Let $L=\mathbb{Q}(\sqrt{2},\sqrt{3},y)$ where $y^2=(9-5\sqrt{3})(2-\sqrt{2})$. Show that $y$ is a primitive element of the extension $L/\mathbb{Q}$. Determine its minimal polynomial and a splitting field $N/K$. Determine Gal$(N/K)$.
I am thinking that we can expand $1,y^2,y^4,y^6$ in terms of the basis $1,\sqrt{2},\sqrt{3},\sqrt{6}$ and get a matrix $A$ where the $i$-th column of $A$ is the coordinate vector of $y^{2^{i-1}}$. Argue that $A$ is invertible. Let $b=(0,1,0,0)^T$ and $c=(0,0,1,0)^T$. Then each of $Ax=b$ and $Ax=c$ has unique solution. This proves that $y$ is the primitve element. I wonder if there is a easier way to do this. I tried to mimic the proof for the Primitive Element Theorem but this requires (or does it?) knowledge about the minimal polynomial of $y$ but I do not know how to solve for it.
Also, I am guessing the minimal polynomial of $y$ over $\mathbb{Q}$ is of degree 8 since $X^2-(9-5\sqrt{3})(2-\sqrt{2})$ is over $\mathbb{Q}(\sqrt{2},\sqrt{3})$ irreducible by Eisenstein ($\sqrt{3}$ is prime in $\mathbb{Z}[\sqrt{2},\sqrt{3}]$.) If that is correct, then say we get from the above that some monic polynomial $f(X)$ in $\mathbb{Q}[X]$ such that $f(y)=\sqrt{2}$. Then we take $f^2-2$. Since $f$ is of degree $4$, then $f^2-2$ is the minimal polynomial. Is it correct? But then, how do we compute Gal$(N/K)$?
$$2-{y^2\over9-5\sqrt3}=\sqrt2$$
$$2=4-{4y^2\over9-5\sqrt3}+{y^4\over(9-5\sqrt3)^2}$$
$$4y^2(9-5\sqrt3)=2(9-5\sqrt3)^2+y^4$$
$$y^4-36y^2+162+150=(-20y^2+180)\sqrt3$$
Now square both sides to get a polynomial of degree 8 for $y$.