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Is there a characterization of fields $k$ such that, if $f\in k[x]$ splits over $k$, so does $f'$?

This is trivially true is $k$ is algebraically closed, and the answer to this question show that it is true for real closed $k$.

I can see that it is also true for $k=GF(2)$, since $f$ must be of the form $x^m(x-1)^n$, which has derivative of the same form or equal to $0$. This is obvious unless $n$ and $m$ are both odd, but in that case $f'=x^{m-1}(x-1)^n+x^m(x-1)^{n-1}=x^{m-1}(x-1)^{n-1}(x-1+x)=x^{m-1}(x-1)^{n-1}$. So this property hold for more than just real closed and algebraically closed fields.

On the other hand, this property is not true for all fields. In particular, it is not true for $k=\Bbb{Q}$ since $f=x^3-x=(x-1)(x-0)(x-(-1))$ splits over $\Bbb{Q}$, but its derivative $f'=3x^2-1$ is irreducible over $\Bbb{Q}$.

So I'm curious whether there is a characterization of such $k$.

Generalization: More generally, consider a tower $R\le k\le F$ with $R$ an integral domain and $k$ and $F$ fields. Is there a characterization of towers with the property that, if $f\in R[x]$ splits over $k$, $f'$ splits over $F$?

C Monsour
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  • A little calculation shows that if you restrict to just cubic $f$ (and assume $k$ has characteristic different from $2$), this is equivalent to asking that any sum of two squares is a square in $k$. (And in characteristic $2$, you get that every element of $k$ is a square.) Interesting... – Eric Wofsey Apr 04 '21 at 23:56
  • Unsurprisingly, though, this condition is not sufficient for higher degree $f$ (for instance, for $f(x)=x(x+1)(x+3)(x-1)$, $f'$ does not split over the quadratic closure of $\mathbb{Q}$). – Eric Wofsey Apr 05 '21 at 00:19
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    Another interesting observation: if you start with a field $k_0$, the smallest extension of $k_0$ with your property will be normal over $k_0$. So for instance, the smallest such extension of $\mathbb{Q}$ is strictly smaller than the real algebraic numbers. (Is it equal to the largest normal subextension of the real algebraic numbers, i.e. the set of real algebraic numbers whose Galois conjugates are all real?) – Eric Wofsey Apr 05 '21 at 00:34
  • @EricWofsey by your first comment, any finite field other than $GF(2^n)$ will fail this condition, right? – Alvaro Martinez Apr 05 '21 at 00:49
  • @AlvaroMartinez Correct – C Monsour Apr 05 '21 at 00:49
  • Er, sorry, my first comment should also have required that $k$ has characteristic different from $3$. In characteristic $3$, of course, the condition becomes trivial for cubic $f$. – Eric Wofsey Apr 05 '21 at 00:54
  • Nonetheless, it seems the operative thing is having characteristic different from 2. I spent a few minutes trying to find counterexamples over $GF(4)$ without success. – C Monsour Apr 05 '21 at 01:04
  • I'm not so sure about my computation for $GF(2^n)$ – Alvaro Martinez Apr 05 '21 at 01:11
  • A short Sage computation shows (after a reduction to finitely many cases as in the post) that $GF(4)$ satisfies this property, and counterexamples are not hard to find for $GF(2^n)$ when $n>2$. – Alvaro Martinez Apr 05 '21 at 03:59
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    For $GF(3)$ there are counterexamples: Over $GF(3)$, $x^4+x^3-x^2-x=x(x-1)(x+1)^2$ has derivative $x^3+x-1=(x+1)(x^2-x-1)$ which thus fails to split, as $x^2-x-1$ is irreducible. I'm guessing one can find similar counterexamples over $GF(3^n)$ so it appears $GF(2)$ and $GF(4)$ are the only finite fields with this property. – C Monsour Apr 05 '21 at 08:43
  • @EricWofsey Your comment I upvoted deserves to be posted as a partial answer, without noting the smallest field is normal it was not obvious at all that there were fields of characteristic $0$ neither algebraically nor real closed. – reuns Apr 05 '21 at 09:55

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