Let $f\in \mathbb Q[x]$ be an irreducible monic polynomial of degree $n$ and let $\alpha,\beta\in \overline{\mathbb Q}$ be two distinct roots of $f$. Is it possible to find a lower bound on the degree of $\alpha-\beta$? By heart, my claim is that $$ [\mathbb Q(\alpha-\beta):\mathbb Q]\geq \frac n2 $$ The original question Bound for the degree concerned the same claim for arbitrary fields. If the claim is false, can someone find a bound, if it exists?

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    $\begingroup$ At least a start: Are you aware of the fact that $\alpha - \beta \notin \mathbb Q$? This is easy to show and gives you the the bound $\geq 2$. $\endgroup$ – MooS Jun 30 '16 at 15:51
  • $\begingroup$ Just adding that equality is possible. $\endgroup$ – Hmm. Jun 30 '16 at 15:57
  • $\begingroup$ @Hmm Have you got an example with equality, when $n$ has an odd prime factor? $\endgroup$ – Jyrki Lahtonen Jun 30 '16 at 21:16
  • $\begingroup$ @JyrkiLahtonen, for $\alpha-\beta$, I do not. For $\alpha+\beta$ I do, consider the cyclotomic fields. $\endgroup$ – Hmm. Jun 30 '16 at 21:17
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    $\begingroup$ @JyrkiLahtonen :) But even the difference is not too bad. It's just $i\text {sin}\biggr(\dfrac{2\pi}{n}\biggl)$, whose degree is, if I'm not mistaken, is $\dfrac{\varphi(n)}{2}$ if $(n,8)=4$. Let me check.. $\endgroup$ – Hmm. Jun 30 '16 at 21:24

It is possible that $\alpha-\beta$ is algebraic of degree $<n/2$.

As an example I proffer $$ \alpha=\sqrt5+\sqrt3+\sqrt2,\quad\beta=\sqrt5+\sqrt3-\sqrt2. $$ Here $\alpha$ and $\beta$ are both conjugate primitive elements of $\Bbb{Q}(\sqrt5,\sqrt3,\sqrt2)$ - a degree eight extension. Yet $\alpha-\beta=2\sqrt2$ is a root of a quadratic.

It is hopefully clear how to extend the above example to a case where $f(x)$ has degree $2^\ell$ for arbitrary positive integer $\ell$ such that $\alpha-\beta$ generates a quadratic extension only.

As MooS pointed out $\alpha-\beta$ cannot be rational, so $[\Bbb{Q}(\alpha-\beta):\Bbb{Q}]=2$ is as low as it can go. For the sake of completeness let me recap an argument. If $\alpha-\beta=q\in\Bbb{Q}$, then $\beta=\alpha-q$. Therefore $\alpha$ is a zero of two monic polynomials with rational coefficients, $f(x)$ and $f(x+q)$. Because we are in characteristic zero $f(x)$ and $f(x+q)$ are distinct (look at the coefficients of degree $n-1$ terms). Therefore their greatest common divisor has a lower degree, and must be non-trivial given that it has $\alpha$ as a root.

See the linked question to learn why the assumption about characteristic zero is essential.

  • $\begingroup$ Just adding a link to a proof of the fact that $\alpha$ and $\beta$ are conjugates- math.stackexchange.com/questions/1313897/…. Hope you do not mind... $\endgroup$ – Hmm. Jun 30 '16 at 21:09
  • $\begingroup$ Of course, not @Hmm. I was about to start looking for one. $\endgroup$ – Jyrki Lahtonen Jun 30 '16 at 21:10
  • $\begingroup$ For a proof of conjugacy of sums of signed square roots of primes see this thread and the material linked to there. $\endgroup$ – Jyrki Lahtonen Jun 30 '16 at 21:12

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