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Let $n > 1$ be a fixed integer. Does there exist a field $F$ with the following properties?

  1. $F$ is not algebraically closed.
  2. Every polynomial $f(x) \in F[X]$ of degree $n$ is reducible.

I cannot think of any such field, even for $n=2$. All I can prove is that $F$ must be infinite.

  • 9
    how about $F=\mathbb{R}$ and $n=3$?2012-09-28
  • 1
    Take $\mathbb{Q}$ and adjoin all square roots.2012-09-28
  • 0
    It would seem that the intent of the problem is "every polynomial is reducible *into linear factors*". In this case $n=3$ and $\mathbb{R}$ would *not* be an example (because of irreducible quadratics).2012-09-28
  • 0
    Any element of this field lives in an extension of even degree over $\mathbb{Q}$. So in particular, $\sqrt[3]{2}$ isn't in there.2012-09-28
  • 2
    @Bill: That wouldn't work. You added $\sqrt 2$ but not $\sqrt[4]2$. So the polynomial $x^2-\sqrt 2$ has is not reducible. You need to *close* under square roots.2012-09-28
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    Your claim that $F$ must be finite is correct. If $\operatorname{char}F\ne 2$, then $x\mapsto x^2$ is not injective because $\pm1\mapsto 1$. Therefore, if $F$ is finite, it is also not surjective, hence there exists $c\in F$ such that $x^2=c$ has no solution in $F$. If $\operatorname{char} F=2$, then $x\mapsto x^2+x$ is not injective because $0\mapsto0$ and $1\mapsto 0$, hence for finite $F$ there is a $c$ such that $x^2+x=c$ has no solution in $F$.2012-09-28

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