# Construction of algebraic closure

I am recently studying abstract algebra and I found a great difficulty on understanding algebraic closure. By definition, an algebraic closure $\bar F$ of $F$ is an algebraic extension of $F$ which is algebraically closed. By Kronecker's Theorem, every polynomial in $F[x]$ has a root in an extension field. Let $K_f$ be the extension field that contains zeros of $f(x)\in F[x]$ and let $K=\bigcup_{f\in F[x]}K_f$

Now, any algebraic element of $F$ is contained in $K$ so any algebraic extension of $F$ is a subset of $K$. We have $\bar F\subseteq K$. Let $\alpha\in K$ and so $\alpha\in K_f$ for some $f(x)\in F[x]$. If $\alpha \notin \bar F$, then upon reducing the polynomial $f$, it will not have a root in $\bar F$, contradicting the definition of $\bar F$. So $K\subseteq \bar F$

Does it mean that $K=\bar F$ ?

• The problem is to have all of the $K_f$ inside some field containing $F$. If so ( like for instance $F= \mathbb{Q}$, and all $K_f \subset \mathbb{C}$) , then yes, $K=\bar F$. Commented Jun 19, 2015 at 6:14
• You may have to be a bit more cautious with expressions like $\bigcup_{f\in F[X]} K_f$: How do you take such a union? These are (probably uncountably many) fields, which do not need to be subsets of some greater subset, so a simple union might not make sense.
– user149890
Commented Jun 19, 2015 at 7:14
• But can we say it in another way round: Since by Zorn's Lemma, $\bar F$ must exists. So according to the above argument, $\cup K_f$ exists? Commented Jun 19, 2015 at 7:32

As @mathmax said, you have to be more careful. This union is a direct limit: take $I=F[X]$, define the relation over $I$: $f\leq g\Leftrightarrow f|g$ and prove that $(I,\leq)$ is a directed set.
Now, for all $f\in I$, let $K_f$ be the splitting field of $f$ (here we need the axiom of choice). Note that if $f\leq g$ then, unless isomorphism, $K_f\subseteq K_g$. Let $i_{g,f}:K_f\rightarrow K_g$ be the canonical inclusion. Prove that $(K_f,i_{g,f})$ form a direct system and its direct limit $\displaystyle\lim_{\longrightarrow}K_f$ can be identified with $\bigcup_{f\in I}K_f$.
And more precisely, $\displaystyle\lim_{\longrightarrow}K_f\simeq\overline{F}$.