# Primitive element and Galois group Computation

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$(N/K)$?

• Just a little observation: $\sqrt{3}$ is not prime in $\mathbb{Q}(\sqrt{2},\sqrt{3})$, because it's a non-zero element of a field. It is anyways true that $[L:Q]=8$, because primality of $\sqrt{3}$ in $\mathbb{Z}[\sqrt{2},\sqrt{3}]$ holds. – user228113 Apr 19 '15 at 7:55
• Edited my first comment: obviously $\mathbb{Z}[\sqrt{2},\sqrt{3}]/(\sqrt{3})\cong\mathbb{Z}[\sqrt{2}]$. – user228113 Apr 19 '15 at 8:02
• Thanks! I was supposed to use $\sqrt{3}$ prime in $\mathbb{Z}[\sqrt{2},\sqrt{3}]$ – user_m Apr 19 '15 at 8:57

$$2-{y^2\over9-5\sqrt3}=\sqrt2$$

$$2=4-{4y^2\over9-5\sqrt3}+{y^4\over(9-5\sqrt3)^2}$$

$$4y^2(9-5\sqrt3)=2(9-5\sqrt3)^2+y^4$$

$$y^4-36y^2+162+150=(-20y^2+180)\sqrt3$$

Now square both sides to get a polynomial of degree 8 for $y$.

• Thanks! Do you how to compute the Galois group? – user_m Apr 19 '15 at 7:58
• I have figured that out myself! – user_m Apr 22 '15 at 4:19

To find the Galois group of the splitting field $N$ over $\mathbb{Q}$, let $f$ be the minimal polynomial of $y$. There is a calculation of $\text{Aut}(\mathbb{Q}(y)/\mathbb{Q})$ here at The Galois group for the following field extension showing that $\text{Aut}(\mathbb{Q}(y)/\mathbb{Q})=Q_8$. Since $|\text{Aut}(\mathbb{Q}(y)/\mathbb{Q})|=[\mathbb{Q}(y):\mathbb{Q}]=8$, $\mathbb{Q}(y)/\mathbb{Q}$ is Galois. Since $\mathbb{Q}(y)$ is the stem field and $\mathbb{Q}(y)/\mathbb{Q}$ is normal, it is the splitting field of $f$ over $\mathbb{Q}$. Thus $\text{Gal}(\mathbb{Q}(y)/\mathbb{Q})=Q_8$