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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.

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how about $F=\mathbb{R}$ and $n=3$? – user8268 Sep 28 '12 at 14:53
Take $\mathbb{Q}$ and adjoin all square roots. – Bill Cook Sep 28 '12 at 14:53
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). – Bill Cook Sep 28 '12 at 14:55
@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. – Asaf Karagila Sep 28 '12 at 15:04
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 $F$. – Hagen von Eitzen Sep 28 '12 at 15:15

Here's an example of historical importance for $n=2$:

Let $F$ be the subfield of $\mathbb C$ consisting of all numbers constructible with ruler and compass if $0$, $1$ and $i$ are given. ($F$ is a field because adding, negating, multiplying, taking reciprocals can be done with ruler and compass). Since it is possible to take square roots and solve quadratics with ruler and compass, we see that all quadratic polynomials are reducible. On the other hand, the old classic problem of doubling the cube (i.e. finding a root of $x^3-2=0$) is not solvable with ruler and compass, hence $F\ne \overline F$.

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Over a finite field $\mathbb F_p$, fix an algebraic closure $\bar{\mathbb F}_p$ and consider the (increasing) union $F$ of the subextensions ${\mathbb F}_{p^{n^d}}$, $d\ge 0$. Let $f(x)\in F[x]$ of degree $n$. Then for some $d\ge 0$, $$f(x)\in F_d:=\mathbb F_{p^{n^d}}[x].$$ If $f(X)$ is irreducible over $F$, then it is irreducible over $F_d$. The extension of $F_d$ generated by a root $\alpha$ of $f(x)$ in $\bar{\mathbb F}_p$ has degree $n$ over $F_d$, so $\alpha\in F_{d+1}\subseteq F$. Thus $f(x)$ is reducible in $F[x]$. Finally, $F\ne \bar{\mathbb F}_p$: take a number $\ell$ prime to $n$, then $\mathbb F_{p^\ell}\cap F=\mathbb F_p$. In particular, $\mathbb F_{p^\ell}\not\subseteq F$.

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This doesn't have any added value. – user18119 Aug 8 '13 at 22:37
Unless there is a very good reason, we prefer not to arbitrarily delete good content: the answers could well help other users in the future. – robjohn Aug 9 '13 at 23:23
@robjohn: Dear moderators, I think I explained in my previous comment why I want to delete my answer. I don't understand why you say "we prefer not to arbitrarily delete...", as it is me who delete my answer. Do my answers belong to MSE ? – user18119 Aug 10 '13 at 15:12
Please see Robert Cartaino's answer. This section of the FAQ may also be helpful. – robjohn Aug 10 '13 at 22:03

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