You may have seen that I posted this proof with some questions earlier today. But I found the answer to most of them. Now I have just one question regarding this proof, so I thought it would be better to delete the earlier post and create a new one, since the earlier one didn't get any answers.

The proof looks big, and the picture is big, but the author says that it is written in very much detail. My question will revolve around the red line, and I will post it after the picture:

enter image description here

My question is just how does he knows that $\Omega$ has enough elements to do what he does the over the red line? I get why it has enough elements to contain every algebraic extension of F, since an element of an algebraic extension of F, has to be a zero of a polynomial in $F[x]$, and he has made it so that we have enough elements for all possible zeros of all possible polynomials in $F[x]$.

But in the last part of the proof. How does he know that $\Omega$ has enough elements to construct $\bar{F}(\omega)$? I mean, he said that $\Omega$ was strictly bigger than A, so we know that we have more elements than we need for all possible algebraic extensions of F. But "strictly more" and "many more" is not precise?

Basically: A was created so that it has enough elements for all algebraic extensions of F, but how is it certain that $\Omega$ has enough elements for even one algebraic extension of $\bar{F}$ (the extension $\bar{F}(\omega)$)?

I must admit that I don't even know why the statement "Since $\Omega$ has many more elements than $\bar{F}$" is true. I know that Zorn's lemma uses the word maximal element, but by definition all we require is that a maximal element is not the entire set, so what is from stopping it from just being the entire set -1 element, er the entire set-"a very small subset"?

UPDATE: I thought of an argument, do you think this works? Wikipedia talks about cantors theorem and a strict subset by taking the set of all subsets, is what I write under correct, and can be put in the proof?

We assume that $\Omega$ is created by taking the union of "all subsets of A" and "one more element" . Then we have all the A elements embedded in $\Omega$ as the singletons. Since $\bar{F}$ is an algebraic extension it is represented by these singletons.

Now we do as in the proof assume that $f(x)$ is a nonconstant polynomial in $\bar{F}[x]$, by Kroneker's theorem, there exists an extension field K, with an element s. This element is algebraic over $\bar{F}$ since it is a zero of a polynomial in $\bar{F}[x]$. So the set $F(s)$ is created by $k_0+...+k_{n-1}s^{n-1}$ for $k_i \in \bar{F}$. We can associate s with the "one more element", and since $\bar{F}$ is an algebraic extension, every k is associated with the singletons, and every $k_0+...+k_{n-1}s^{n-1}$ is naturally associated with the subset that is the union of the singletons that each k is associated with.

Hence we have enough elements. Does this work?

  • $\begingroup$ Sorry for removing your tags; but this is not really a question about how Zorn's lemma is used or its necessity in the proof, and it's certainly not a set theoretic question. It's a question about a proof in algebra, that every field has an algebraic closure. $\endgroup$ – Asaf Karagila Apr 29 '15 at 22:09

If $f(x)$ is a nonconstant polynomial in $\bar{F}[x]$ having no roots, the field $K=\bar{F}[x]/(f(x))$ is a proper algebraic extension of $\bar{F}$ where $f(x)$ has a root.

Now $|K\setminus\bar{F}|\le|K|<|\Omega|$ and $|\Omega\setminus\bar{F}|=|\Omega|$ by cardinal arithmetic. So you can find an injective map $$ \varphi\colon K\to\Omega $$ such that $\varphi(a)=a$ for $a\in\bar{F}$. You can transport the structure of $K$ on $\varphi(K)$ which so becomes a proper field extension of $\bar{F}$ contained in $\Omega$, contradicting the maximality of $\bar{F}$.


$\overline{F}$ is, in particular, an algebraic extension of $F$ by definition. As such, it will be in bijection with a subset of $A$: to each element $x$ of $\overline{F}$ we have its minimal polynomial $p_x$ over $F$, and then we can use the Axiom of Choice to choose appropriate $i$ to index the roots of $p_x$ that occur in $\overline{F}$.

Next, we're given that $\Omega$ is "strictly larger" than $A$, by which I assume the author means "has cardinality strictly larger than $|A|$".

So because $\overline{F}$ has cardinality at most $|A|$, we know that there cannot be a bijection from $\overline{F}$ onto $\Omega$. As such, no map with domain $\overline{F}$ can surject onto $\Omega$, and so in particular the inclusion map $\iota: \overline{F}\hookrightarrow \Omega$ is not surjective, allowing us to choose $\omega\in \Omega$.

To actually form the proposed field, we take $\overline{F}(\omega)=\overline{F}[\omega]/\langle f\rangle$, which is a field by the fact that $f$ has no zeroes, and hence $\langle f\rangle$ is maximal.

To see that $\Omega$ is large enough to contain $\overline{F}(\omega)$, we note that $[\overline{F}(\omega):\overline{F}]=\deg f$ so $|\overline{F}(\omega)|=|\overline{F}|^{\deg f}$. Now, $\overline{F}$ won't be finite, but even if it were $\overline{F}(\omega)$ would be finite. Otherwise $|\overline{F}|=|\overline{F}(\omega)|$. $\Omega$ is infinite, so we still have $|\Omega|>|\overline{F}(\omega)|$.

Because $|\Omega|>|\overline{F}(\omega)|$, we know that $|\overline{F}(\omega)\setminus \overline{F}|<|\Omega\setminus \overline{F}|$, so there is an injection of $\overline{F}(\omega)$ into $\Omega$ that fixes $\overline{F}$. Declaring this injection to be a ring homomorphism gives the image the desired structure and field extension of $\overline{F}$ in $\Omega$.

EDIT: Responding to your update, $\mathcal{P}(A)$ is already enough. Cantor's Theorem says that $|A|<|\mathcal{P}(A)|$, so from what I've written above we know that setting $\Omega$ to be $\mathcal{P}(A)\cup F$ is enough to get "enough elements". The problem with the way you try assigning elements of $F(s)$ with elements in $\mathcal{P}(A)$ is that your method doesn't retain the ordering of the $k_i$'s. Of course, as shown above you don't actually need to actually come up with an injection of $F(s)$ to know there is one.

  • $\begingroup$ Thank you very much for the answer. I see that it is a long and good answer, but I will need some time to understand it. I hope it is ok I ask about the update since you commented on it, sorry for not asking about your original answer. If I instead do take the cantor set twice, and I identify double singletons $\{\{\omega\}\}$ with elements in the extension fields, and I identify $\{\{\omega_0\},\{\omega_0,\omega_1\},\{\omega_0,\omega_1,\omega_2\}.....\}$ with $k_0+...+k_{n-1}s^{n-1}$, will it work then? Then I have taken care of the ordering? $\endgroup$ – user119615 Apr 30 '15 at 1:50
  • $\begingroup$ And last I wonder if we can use this trick in order to only have to work with the set A, and not with $\Omega$ at all: When we come to the paragraph I have in red we assume that we are working somewhere else, we are not in A(and not in $\Omega$ since we have not defined it). But we show that here $\bar{F}$ still has an algebraic extension(like he does), and now, this extension must then be isomorphic to an element in S(and then this is bigger than the maximal element, so we have a contradiction), since it is an algrebraic extension of F. So can we this way get away from working with $\Omega$? $\endgroup$ – user119615 Apr 30 '15 at 1:54
  • $\begingroup$ @user119615 I assume you meant $k_i$ when you say $\omega_i$. Your mapping in this case isn't injective; $1+s+\cdots+s^{i-1}+s^{i+1}+\cdots+s^{n-1}$ has the same image $\{\{\{1\}\},\{\{0\},\{1\}\}\}$ (using double singletons doesn't really matter here, we'd have the same with triple singletons, quadruple singletons, etc). The problem with defining an explicit map of $\overline{F}(\omega)$ into $\Omega$ (or subset of $\Omega$, however you're looking at it) is that you don't know what $\overline{F}(\omega)$ looks like. For all we know, $\overline{F}$ actually contains $A$, which... $\endgroup$ – Hayden Apr 30 '15 at 9:08
  • $\begingroup$ ...is a real possibility. This is the main reason why we can't define $\Omega$ to be exactly what we need to make sure the construction of $\overline{F}(\omega)$ goes through; we don't have any control over what $\overline{F}$ is, so those elements you added to be $\omega$ or the polynomials in $\omega$ might have already been used. You could add tons of conditions to what $S$ is, but this is more cumbersome than just picking $\Omega$ to be so large that it doesn't matter. $\endgroup$ – Hayden Apr 30 '15 at 9:08
  • $\begingroup$ If we only took $S$ to be the collection of algebraic extensions of $F$ contained in in $A$, then we'd run into the problem of needing to put more conditions on $S$ again: $A$ is going to be infinite no matter what, and when $F$ is infinite, we'll have that $F$ and $A$ have the same cardinality. Given any element $\omega$ that we want to be the root of an irreducible polynomial in $F[x]$, this will also be the same size as $A$, so we could (even after deciding on an explicit injection of $F$ into $A$) define $F[x]$ in such a way that it is the whole of $A$. $\endgroup$ – Hayden Apr 30 '15 at 9:12

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