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QUESTION: Suppose K/F is a non-trivial finite degree purely inseparable extension. Prove that there is a purely inseparable degree p extension of K.

I know, or can prove, that [K:F] is a power of p as well as the standard equivalences that usually come when defining purely inseparable extensions. However, this problem is stumping me. Can someone help me see why this purely inseparable extension being non-trivial and finite degree means K is not algebraically closed? And furthermore why there is an extension of K of degree p?

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This is a nice problem: the statement is a bit weird and the solution , while not too hard, checks the understanding of and ability to use basic results on fields of characteristic $p$. – Georges Elencwajg Oct 6 '12 at 7:58

If $K/F$ is purely inseparable then any $\alpha\in K-F$ has minimal polynomial of the form $X^{p^n}-a$ for some element $a\in F$ of the base field. Fixing $\alpha$ and $a$, can $X^{p^r}-a$ have a root in $K$ for every $r$?

If $\alpha\in K-F$ has minimal polynomial $X^{p^n}-a$ over $F$, then not every one of $a,\sqrt[p]{a},\sqrt[p^2]{a},\sqrt[p^3]{a},\cdots$, considered as elements of the algebraic closure of $K$, say, can be contained in $K$. Suppose $\beta=\sqrt[p^n]{a}$ is the last such term of the sequence that is contained in $K$. Consider $K(\sqrt[p]{\beta})/K$.

(I am pretty clumsy with field theory so I hope this checks out.)

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Yes, this is correct. It rests on the following nontrivial and basic theorem (due to Abel, I think) on fields of characteristic $p$: If an element $a\in F$ is not of the form $a=b^p, b\in F$, then all the polynomials $X^{p^r}-a, r\in \mathbb N$ are irreducible in $F[X]$. A paraphrase is that the perfect (or a fortiori algebraic) closure of an imperfect field is always infinite-dimensional (imperfect means that the Frobenius field endomorphism $F\to F: x\mapsto x^p$ is not surjective). – Georges Elencwajg Oct 6 '12 at 8:06

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