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 $Aut(L/\mathbb{Q})=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$?

  • $\begingroup$ 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? $\endgroup$ – Jack D'Aurizio Jul 20 '15 at 11:15
  • 2
    $\begingroup$ 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... $\endgroup$ – sss89 Jul 20 '15 at 21:10
  • $\begingroup$ Related: math.stackexchange.com/questions/649466 $\endgroup$ – 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 $\sigma(y)y=1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.