Degree of the difference of two roots 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?
 A: 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.
