Basis of $\mathbb{Q}(i\sqrt{2},\sqrt{3},\beta)/\mathbb{Q}$, being $\beta$ root of $x^3+5x-3$

Question

I have to find a basis of $\mathbb{Q}(i\sqrt{2},\sqrt{3},\beta)/\mathbb{Q}$, being $\beta$ root of $x^3+5x-3$. My first try was to try to see if $x^3+5x-3$ was irreducible. It is in fact, irreducible over $\mathbb{Q}$, as the only possible rational roots of this polynomial would be $\pm1$, or $\pm 3$ (as it is monic), and none of these is a root. So, we could say that this polynomial is the minimal polynomial of $\beta$ in $\mathbb{Q}$, so $[\mathbb{Q}(\beta):\mathbb{Q}]=3$; evidently, $[\mathbb{Q}(i\sqrt2):\mathbb{Q}]=2$ (as it is root of $x^2+2$, irreducible over $\mathbb{Q}$), and $[\mathbb{Q}(\sqrt3):\mathbb{Q}]=2$ (as it is root of $x^2-3$, irreducible over $\mathbb{Q}$). We then know that $[\mathbb{Q}(i\sqrt2, \sqrt3):\mathbb{Q}(i\sqrt2)]$ is lesser or equal than $2$, but it can't be $1$ (as if it is one, $\mathbb{Q}(i\sqrt2, \sqrt3)=\mathbb{Q}(i\sqrt2)$, and then $\sqrt3\in\mathbb{Q}(i\sqrt2)$, which is obviously false), so $[\mathbb{Q}(i\sqrt2, \sqrt3):\mathbb{Q}(i\sqrt2)]=2$. So, if I determine $[\mathbb{Q}(i\sqrt2, \sqrt3,\beta):\mathbb{Q}(i\sqrt2,\sqrt{3})]$, I would know $[\mathbb{Q}(i\sqrt2, \sqrt3,\beta):\mathbb{Q}]$ (as it is equal to $[\mathbb{Q}(i\sqrt2):\mathbb{Q}]\cdot [\mathbb{Q}(i\sqrt2,\sqrt3):\mathbb{Q}(i\sqrt2)]\cdot[\mathbb{Q}(i\sqrt2, \sqrt3,\beta):\mathbb{Q}(i\sqrt2,\sqrt{3})] $). Nevertheless, I get lost here... I don't know if I am doing it right... If someone could make sure I am not making mistakes, and in the case I am in the right path, give me some hints to finish it correctly... Thanks a lot! (I also post a diagram of all the different extensions involved)

Answer 1

Hint: What is $[\Bbb{Q}(\beta):\Bbb{Q}]$? Then consider the fact that $$[\mathbb{Q}(i\sqrt2, \sqrt3,\beta):\mathbb{Q}(i\sqrt2,\sqrt{3})]\cdot[\mathbb{Q}(i\sqrt2,\sqrt3):\mathbb{Q}]=[\mathbb{Q}(i\sqrt2, \sqrt3,\beta):\mathbb{Q}(\beta)]\cdot[\mathbb{Q}(\beta):\mathbb{Q}].$$