$[\mathbb{Q}(\alpha, \beta):\mathbb{Q}(\alpha)]\mid [\mathbb{Q}(\beta):\mathbb{Q}]$?

Question

Given $\alpha, \beta$ algebraic numbers over $\mathbb{Q}$, it is known that $d=[\mathbb{Q}(\alpha, \beta):\mathbb{Q}(\alpha)]\le[\mathbb{Q}(\beta):\mathbb{Q}]=b$. It is also true that $d\mid b$ ? If not give a counterexample.

Answer 1

Let $\alpha, \beta$ be two distinct roots of $x^{3} - 2$. Then $[\mathbb{Q}(\beta):\mathbb{Q}] = 3$, whereas $[\mathbb{Q}(\alpha, \beta):\mathbb{Q}(\alpha)] = 2$.

The idea here is that $x^3 - 2$ is irreducible over the rationals, and splits as $(x-\alpha) q(x)$ in $\mathbb{Q}(\alpha)[x]$, where $q(x)$ is irreducible over $\mathbb{Q}(\alpha)$.

Answer 2

$\alpha=\root3\of2$, $\beta=\rho\root3\of2$, where $\rho$ is a primitive cube root of 1. $d=2$, $b=3$.

More generally, given any $n\ge2$, and any $d\lt n$, you can find a polynomial $f$ of degree $n$, irreducible over the rationals, with roots $\alpha,\beta$ in a splitting field, with $\alpha$ of degree $n$, and $\beta$ of degree $d$ over ${\bf Q}(\alpha)$, whether or not $n$ is a multiple of $d$.