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Let $F\subset E$ be an algebraic extension of fields and suppose that every polynomial $f(x) \in F[x]$ has a root in $E$. Is it possible that $E$ is not algebraically closed?

Note that such an example must necessarily be of prime characteristic since over char $0$, $E$ is necessarily algebraically closed. Proof:

Let $g(x) \in E[x]$ be a polynomial with a root $\alpha$ (after fixing an algebraic closure of $E$). Let $f(x)$ be the minimal polynomial of $\alpha$ over $F$ and let $L$ be the splitting field of $f$. By the primitive element theorem, $L=F[\beta]$ and if $h(x)$ is the minimal polynomial for $\beta$, it has some root $\gamma \in E$. Since $F[\beta] = F[\gamma]$, this shows that $L\subset E$.

In particular, if we are looking for counterexamples, we would like $L$ to not be a primitive extension.

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marked as duplicate by Asvin, Jyrki Lahtonen, Mathmo123, Alex Wertheim, Key Flex Aug 9 '18 at 2:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Is this a duplicate? $\endgroup$ – Jyrki Lahtonen Aug 8 '18 at 4:49
  • $\begingroup$ I believe you are correct! $\endgroup$ – Asvin Aug 8 '18 at 6:30