What's the degree of the following field extension?

Question

Let $\alpha$ be a root of $f(t)=t^4+9t^2+15$ - What is $[\Bbb Q(\alpha):\Bbb Q(\alpha ^2+3)\textbf{]}$?

As $f(t)$ is irreducible (Eisenstein $p=3$) $[\Bbb Q(\alpha):\Bbb Q]=4$, so by the tower law $[\Bbb Q(\alpha):\Bbb Q(\alpha ^2+3)\textbf{]}\in \{1,2, 4\}$. It can't be $4$ as $\alpha ^2+3$ is not in $\Bbb Q$ but how do I determine if it is $2$ or $1$? Can't seem to be able to figure it out.

Thanks

Answer 1

You say "it can't be 1 as $\alpha^2 + 3$ is not in $\mathbb Q$". I think you mean that $[\mathbb Q(\alpha^2 + 3) : \mathbb Q]$ cannot be 1 for that reason, and hence $[\mathbb Q(\alpha) : \mathbb Q(\alpha^2 + 3)]$ can't be 4. In order to prove that it can't be 1, you just need to show that the two fields aren't equal.

To do this, observe that $\mathbb Q(\alpha^2 + 3) = \mathbb Q(\alpha^2)$, and $f$ is clearly a quadratic in $\alpha^2$.

Answer 2

$\alpha$ satisfies the polynomial $x^2+3-(\alpha^2+3)$ in the field ${\bf Q}(\alpha^2+3)$.