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?

  • 1
    $\begingroup$ Sounds like overkill. Could you work within the splitting field of $fg$? $\endgroup$ Dec 30, 2017 at 4:22
  • 1
    $\begingroup$ Look in books besides your textbook for other approaches. Most texts do not prove the existence of algebraic closures so early, and the result you describe definitely does not need to go so far as to build an algebraic closure. $\endgroup$
    – KCd
    Dec 30, 2017 at 4:34
  • $\begingroup$ Not sure if it is overkill but I would take $L/F$ the splitting field of $g$ irreducible and $a \in F, g(a) = 0$. Then $b \in L $ is a root of $g$ iff $\sigma : K(a) \to K(b), a \to b$ is an isomorphism. But $a \in F$ and $F$ is normal, thus $\sigma(a)$.. $\endgroup$
    – reuns
    Dec 30, 2017 at 4:36
  • $\begingroup$ The polynomial $g$ that you choose must be irreducible, otherwise consider $\mathbb Q(\sqrt{2})$ over $\mathbb Q$, which is normal, and $x^4-x^2-2 = (x^2-2)(x^2+1)$ has a root in $\mathbb Q(\sqrt{2})$, but doesn't split completely. $\endgroup$
    – Thrash
    Jun 17, 2021 at 12:55

1 Answer 1


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 $b$.)

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    $\begingroup$ So the hardest part is to extend $j: K(\alpha) \to K(\beta)$ to a morphism $L \to j(L)$. No easier way with field norms, resultants ? $\endgroup$
    – reuns
    Dec 30, 2017 at 4:43
  • $\begingroup$ How do you know that $g$ contains a root in $F$? $\endgroup$
    – Sha Vuklia
    May 2, 2019 at 6:19
  • $\begingroup$ @ShaVuklia: That is just by assumption (following the proof strategy proposed in the question): we prove that if $g$ has a root in $F$, then $g$ splits over $F$. $\endgroup$ May 2, 2019 at 6:21
  • $\begingroup$ A formalized version (in Lean) of this proof is now available at github.com/leanprover-community/mathlib4/blob/… $\endgroup$
    – Junyan Xu
    Oct 29, 2023 at 4:22

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