construction of $\bigcup_{i}K_{i}$ for sequence of fields $K_{0}\subseteq K_{1}\subseteq\cdots$? Let $K_{0}\subseteq K_{1}\subseteq\cdots$ be a sequence of fields. 
(Each $K_{i}\subseteq K_{i+1}$ means not set theoretical inclusion but field extension i.e. field homomorphism. )
Then how is the union $\bigcup_{i}K_{i}$, expected to be a field, constructed set theoretically ?
I want to know this because it appears in the proof of the existence of algebraic closure of field.
Here is the definition of field extension:
A trio $(K, L, f)$ is a field extension if $K$ and $L$ are fields, and $f\colon K\to L$ is a field homomorphism. 
 A: It is the colimit of $K_{i}$. 
The quotient set by equivalent relation on disjoint union $\coprod_{i}K_{i}$
$$
(x, i)\sim (y,j)\iff \exists k>i,j \mbox{ s.t. }x=y \mbox{ in }K_{k}
$$
is the desired construction. 
Checking of the well-definedness of “$+$, $\cdot$” is similar to that of stalks of sheaves. 
A: 
$\bigcup_{i}K_{i}$ is a field :

For this, we need to define the addition $+$ and the multiplication $\cdot$ on $\bigcup_{i}K_{i}$. In order to avoid any confusion, call $+_i$ and $\cdot_i$ the operations on $K_i$.
Let $x, y \in \bigcup_i K_i$, then there is $i$ such that $x \in K_i$ and $j$ such that $y \in K_j$. Take any $k \geqslant \max(i, j)$, one sets :
$$x + y := x +_k y \textrm{ and } x \cdot y := x \cdot_k y$$

What if we don't have inclusions but monomorphisms $\varphi_i : K_i \mapsto K_{i+1}$ for each $i$?

For each $i < k$, write $\varphi_{i,k} := \varphi_{k-1} \circ \varphi_{k-2} \circ \dots \circ \varphi_i : K_i \mapsto K_k$. We build the direct limit $K := \varinjlim K_i$ as follow :
$K$ is the quotient of $\bigsqcup_i K_i$ by the equivalence relation $\sim$ bellow.
$$ (x, i) \sim (y,j) :\Longleftrightarrow \exists k, k > \max(i,j) \wedge \varphi_{i,k}(x) = \varphi_{j,k}(y)$$
Defining the addition and multiplication on $K$ :
Let $\bar{x}, \bar{y} \in K$.  


*

*$\bar{x} + \bar{y} := (\varphi_{i,k}(x) + \varphi_{j, k}(y), k)^\sim$ where $\bar{x} = (x, i)^\sim$, $\bar{y} = (y, j)^\sim$ and $k > \max(i, j)$

*$\bar{x} \cdot \bar{y} := (\varphi_{i,k}(x) \cdot \varphi_{j, k}(y), k)^\sim$ where $\bar{x} = (x, i)^\sim$, $\bar{y} = (y, j)^\sim$ and $k > \max(i, j)$
