Proof of Existence of Algebraic Closure: Too simple to be true? Having read the classical proof of the existence of an Algebraic Closure (originally due to Artin), I wondered what is wrong with the following simplification (it must be wrong, otherwise why would we bother with the extra complications in Artin's proof?):
Theorem: Let $K$ be a field, then $K$ has an algebraic closure $\bar{K}$ (i.e an algebraic extension that is algebraically closed). 
"Proof": Define $A=\{ F \supset K | F \text{ is an algebraic extension of } K\}$ and inherit this with the usual partial order of inclusion. One can check that Zorn's lemma applies (union of a nested chain of algebraic extensions is itself algebraic). Thus take $\overline{K}$ to be a maximal element. It must be algebraically closed for otherwise there is an irreducible polynomial with root in some strictly bigger field.  $\blacksquare$
Now here is what I suspect is false about this proof: The definition of $A$ smells like your usual set theory paradoxes like Russell's paradox. In fact one could just as well use the same technique to prove that there exists a "largest set" which of course there does not. I am however under the impression that "most" working mathematicians ignore set theory foundations and just "do" mathematics, so is there a safe way of doing this (i.e: do "concrete everyday mathematics" by avoiding set theory) without getting burnt?
 A: This is the first proof I learnt of the existence of algebraic closures, from Fraleigh's book (an edition from the mid-to-late 80s).  As you note, you have to be a little bit careful with set-theoretic issues, but that is not so serious.  
A: The other answers rescue your quoted proof of the existence of algebraic closure by observing that there is an explicit bound you can place on the cardinality of the algebraic closure. Here is an alternative proof that side-steps the set-theoretic issue altogether:
Let $S$ be the set of all irreducible polynomials with coefficients in $K$, and let $$R := K\left[\{x_\alpha\}_{\alpha \in S}\right]/\left(\{f(x_\alpha)\}_{\alpha \in S}\right).$$ Then $R$ has a maximal ideal, so we can define $F := R/\mathfrak m$ for $\mathfrak m$ such a maximal ideal, which makes $F$ into an algebraic extension of $K$ in which every polynomial in $S$ has a root.
At this point, it is true that $F$ is algebraically closed, but not easy to prove. You can avoid this additional piece of machinery by simply defining $F_1$ to be the field obtained above, and iteratively defining $F_2, F_3, \ldots$ by the same procedure. A priori, each $f \in S$ can readily be shown to split in $F_i$ for $i \ge \deg f$, and so $F := \displaystyle\bigcup_i F_i$ is an algebraic closure for $K$.
A: You are right. $A$ is a proper class. The reason is simple, by considering all possible fields which are algebraic extensions we immediately have a proper class of sets.
However one can easily observe that if $F$ is a field, then there is a map from $F[x]$ onto any algebraic extension, therefore it suffices to consider algebraic extensions whose underlying set is a partition of $F[x]$.
In either case, one can show that despite the fact that $A$ is a proper class, it is "locally a set", in the sense that below each field there is only set-many fields; and that every chain has size no larger than $|F|+\aleph_0$.
A: To answer your other question, of "how to do concrete everyday mathematics, avoiding set theory, without getting burnt":
In NBG set theory, a class (where "class" means "collection of sets") is a set if and only if it is "small"—which is to say, strictly smaller than the class of all sets. In addition, every theorem of NBG that doesn't use the word "class" is a theorem of ZFC.
So, here's how to avoid getting burnt. Think of any class. (A class can be defined by any predicate that only talks about sets, not classes.) If you can show that your class is small, that means it's a set. If you make sure all of your sets are created this way, you end up with a theorem of ZFC.
How do you know that your class is small? These axioms should do the trick:


*

*The class of all integers is small.

*Given any small class, the class of all of its subclasses is a small class.

*Given any small class, the union of its elements is a small class.


Intuition should give you everything else you need.
A: This is a comment on a common error made in proves of existence of algebraic closures sin Artin's method (I don't have Artin's reference at hand, but I am sure he had a flawless argument)
The construction as suggested by @Dustan Levenstein is on the right track however, the implication from the arguments laid down there, specifically that the fields $K_n$ constructed iteratively satisfy $K=K_0\subset K_1\subset\ldots K_n\subset K_{n+1}\subset\ldots$ is not correct! Each $K_n$ is embedded isomorphically into a quotient field $K_{n+1}$,  but $K_n$ is not a subset of $K_{n+1}$.
To fix that little nuance, it seems one needs to take the colimit also known as (direct limit) of the $K_n$'s.
This a learnt from a note by Kevin Buzzard. The construction outlined  by @Dustan Levenstein produces a sequence of fields $\{K_n:n\in\mathbb{Z}_+\}$ with $K_0=K$ (the original field) and (field) homomorphisms $F_n\stackrel{\pi_n}{\longrightarrow} F_{n+1}$ such that $\pi_{n}(F_n)$ and $F_n$ are isomorphic (as fields). Let $L:=\bigsqcup_n K_n$ the disjoint union of the $F_n's$. In $L$ considers the equivalence relation $x Ry$ iff there is $z\in \overline{K}$ such that
\begin{aligned}
z&=\pi_{n,m}(x)=\pi_{m-1}\circ\ldots\circ\pi_n(x)\\
&=\pi_{\ell,m}(y)=\pi_{m-1}\circ\ldots\circ\pi_\ell(y)
\end{aligned}
where $\ell,n\leq m$ ($x\in K_n$ and $y\in K_\ell$ for some $n,\ell\in\mathbb{Z}_+$. The quotient space $\overline{K}:=L/R$ will be the desired object with the algebraic operations transported from the spaces $K_n$ to $L$: For $[x],\,[y]\in L$ suppose $x\in K_m$ and $y\in K_n$ with $m\leq m$. Define

*

*$[x]+[y]=[\pi_{m,n}(x)+y]$

*$[x]\cdot[y]=[\pi_{m,n}(x)\cdot y]$
where $ \pi_{m,n}=\pi_{m-1}\circ\ldots\circ\pi_n$. This can be show to define a field structure in the colimit $\overline{K}$ and that $K$ is embedded in $\overline{K}$ (say as $K\simeq K'\subset\overline{K}$), and $\overline{K}$ is algebraically closed (every non constant polynomial in $\overline{K}[x]$ has a root in $\overline{K}$). Now, consider the set $A$ of all algebraic elements in $\overline{K}$ over the $K'$. This is the algebraic closure of $K$ (or rather $K'$), that is $A$ is algebraically closed and every element of $A$ is algebraic over $K$.
A: In fact, there is a slightly goosed version (by Jelonek) which avoids the set-theoretic issues. Poles it know from set theory...
