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 $alphain F$ 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 $Fsubset Omega$ would make sense.
Then choose any $gamma in E$ and rename every element in $F(gamma)$ with different $omegain Omegabackslash F$. 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 $Omegabackslash (Fcup F(omega))$. 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 $Esubset Omega$.
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?