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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 algraically 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.

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You are overthinking this. You are correct that $S$ is just a set. An element of $\mathcal{A}$ is a subset $K$ of $S$ together with a field structure on $K$ such that this field structure makes $K$ an algebraic field extension of $F$. A subset of $S$ on its own does not automatically get a field structure, but specifying an element of $\mathcal{A}$ involves choosing one particular field structure to put on your subset.

The whole point of the Zorn's lemma argument is then to avoid the mess you describe about figuring out how to compatibly embed all possible algebraic extensions into $S$. You just pick a maximal algebraic extension of $F$ whose underlying set is subset of $S$. Then, as argued in the proof, maximality guarantees that this field is algebraically closed.

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  • $\begingroup$ In the grand scheme of things, I think that overthinking stuff is the boon and bane of almost all mathematicians. $\endgroup$ – Asaf Karagila Aug 1 '18 at 15:24
  • $\begingroup$ Right, I think I've got it now... And Asaf's argument says there are only so many algebraic extensions of $F$ available, so they can all be embedded into $S$ without any trouble. And then Zorn's lemma takes over. Does that sound about right? $\endgroup$ – Brahadeesh Aug 1 '18 at 15:31
  • $\begingroup$ @Brahadeesh: Sort of but not exactly. It's not actually important that all the algebraic extensions can be embedded into $S$ and that there are only so many of them. The proof never uses any such statement. Rather, it just directly shows that a maximal element of $\mathcal{A}$ must be algebraically closed. The argument for this is a cardinality argument which is related to what Asaf said but not quite the same thing. $\endgroup$ – Eric Wofsey Aug 1 '18 at 15:33
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    $\begingroup$ Asaf's idea is more of a motivation for why you would want to use a construction like this and expect that it works. It's not directly relevant to proving that it works. $\endgroup$ – Eric Wofsey Aug 1 '18 at 15:35
  • $\begingroup$ Oh! Yes, I see it now. Thank you! $\endgroup$ – Brahadeesh Aug 1 '18 at 15:36
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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


  1. 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 $F$.

  2. 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.

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  • $\begingroup$ Can you elaborate on how there are only "set many" different ways to give $F$ an algebraic extension? $\endgroup$ – Brahadeesh Aug 1 '18 at 10:39
  • $\begingroup$ If F is infinite, an algebraic extension does not increase the cardinality of the resulting field. On a fixed set there are only set many field structure since those are subsets of some fixed set, there are only set many of then. $\endgroup$ – Asaf Karagila Aug 1 '18 at 10:46
  • $\begingroup$ This is an excellent explanation of the reason why the proof is set up the way it is, but I think OP's actual question was just about what the definition of $\mathcal{A}$ even means. $\endgroup$ – Eric Wofsey Aug 1 '18 at 15:30
  • $\begingroup$ @Brahadeesh: I forgot to thank you for this, by the way! Thanks! :) $\endgroup$ – Asaf Karagila Aug 7 '18 at 6:06
  • $\begingroup$ @AsafKaragila you’re welcome :) $\endgroup$ – Brahadeesh Aug 7 '18 at 6:08

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