing $K$ and so it does not have any field structure defined on it a priori, how can we talk about field extensions of $F$ contained in $S$?
The rough idea that I am currently keeping in mind is that we take an algebraic extension $K$ of $F$ and then use a bijection of this set onto a subset of $S$ to define the addition and multiplication on this subset, in the same way we do later on for $L$. And then we possibly do this process for every algebraic extension of $K$.
But then what I don't see at all is how to ensure that the different mappings are compatible with each other. And, if I have two isomorphic algebraic extensions of $K$, do I then map it to the same subset of $S$ or something?
It would be really helpful if someone could guide me in filling out the details here.
Best Answer
Any countable set can be endowed with a field structure using a bijection between the set and $\Bbb Q$ and what is known as transport of structure. Similarly, any set of size $2^{\aleph_0}$ can be given a field structure isomorphic to $\Bbb R$ or $\Bbb C$ or any other field of that cardinality.
This is similar to the way that any location on your hard drive can contain the data of this page. The set theoretic universe is just a large pagefile, so to speak. So when you talk about a field, you allocate a suitable set and endow it with the wanted structure.
In turn, this flexibility also imposes a problem. If $F$ is a field, then any set containing $F$ can be an extension of $F$ (or a subset of an extension). So when you come to apply Zorn's lemma, you have a proper class of fields to work with. But Zorn's lemma applies to sets, not to classes. So you need to somehow restrict this.
But luckily, there are only "set many" different ways to give $F$ an algebraic extension,1 since all those must have the same cardinality (plus or minus a countable set) as $F$. Therefore, we can fix a large enough set and require that all the algebraic extensions will be taken as subsets of that set, and in a coherent way.2
The reason being that all algebraic extensions can be given to a set $A$ with the same cardinality as $F\cup\Bbb N$. Each such structure is made of addition and multiplication, both of which are subsets of $A\times A$. Since there are only set many subsets to $A\times A$, there are only set many ways to give $A$ a field structure, let alone an algebraic extension of $F$.
This can be done by taking isomorphism classes and lifting the embeddings to the equivalence classes. Of course there is no a priori way of choosing representatives so that the embeddings correspond to inclusions, but we can prove using Zorn's lemma there is a maximum equivalence class which corresponds to the algebraic closure. After choosing that one, we can then choose the algebraic extensions to be ordered by inclusions as subfields of the closure.