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Im trying to prove the statement of the title:

If every polynomial in $F[x]$ splits then $F$ has no nontrivial algebraic extension

I was thinking about arguing as follows: if there existed an algebraic extension $K=F(\alpha)$ then $\alpha$ would have to be transcendental. But i)I don't know if this would prove the claim ii)I don't even know how to continue along that line of reasoning

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    $\begingroup$ That's an oxymoron: an algebraic extension cannot be generated by a transcendental element. $\endgroup$ – DonAntonio Apr 12 '16 at 16:55
  • $\begingroup$ Said another way, "algebraic extension" means every element is algebraic, (in particular $\alpha$.) Trancendental means "not algebraic." $\endgroup$ – rschwieb Apr 12 '16 at 17:01
  • $\begingroup$ yeah, sorry i meant to say it would be proof by contradiction $\endgroup$ – sh donny Apr 12 '16 at 17:03
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Let $E$ be an algebraic extension of $F$ and let $\alpha \in E$.

Then $F(\alpha) \cong F[x]/(f(x))$, where $f(x)$ is the minimal polynomial of $\alpha$.

Since $f$ is irreducible, it must have degree $1$ and so $\alpha \in F$.

Thus $E \subseteq F$ and so $E=F$.

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