Let $F$ be an extension of $K$ (they are both fields). I know that if $F$ has finite degree over $K$, then the following things are equivalent:

1) $F$ is such that every irreducible polynomial in $K[x]$ has a root in $F$

2) If $T$ is an extension of $F$ and $\alpha$ an automorphism of $T$ such that $\alpha(b)=b$ for each $b\in K$, then $\alpha(F)=F$

The implications are true if $F$ is an algebraic extension of $K$. Do they hold also in the general case?

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    $\begingroup$ Despite what you think 1) and 2) are always equivalent if $F$ is algebraic over $K$, finiteness of degree only confusing the issue: Lang, Algebra, Ch.V, §3, Thm. 3.3. If the extension $K\subset F$ is not algebraic "normal" is not defined. $\endgroup$ – Georges Elencwajg May 28 '16 at 19:39
  • $\begingroup$ I've modified the question, since I for "normal" extension I meant something always defined. $\endgroup$ – W4cc0 May 29 '16 at 17:19
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    $\begingroup$ The current version is false. Consider the case $K=\Bbb{Q}$, $F=K(\sqrt3)$. Condition 2 holds for this pair. But condition 1 fails, because the irreducible polynomial $x^2-2\in K[x]$ has no roots in $F$. Did you mean that $F/K$ should be normal? Or, if an irreducible polynomial $p(x)\in K[x]$ has a zero in $F$ it splits completely over $F$ (instead of 1). $\endgroup$ – Jyrki Lahtonen May 29 '16 at 17:26
  • $\begingroup$ Thanks for the first counterexample. I mean that $F/K$ should be normal. Do you know also some example contradicting the other implication? $\endgroup$ – W4cc0 May 31 '16 at 21:29

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