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Let $K$ be a perfect field and let $f\in K[x]$ be a monic irreducible polynomial of degree $n$. Denote by $\alpha,\beta$ two distinct roots of $f$. Is the following bound true? $$ [K(\alpha-\beta):K]\geq \frac n2 $$ If not, does someone know a similar bound (if it exists)?

Joel92
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    A general result is that $\alpha-\beta$ generates all of $K(\alpha,\beta)$ unless $\alpha-\beta=\alpha'-\beta'$ for some zeros $\alpha'\neq\alpha$ and $\beta'\neq\beta$ of $f$. This forces $[K(\alpha-\beta):K]\ge n$ often enough. – Jyrki Lahtonen Jun 30 '16 at 13:39
  • Yes, it's the proof of the primitive element. – Joel92 Jun 30 '16 at 13:43
  • Trying to come up with a characteristic zero example :-( If you think about it inside the splitting field of $f$, you see that $\alpha-\beta$ needs a very large stabilizer inside the Galois group. – Jyrki Lahtonen Jun 30 '16 at 13:46
  • Should I write another question in order to obtain a conterexample in $\mathbb Q$? – Joel92 Jun 30 '16 at 14:05
  • @Joel92, yes I think that would be best. Oh and no consecutive roots this time, so that's out ... – Hmm. Jun 30 '16 at 14:21
  • http://math.stackexchange.com/questions/1844835/degree-of-the-differents-of-two-roots – Joel92 Jun 30 '16 at 14:38

1 Answers1

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This bound does not hold. Consider the Artin-Schreier polynomial, $$f(X)=X^p-X+1\in\mathbb{F}_p[X]$$

Note that $f(\alpha)=0\implies f(\alpha+1)=0$. Take $\beta=\alpha+1$ to obtain a contradiction to the proposed bound.

Hmm.
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