# Determine the Galois group of $\mathbb{Q}(\sqrt{a+b\sqrt{d}})$

Suppose $L=\mathbb{Q}(\sqrt{a+b\sqrt{d}})$,($d$ and $a+b\sqrt{d}$ are square free algebraic integers) when is $L/\mathbb{Q}$ a normal extension? When does $Aut(L/\mathbb{Q})=\mathbb{Z}/4\mathbb{Z}$ or $Aut(L/\mathbb{Q})=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$?

• Assuming that $d$ is a square-free integer, what does it mean that $a+b\sqrt{d}$ is a square-free integer, provided that it is not an integer? – Jack D'Aurizio Jul 20 '15 at 11:15
• Since $\sqrt{a-b\sqrt{d}} = \sqrt{a^2-db^2}/\sqrt{a+b\sqrt{d}}$ it is sufficient condition for the normality that $a^2-db^2$ is a square. I think it is necessary as well, but don't have proof... – sss89 Jul 20 '15 at 21:10
• – Watson Dec 4 '18 at 10:28

The extension $L/\mathbb{Q}$ have at least two automorphisms coming from The Galois action of $L/\mathbb{Q}[\sqrt{d}]$. If $\sigma$ is the automorphism of $\mathbb{Q}[\sqrt{d}]$ sending $\sqrt{d}$ to $-\sqrt{d}$, then $L/\mathbb{Q}$ is Galois iff $\sigma$ can be extended to $L$.
Since $\sigma$ sends the polynomial $f(x)=x^2+(a+b\sqrt{d})$ to $g(x)=x^2+(a-b\sqrt{d})$, it can be extended to $L\cong \mathbb{Q}[\sqrt{d}][x]/f(x)$ iff it contains a root for $g(x)$. So $L/\mathbb{Q}$ is Galois iff $\sqrt{a-b\sqrt{d}}=x+y\sqrt{a+b\sqrt{d}}\in L$, namely $x^2+y^2(a+b\sqrt{d})=a-b\sqrt{d}$ and $2xy\sqrt{a+b\sqrt{d}}=0$ where $x,y\in\mathbb{Q}[\sqrt{d}]$. The second equation shows that $x=0$ or $y=0$. If $y=0$, then $x^2=a-b\sqrt{d}$, but then applying $\sigma$ we get that $(\sigma{x})^2=a+b\sqrt{d}$ contradicting the assumption that it is not a square. It follows that $x=0$ and that $y^2=\frac {a-b\sqrt{d}}{a+b\sqrt{d}}$. Reduce this equation to two quadratic equation over $\mathbb{Q}$ and solve them. Note in particular that applying $\sigma$ you get $\sigma(y)^2=\frac {a+b\sqrt{d}}{a-b\sqrt{d}}=y^{-2}$ so that $(\sigma(y)y)^2=1$, namely $\sigma(y)y=\pm 1$.
For determining the group, note that the Galois automorphisms of $L/\mathbb{Q}$ are of order 1 and 2, so if the group is cyclic of order 4, then the extension of $\sigma$ must be of order 4. You already know that $\sigma(\sqrt{a+b\sqrt{d}})=y \sqrt{a+b\sqrt{d}}$, so $\sigma^2(\sqrt{a+b\sqrt{d}})=\sigma(y)y \sqrt{a+b\sqrt{d}}$. It follows that it is cyclic if $\sigma(y)y=-1$ and it is the Klein four group if $\sigma(y)y=1$.