For a countable field $K$, can an algebraic closure be constructed by successively constructing splitting fields of the polynomials in $K[x]$? $K$ countable $\implies K[x]$ countable.
Enumerate the polynomials $f_1, f_2, f_3, \dots$ in $K[x]$.
Let $K_0 = K$.
Iteratively, construct splitting fields $K_i$ of $f_i$ viewed in $K_{i - 1}[x]$.
$K_0 \subseteq K_1 \subseteq K_2 \subseteq \cdots$ is a tower of field extensions.
Take $L = \bigcup\limits_{i=0}^{\infty} K_{i} $. Then we can readily verify that $L$ is a field, $L:K$ is algebraic, and every irreducible $f \in K[x]$ splits over $L$.  So $L$ is an algebraic closure of $K$.
But is the construction valid?
This construction is part of an exercise in the book on Galois theory by Garling. The exercise asks if the construction is less fallacious than the 'fallacious proof' of the existence of an algebraic closure. Unlike the 'fallacious proof', there is not the problem of comparing extensions $(i, K, M)$, where $i$ is a monomorphism from $K$ into $M$, instead of comparing actual subsets. But apparently there is a problem with the construction.
What part of the construction is fallacious?
 A: I think the issue is that we don't really have set inclusions. When you construct the splitting field $K_i$ of $f_i$ over $K_{i-1}$, you don't construct a superset $K_i\supseteq K_{i-1}$, but only a field $K_i$ with an injective homomorphism $\iota\colon K_{i-1}\hookrightarrow K_i$. Usually, this doesn't matter, because we can "identify" $K_{i-1}$ with $\iota(K_{i-1})$ (these are isomorphic via $\iota$) and $\iota(K_{i-1})\subseteq K_i$.
By doing this, we can, for any $n\in\mathbb{N}$, act as if we have set inclusions $K_0\subseteq K_1\subseteq K_2\subseteq...\subseteq K_n$ by identifying the fields with their images under the inclusions iteratively. The catch is that, a priori, we can't quite do this for all $K_i,\,i\in\mathbb{N}$ at once, because it appears we don't yet have a field in which they all embed.
The good news is that we salvage the proof. We can't take the union, but since the sequence of sets together with the injective homomorphisms is close enough to sets with inclusions, we can construct something that isn't a union, but behaves like a union in the ways we want. This thing we are looking for is the corresponding direct limit.
