Can I prove that a splitting field is normal without using zorn lemma There is a theorem ：
If $K \in F$ and $F$ is a splitting field of a polynomial  in $K[x]$,then F is a normal extension over $K$.  
For proving this I choose a polynomial $g \in K[x]$ which has a root in $F$ and I want to prove that g splits completely over $F$.  
In my textbook,the proof of this follows the existence of an algebraic closure，which seems like can not be proved without the using of zorn lemma.  
However the property I want to get  looks like have nothing to do with the set-theoretic difficulty.
So is there any proof more direct？
 A: The argument doesn't actually require an algebraic closure; you can just replace the algebraic closure by sufficiently large finite extensions in each step of the argument.
Specifically, say $F$ is the splitting field of $f$ over $K$ and let $L$ be an extension of $F$ over which $g$ splits.  Let $\alpha\in F$ be a root of $g$ and let $\beta\in L$ be another root of $g$; we wish to show $\beta\in F$.  There is an embedding $i:K(\alpha)\to L$ which fixes $K$ and sends $\alpha$ to $\beta$.  This embedding can then be extended to an embedding $j:F\to L'$ for some finite extension $L'$ of $L$ (in fact, you can prove that you can take $L'=L$, but that is not needed for this argument).  But then both $j(F)$ and $F$ are splitting fields of $f$ over $K$ inside $L'$, so $j(F)=F$.  Since $j(\alpha)=\beta$, this implies $\beta\in F$, as desired.
(In case you are skeptical of the existence of $L'$, here is the general lemma I am using.  If $i:k\to L$ is a field embedding and $F$ is a finite extension of $k$, then there exists a finite extension $L'$ of $L$ and an embedding $j:F\to L'$ extending $i$.  Indeed, $F$ is generated over $k$ by finitely many elements, and so by extending one element at a time, we may assume $F=k(a)$ for a single element $a$.  Now let $h$ be the minimal polynomial of $a$ over $k$ and let $L'$ be an extension of $L$ obtained by adjoining a root $b$ of the polynomial obtained by applying $i$ to all the coefficients of $h$.  There is then a unique extension of $i$ to an embedding $j:F\to L'$ which sends $a$ to $b$.)
