Update My previous proof was incorrect. This updated proof is inspired by the comment by 'MartianInvader'.

Problem I can prove the statement 'Every algebraic extension $L:K$ has a normal closure $F:L$' by taking an algebraic closure $\bar{L}$ of $L$ (existence needs AC), and letting $F = L(S)$ where $S$ is the set of $\alpha \in \bar{L}$ whose minimal polynomial over $K$ is precisely that of some $\beta \in L$. The $F$ produced this way is normal because it is a splitting field extension of the set of minimal polynomials of $\alpha \in L$ over $K$.

The question is: Is there a proof without taking the algebraic closure?

Note $F:L$ is called a normal closure of an algebraic extension $L:K$ if $F:K$ is normal (meaning it is algebraic and the minimal polynomial of any $\alpha \in F$ over $K$ splits) and there are no intermediate fields $M$ between $F,L$ such that $M:K$ is normal (so $F$ is the 'smallest' such).

Also, an extension $F:K$ is normal iff it is a splitting field extension of some subset of $K[x]$.

  • $\begingroup$ Instead of applying Zorn's lemma, can you just take the intersection? $\endgroup$ – MartianInvader Sep 12 '13 at 22:27
  • $\begingroup$ The existence of the algebraic closure $\bar L$ requires the axiom of choice. $\endgroup$ – John Gowers Sep 12 '13 at 22:28
  • $\begingroup$ @MartianInvader The existence of an algebraic closure is proved using AC, and I strongly believe that in fact the existence of algebraic closures for any field is in fact equivalent to AC. $\endgroup$ – Matemáticos Chibchas Sep 12 '13 at 22:29
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    $\begingroup$ In fact 'algebraic closures exist for any field $\Rightarrow$ AC' is an open problem. See this MO post: mathoverflow.net/questions/46566/… $\endgroup$ – John Gowers Sep 12 '13 at 22:32
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    $\begingroup$ @MatemáticosChibchas: It isn't open at all. We know that the ultrafilter theorem (also Boolean Prime Ideal theorem, and many others) imply the existence and uniqueness (up to isomorphism of course) of algebraic closures. We also know of it to be weaker than the axiom of choice. Therefore assuming the existence, or uniqueness, or both, of algebraic closures will never imply the axiom of choice. Unless ZF is inconsistent, of course. $\endgroup$ – Asaf Karagila Sep 13 '13 at 5:25

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