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It is well-known that, in a real-closed field $K$, every polynomial of degree $>2$ is reducible in $K$. But in this case the characteristic of $K$ is zero.

My question is: there exists a non-perfect field $F$ and a positive integer $n=n(F)$ such that all polynomials of degree $>n$ are reducible in $F$.

Edit: polynomials of degree $>n$ coprime with the characteristic $p>0$ of $F$

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    $\begingroup$ Will Sawin's answer being (duly) accepted, you should ask the follow-up ("edit") question separately. Nevertheless, it seems that any non-algebraically-closed, separably closed field would work (every polynomial of degree $>1$ coprime to $p$ is reducible). $\endgroup$
    – YCor
    Commented Jun 27 at 14:31
  • $\begingroup$ PS I see afterwards that this is already observed in comments to that question. So there is no follow-up question anymore. $\endgroup$
    – YCor
    Commented Jun 27 at 14:34
  • $\begingroup$ @WillSawin's answer in the comments to the edited question. $\endgroup$
    – LSpice
    Commented Jun 27 at 20:30

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No. Since $F$ is not perfect, there exists $a\in F$ not a $p$th power. Then I claim $x^{p^k}-a$ is irreducible for all $k$, which contradicts the claim for any $n$.

To check this, observe that over the algebraic closure, this polynomial factors as $(x-a^{1/p^k})^{p^k}$. So any divisor must have the form $(x-a^{1/p^k})^d$ for some $d$. Since this is true over the algebraic closure it must be true over $F$. But if a divisor has the form $(x-a^{1/p^k})^d$ then $a^{ d/p^k} \in F$ which implies $a^{ \gcd(d,p^k)/p^k}\in F$ by Euclid's algorithm. But for $1\leq d \leq p^{k-1}$, $\gcd(d,p^k)$ divides $p^{k-1}$, contradicting the assumption that a $p$th root of $a$ does not lie in $F$.

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  • $\begingroup$ what about polynomials of degree $m$ such that $gcd(m,p)=1$? $\endgroup$
    – Medo
    Commented Jun 27 at 13:42
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    $\begingroup$ @Medo Over a separably closed field, every irreducible polynomial has degree a power of $p$ and hence every polynomial whose degree is not a power of $p$ is reducible. $\endgroup$
    – Will Sawin
    Commented Jun 27 at 13:44
  • $\begingroup$ there exists a separably closed field that is not perfect? $\endgroup$
    – Medo
    Commented Jun 27 at 13:55
  • $\begingroup$ @Medo The separable closure of any imperfect field will do the trick. $\endgroup$
    – Will Sawin
    Commented Jun 27 at 14:17
  • $\begingroup$ Or, to say it differently, a perfect, separably closed field is algebraically closed. \\ Irreducibility for $f_k(x) = x^{p^k} - a$ also follows from (1) $f_1(x)$ is not a perfect power because $a$ is not a $p$th power, (2) $f_1(x) = g(x)h(x)$ with $g(x)$ and $h(x)$ coprime implies $0 = g'(x)h(x) + g(x)h'(x)$ implies $g'(x)h(x) = -g(x)h'(x)$ is a common multiple of $g(x)$ and $h(x)$, which is impossible since it's non-$0$ of degree less than $\deg g(x) + \deg h(x)$, so (3) $F[\sqrt[p^k]a]/F$ has degree $p^k$ by induction on $k$, so $f_k(x)$ is the minimal polynomial of $\sqrt[p^k]a$. $\endgroup$
    – LSpice
    Commented Jun 27 at 20:28

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