Timeline for Is there a non-perfect field in which polynomials of large degree are reducible?

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Jun 29 at 14:52	comment	added	LSpice		@Plop, re, yes, that's the power involved in the definition of a perfect field.
Jun 28 at 7:28	comment	added	Plop		The prime $p$ is assumed to be the characteristic of $F$, right?
Jun 27 at 20:28	comment	added	LSpice		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 $\sqrt[p^k]a$.
Jun 27 at 14:17	comment	added	Will Sawin		@Medo The separable closure of any imperfect field will do the trick.
Jun 27 at 13:55	comment	added	Medo		there exists a separably closed field that is not perfect?
Jun 27 at 13:46	vote	accept	Medo
Jun 27 at 13:44	comment	added	Will Sawin		@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.
Jun 27 at 13:42	comment	added	Medo		what about polynomials of degree $m$ such that $gcd(m,p)=1$?
Jun 27 at 13:26	history	answered	Will Sawin	CC BY-SA 4.0