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If $K$ is not perfect then there are inseparable irreducible polynomials.

This is not obvious to me at all. I have tried to reduce the question to

If $K$ is not perfect then there exist irreducible polynomial $p(x)$ such that it is not coprime with $D_xp(x)$.

Since $p(x)$ is irreducible, $p(x)$ must be the common factor. But $D_x p(x)$ has lower degree, hence $D_xp(x)=0$. So $p(x) = q(x^p)$. So now I want to show

If $\exists k \in K$ that $\forall \hat k \in K, k \neq \hat k^p$, then there exist irreducible polynomial $p(x)$ such that it can be written as $p(x) = q(x^p)$.

Then I am stuck..

Community
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1LiterTears
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    Did you think of $P=x^p-k$? (Calling the polynomial $p$ is not such a great idea.) – Marc van Leeuwen Feb 02 '14 at 12:46
  • Not being perfect means that some algebraic extension is not separable, so there is an element such that the simple extension over it is not separable, so there is your polynomial. – user40276 Feb 02 '14 at 13:54
  • Similar question: https://math.stackexchange.com/questions/4107816/an-example-of-a-field-k-of-characteristic-p-and-an-irreducible-inseparable – 19021605 Jun 24 '25 at 11:05

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Hint: Consider $K=\mathbb{F}_2(u)$ and $uT^2-1$. See if you can generalize.

Alex Youcis
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