Failure in natural construction of algebraic closure of a field The question I am posting here is not about "constructing" the algebraic closure, but it is about the failure of the natural way one tries to prove it. The proof is not so simple that an undergraduate will do it independently at first time.
Let the base field be $F$. 
Way 1: Start with an irreducible polynomial $f_1$ over $F$, and construct an extension $E_1$ which contains root of $f_1$. If $E_1$ is algebraiclly closed, we are done, otherwise, if $f_2$ in $F[x]$ is irreducible over $E_1$, obtain algebraic extension $E_2$ of $E_1$ where $f_2$ has root, and continue this process. Then is it not true that $\cup_i E_i$ is an algebraic closure of $F$?
Way 2: (Exercise from book Algebra-by L. Grove) Let $\mathfrak{L}$ be the set of all algebraic extensions of $F$. 
Define partial ordering on $\mathfrak{L}$ by subset-relation. Applyting Zorn's lemma, we find a maximal element in $\mathfrak{L}$ which turns out to be an algebraic closure of $F$. 
However, in Way 2, the Grove says, that this argument is wrong. He says (which I don't understood): suppose , $|\mathfrak{L}|=\alpha$. For each subset of $\mathfrak{L}$, construct an algebraic closure of $F$, and arrive at a contradiction. Thus, the statement let $\mathfrak{L}$ be the set of all algebraic extensions of $F$ in way 2 doesn't make sense. (I didn't arrive at contradiction, can you help me?)
 A: The issue with the first way is first that you have to check that if every polynomial over $F$ has a root in $E$, then every polynomial over $F$ splits in $E$. Then this procedure only works if your field is countable. If it's uncountable, you will need some transfinite recursion of some kind, so you need to well-order the set of irreducible polynomials over $F$ and you need the Axiom of Choice for that.
The issue with the 2nd way is that if $\mathfrak L$ is a set then so is the reunion of the underlying sets of the algebraic extensions. But this would be the set of all sets :
Let $x \notin F$ be a set. Then we have an obvious bijection $f : F \to F_x = (F \setminus \{0_F\}) \cup \{x\}$. We can give a field structure to $F_x$ so that $f$ is a field isomorphism. Thus $f: F \to F_x$ is an algebraic field extension of $F$. When you perform the reunion of all the underlying sets of all the algebraic extensions, $x$ will be in it. Obviously, $F$ will also be included, so you really get the set of all sets.
Resolving this problem is easy, you only need to restrict the possible underlying set to a fixed set of large enough cardinality. In fact you can use $F$ itself unless $F$ is finite, then you can use $\Bbb N$. Once you fix an underlying set, the possible field structures you can put on it form a set, and the possible algebraic extensions with it are again a set.
A: In Way 1, by writing $f_1$ and then $f_2$ you appear to be suggesting there are only countably many (monic) irreducibles in $F[x]$, which need not be true. You can get around that issue by adjoining to $F$ a root to every monic irreducible in $F[x]$ in one step by an algebraic trick using a maximal ideal in a very large polynomial ring. In any case, what you write as "$\bigcup_i E_i$" is meant to be an extension field of $F$ generated by a root of each (monic) irreducible in $F[x]$ and you ask if this is an algebraic closure of $F$.  The answer is yes, and this is a theorem of Robert Gilmer. See here or Theorem 2 here. A similar construction where it is easier to see you get an algebraic closure is just one step is here.
