I am reading the book A First Course in Abstract Algebra written by Fraleigh and I do not really understand the proof of theorem 31.22, that every field $F$ has and algebraic closure $\bar F$.
I notice that many people has asked this question before but I still don't understand. Basically, one need to construct a set $S$ that contains all algebraic extension fields of $F$ and use Zorn's Lemma to assure the maximal element $\bar F$. However we need to assure the set $S$ is a 'legal' set which won't fall into the trap of Russell paradox, as shown below (by Fraleigh):
I am confused with the construction of $\Omega$. I assume that $A$ is the set of all zeros in $F[x]$ and $\Omega=P(A)$ which has $\it{cardinallity}$ strictly greater then $A$. Since every $\alpha\in F$ is a zero of the polynomial $f_\alpha=x-\alpha$, one has $\omega _{f_\alpha1}=\alpha$. I rename $\{\omega _{f_\alpha1}\}\in P(A)=\Omega$ as the element $\alpha$ so $F\subset \Omega$ would make sense.
Then choose any $\gamma \in E$ and rename every element in $F(\gamma)$ with different $\omega\in \Omega\backslash F$. This is how I understand Fraleigh's proof.
My question is: After assigning names to every element in $F(\gamma)$ by $F(\omega)$, then the 'remaining elements' in $\Omega$ would have the size $\Omega\backslash (F\cup F(\omega))$. How can one guarantee that it is still large enough to rename the other extension fields?
Also, in the proof, $\gamma$ is get from $E$. What we know is all elements in $F(\gamma)$ is renamed and contained in $\Omega$. But it does not mean that $E\subset \Omega$.
And thirdly, I suppose the construction is followed in this procedures: Find $\gamma_1$ and make $F(\gamma_1)$ into $\Omega$. Find another $\gamma_2$ and make $F(\gamma_2)$ into $\Omega$, and so on. This is a step by step procedures and I can only make $countably$ many $F(\gamma_i)$ into $\Omega$. How can I know $\Omega$ is actually large enough to contain maybe uncountably many extension fields?
If $\Omega$ has cardinality larger than that of $F$, then all you need to show is that $F\cup F(\omega)$ as the same cardinality of $F$. Then $\Omega\setminus(F\cup F(\omega)$ has the same size as $\Omega$ itself.
If $F$ is finite, then $F(\omega)$ is finite, and $\Omega$ is infinite, so we're good.
If $F$ is infinite, then we really just need to check that $F$ and $F(\omega)$ have the same cardinality, which is again not hard since there is a map from $F[x]$ onto $F(\omega)$, and $F[x]$ and $F$ have the same cardinality.
(Let me add a side remark, that the more I see people struggle with this issue, the more I feel that they may benefit from a transfinite recursion argument; where you well-order $F[x]$, and one by one you ensure that each polynomial has been added a zero.)