Basis of the field $E$=$mathbb{Q}(sqrt{6}i-sqrt{5})$.

Let’s take a field $E$=$mathbb{Q}(sqrt{6}i-sqrt{5})$.

I wanted to find a basis of $E$ as a $mathbb{Q}$-vectorspace.
My first thought was to use the structure theorem for simply field extension. Therefore I needed to find an element $uin E$ so that $u$ is algebraïc over $E$ and a $f$ so that $f(u)=0$. Then would $u,u^{2},…,u^{deg(f)-1}$ be a basis. I just don’t know how I can do this correctly.

Second I wanted find a polynomial $fin mathbb{Q}[X]$ so that $E$ is the splitting field of $f$ over $mathbb{Q}$.
Because I’m stuck with my first thought I don’t understand how I can find such an $f$.

EDIT:

FIRTS

• let’s name $alpha$=$sqrt{6}i-sqrt{5}$. We’re looking for a polynomal $fin mathbb{Q}$ so that $f(alpha)$=0.

$alpha^{2}$=$-1-2isqrt{30}$

$alpha^{3}$=$-isqrt{6}+sqrt{5}+2sqrt{+}sqrt{30}+2isqrt{5}sqrt{0}$

$alpha^{4}$=$1+4isqrt{30}-120$=$-119+4isqrt{30}$

Then follows that $alpha^{4}+2alpha^{2}+121=0$

we name $f(x)$=$x^{4}+2x^{2}+121$

Because $f$ isn’t $0$ and has no points $x$ so $f(x)=0$ , is $f$ irriducibel.
This means that $f$ is the minimal polynomal of $alpha$ over $E$. Now we use the theorem of structure for simply field extension and it follows that {$1,alpha,alpha^{2},alpha^{3}$} is a basis.

SECOND

• I’m still stuck proving this